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Static And Kinetic Friction

Coulomb, a famous scientist in the eighteenth century was the first to mention static and kinetic (or sliding) friction, separately. Let a wooden block of weight W rest on a horizontal table.

• Weight W of the block acts on the table vertically downwards; by Newton’s third law of motion, the table also exerts an equal and opposite reaction force R. Force R is called normal force or normal reaction.
• If no other force acts on the system, weight and normal force balance each other and the block rests on the table.
• Let a horizontal force F be applied on the block. If F is below a certain magnitude, the block remains at rest. At the surface of contact, and opposite to the direction of F, a frictional force f develops, which opposes the motion of the block.

Read and Learn More: Class 11 Physics Notes

• In the table below, the variation of the frictional force f with the change in the magnitude of the applied force F, and the corresponding state of motion of the block, are illustrated.

WBBSE Class 11 Static and Kinetic Friction Notes

A Study Of The Table Above Brings Out The Following Important Properties Related To Friction:

1. When no external force acts on the block, the block remains stationary and there is no existence of friction [observation number 1].
2. Observations 2 to 4 point out an important fact as the applied force is increased, the frictional force also increases automatically to balance it. The block, therefore, remains at rest.
   • Frictional force by itself, increases by the exact amount required to balance the applied force, before the motion starts. Hence, before any relative motion between the surfaces begins, the frictional force is a self-adjusting force. The frictional force acting under this condition is called static friction.
   • Static Friction Definition: When two surfaces are relatively at rest but one is trying to slide over the other, then the force that comes into existence between the two surfaces and tries to oppose the motion is called static friction.
3. As the applied horizontal force is increased, the frictional force also increases gradually to its maximum limit. When the applied force exceeds this limit, the block can no longer be at rest [observation number 4]. This maximum value of static friction is called limiting friction or the limiting value of static friction.
   • Limiting Friction Definition: The maximum possible magnitude of static friction is called limiting friction.
4. As soon as the magnitude of the applied horizontal force exceeds the limiting friction, the horizontal resultant force is no longer zero. Because of this resultant force, the block starts moving over the table with acceleration. This motion is termed as sliding.
   • During sliding, the frictional force that comes into play to oppose the motion is called sliding or kinetic friction. An important property of sliding friction is that the magnitude of this frictional force falls a little as soon as motion sets in. Hence, the magnitude of sliding friction is less than that of limiting friction for a given pair of surfaces [observation number 5].
   • Sliding Or Kinetic Friction Definition: When two surfaces slide against each other, the force developed at their surfaces of contact, opposing the relative motion, is called kinetic or sliding friction.
5. To keep the block moving with uniform velocity, the resultant of the horizontal forces must be zero (observation 6). To achieve this, the applied force is reduced so that it becomes equal to the kinetic friction.

Understanding Static Friction vs. Kinetic Friction

It is a common experience that, to move a heavy stone over the ground, a large force has to be applied. However, once the stone is moving, a comparatively smaller force (push) is required to maintain its motion.

• Kinetic friction between two surfaces is independent of the speed of their relative motion, and therefore it is nearly a constant for a specific pair of surfaces.
• The graph represents the force of friction acting between two surfaces in contact with each other, against the applied force that acts parallel to the surface of contact.
• The static friction is initially equal to the applied force. This is represented by the line OA. As the magnitude of the applied force becomes equal to OD, the block is on the verge of moving.
• So AD is the magnitude of limiting friction. On increasing the applied force further, the block begins to move and kinetic friction comes into play. Its value is slightly less than that of the limiting friction.

In the graph, EG represents the magnitude of this kinetic friction which is approximately a constant.

Laws Of Motion – Friction

Laws Of Friction

Friction, acting between two surfaces in contact, follows certain laws. The eminent fifteenth-century artist and scientist, Leonardo Da Vinci was the first to establish these laws experimentally.

Laws Of Static Friction

1. The force of static friction between two surfaces in contact always acts against the force that attempts to cause relative motion.
2. The force of limiting friction between two surfaces in contact is directly proportional to the normal reaction.
3. When normal force remains constant, the force of friction between the surfaces is independent of the area of contact.

Laws Of Kinetic Friction

1. The force of kinetic friction between two surfaces in contact always acts against their relative motion.
2. The force of kinetic friction between two surfaces in contact is directly proportional to the normal force.
3. When normal force remains constant, the force of friction between the surfaces is independent of the relative velocity of the surfaces.

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Coefficient Of Static Friction: On application of a horizontal force on a wooden block kept on a table, the following forces act on the block:

1. Normal force (R) acting vertically upward
2. Weight (W) of the block acting vertically downward
3. Static friction opposite to the direction of the applied force

Let the maximum value of static friction or limiting friction be denoted by f_s. When the block is just about to move, let the force applied be \(-\vec{F}\) .Then, \(\overrightarrow{f_s}=-\vec{F}\)

Applications of Static and Kinetic Friction

According to the laws of static friction, if a body is in contact with a plane, limiting friction (f_s) is directly proportional to the normal reaction (R). With the increase of normal reaction, the number of contact points also increases. So, this increases the adhesion within the area covered by the block. This in turn increases the force of friction.

Hence, f_s ∝R or, f_s = μR

Where μ is a constant for that pair of surfaces. This constant is termed as the coefficient of static friction or simply, the coefficient of friction. The coefficient of friction depends on the materials of the two surfaces in contact and their smoothness.

The coefficient of friction does not depend on the area of the surfaces of contact.

∴ \(\mu=\frac{f_s}{R}=\frac{\text { limiting frictional force }}{\text { normal force }}\)

Definition Of Coefficient Of Static Friction is the ratio between the limiting friction and the normal force between the two surfaces in contact.

• The coefficient of friction (μ) being a ratio between two forces, is a dimensionless quantity and has no unit. The value of μ is generally less than 1. In special cases, μ can be equal to or even greater than 1.
• μ may rise up to 10, for two scientifically cleaned metal surfaces, kept in a vacuum. When highly polished and clear metal surfaces are brought together in a vacuum, they instantly form a single piece of metal and cannot slide over each other due to a sudden increase in frictional force.
• This happens because a large number of atoms (lying on both surfaces) come in contact in this case. This results in a stronger adhesive force which increases friction. For example, μ ≈ 1.6 for two copper surfaces, but μ ≈ 1 if the surfaces are made of glass.

Coefficient of Static and Kinetic Friction

Coefficient Of Kinetic Friction: As a wooden block slides over a table, two reaction forces act on the block

1. Normal force (R) and
2. Force of kinetic friction (f_k).

According to the law of kinetic friction, the forces stated above are directly proportional to each other, that is, \(f_k \propto R \quad \text { or, } f_k=\dot{\mu}^{\prime} R \quad \text { and } \mu^{\prime}=\frac{f_k}{R}\)

The constant of proportionality μ’ is called the coefficient of kinetic friction.
This constant μ’ is a ratio between two forces. So, it is dimensionless and has no unit.

As kinetic friction is less than limiting friction, the coefficient of kinetic friction is less than that of static friction μ, i.e., μ’ < μ for a given pair of surfaces. It decreases further if the relative velocity between the surfaces becomes very high. However, for rough estimates, μ’ is often taken to be equal to μ.

Coefficient Of Kinetic Friction Definition: For two surfaces in contact, the ratio of the kinetic friction to the normal force, is called the coefficient of kinetic friction.

Motion Over A Rough Surface: To move a body over a rough surface, the applied force needs to be greater than the force of friction between the surfaces in contact.

Let the acceleration produced on application of a force F on mass m, be a. As the surface is not smooth, the motion of the body is opposed by kinetic friction. Let the normal force of the plane be R, and the coefficient of kinetic friction be μ’. Hence, the force of kinetic friction =μ’R.

∴ \(F-\mu^{\prime} R=m a \quad \text { or, } a=\frac{F-\mu^{\prime} R}{m}\)

Class 11 Physics  Laws Of Motion  Friction

Laws Of Friction Numerical Examples

Short Answer Questions on Static and Kinetic Friction

Example 1. To set a body of mass 5 kg in motion over a horizontal surface, a minimum force of 30 N has to be applied. What is the value of the coefficient of friction?
Solution:

Force of limiting friction, f = 30 N; weight of the body, mg =5 x 9.8 = 49 N

Normal force R by the plane is equal to the weight of the body as it is on a horizontal plane.

∴ R = mg = 49 N

Hence, μ = \(\mu=\frac{f}{R}=\frac{30}{49}\) = 0.612.

Example 2. An iron block of mass 10 kg is kept on a horizontal floor. The block is pulled by a rope at an angle 30° with the floor. What should be the minimum force necessary to set the block in motion? Given μ = 0.5
Solution:

Let the minimum force applied by the rope be T which makes an angle θ with the horizontal.

The horizontal and vertical components of T are T cosθ and T sinθ respectively. T sinθ acts in the direction of the normal force (R) as shown in the diagram. In this problem, θ = 30°.

As the block is about to move, the frictional force becomes equal to the force of limiting friction. The resultant of forces acting on the body is zero, as the block is still stationary.

According to,

net horizontal force = Tcosθ -μR

net vertical force = R + Tsinθ- W

As the block is stationary,

∴ Tcosθ-μR = 0 or, μR = Tcosθ…(1)

and R+ Tsinθ – W = 0 or, R = W- Tsinθ…(2)

Dividing (1) by (2), we get, \(\mu=\frac{T \cos \theta}{W-T \sin \theta}\)

or, \(T=\frac{\mu W}{\mu \sin \theta+\cos \theta}=\frac{0.5 \times 10 \times 9.8}{0.5 \times \frac{1}{2}+\frac{\sqrt{3}}{2}}\)

= \(\frac{49}{0.25+0.866}=43.9 \mathrm{~N}\)

Example 3. A body moving on the surface of the earth at 14 m • s^-1, comes to rest due to friction after covering 50 m. Find the coefficient of friction between the body and the earth’s surface. Given, acceleration due to gravity = 9.8 m • s^-2.
Solution:

Initial velocity, u = 14m · s^-1; final velocity, v = 0; displacement, s = 50 m and mass of the body = m.

If retardation is a, then using the formula v² = u²-2as,

a = \(\frac{u^2}{2 s}=\frac{14 \times 14}{2 \times 50}=1.96 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

∴ Kinetic friction, f_k = ma = 1.96 m N and normal force, R = mg = 9.8m N

∴ Coefficient of friction, \(\mu^{\prime}=\frac{f_k}{R}=\frac{1.96 m}{9.8 m}=0.2\)

Example 4. A man holds a book vertically between his two palms so that it does not fall. The mass of the book is 1 kg and j the force exerted by each palm is 2.5 kg x g. Find the coefficient of friction between the book and the palm.
Solution:

The weight of the book is 1 kg x g, which acts vertically downwards.

As two surfaces of the book are passed by two hands, the downward force on each surface of the book = \(\frac{1 \mathrm{~kg} \times g}{2}=0.5 \mathrm{~kg} \times g.\)

As the book is at rest, the frictional force acts upwards to balance this downward force of 0.5 kg x g on each surface.

So the limiting force of friction, f = 0.5 kg x g.

Force exerted by each hand = normal force (R) = 2.5 kg x g

∴ Coefficient of friction, \(\mu=\frac{f}{R}=\frac{0.5 \mathrm{~kg} \times g}{2.5 \mathrm{~kg} \times g}=0.2\)

Example 5. A block of mass 0.1 kg is kept pressed onto a wall by applying a horizontal force of 5 N. If the coefficient of friction between the block and the wall is 0.5, find the force of friction on the block.
Solution:

Downward force on the block = weight of the block = mg = 0.1 x 9.8 = 0.98 N

Since the block is at rest, the downward force must be balanced by the upward frictional force.

∴ The frictional force on the block = 0.98 N.

Example 6. Part of a uniform chain of length L is hanging out of a table. If the coefficient of friction between the chain and the table is μ, estimate the maximum length of the chain that can hang without slipping.
Solution:

Let the maximum length of the chain that can hang out without slipping be l.

Length of the chain on the table = L-l.

If m is mass per unit length of the chain, the weight of the chain resting on the table =(L- l)mg

∴ The normal force of the table, R =(L-l)mg

The downward force on the hanging part = weight of the hanging part = lmg

To avoid slipping this downward force has to be balanced by the limiting friction f. i.e., f = lmg

Hence, coefficient of friction, \(\mu=\frac{f}{R}=\frac{l m g}{(L-l) m g}=\frac{l}{L-l}\)

or, \(l=\mu L-\mu l \quad \text { or, } l(1+\mu)=\mu L\)

∴ \(l=\frac{\mu L}{1+\mu} \text {. }\)

Example 7. A tram is moving with an acceleration of 49 cm · s^-2 using 50 % of its motor power; the remaining 50 % is used up to overcome friction. Find the coefficient of friction between the wheel and the tram line.
Solution:

Let the power of the motor be P, and the mass of the tram be m. Let the distance covered by the tram in time t be s.

As per given conditions, work against friction per second, \(\frac{\mu m g \times s}{t}=P \times 50 \%=\frac{P}{2}\)…(1)

and rate of work done for the accelerated motion, \(\frac{m a \times s}{t}=P \times 50 \%=\frac{P}{2}\) [where a = acceleration]…(2)

Now dividing (1) by (2) we get,

∴ \(\frac{\mu g}{a}=1 \text { or, } \mu=\frac{a}{g}=\frac{49}{980}=0.05 \text {. }\)

Class 11 Physics Laws Of Motion  – Angle Of Friction Cone Of Friction

Laws Governing Static and Kinetic Friction

Angle Of Friction: Let us consider a body placed on a rough surface. Now if a horizontal force is applied on the body, a frictional force comes into play. If the force applied on the body is increased, the friction also increases until it becomes equal to the limiting friction.

Let the limiting friction be denoted by \(\vec{f}\), the normal force acting on the body be denoted by \(\vec{R}\) and the resultant of the two forces \(\vec{R}\) and \(\vec{f}\) be denoted by \(\vec{Q}\). Now if \(\vec{Q}\) makes an angle A with \(\vec{R}\), then A is called the angle of friction.

Angle Of Friction Definition: The resultant of the limiting friction and the normal force between two surfaces in contact, makes an angle with the normal force. This angle is called the angle of friction.

From, R = Q cosλ, f = Q sinλ.

Again, \(\mu=\frac{f}{R}=\frac{Q \sin \lambda}{Q \cos \lambda}=\tan \lambda\)

Hence, the tangent of the angle of friction is equal to the coefficient of friction.

Also, \(R^2+f^2=Q^2\left(\cos ^2 \lambda+\sin ^2 \lambda\right)=Q^2\)

or, \(R^2+\mu^2 R^2=Q^2 \quad \text { or, } Q=R \sqrt{1+\mu^2}\)

Real-Life Examples of Static and Kinetic Friction

Cone Of Friction: Let a force be applied on a body at rest on a plane. The force of friction acts on the body exactly in the opposite direction. The applied force acts at point O of the body.

• If the direction of the applied force along the plane changes, the direction of the force of limiting friction f will accordingly change, thereby changing the direction of the resultant Q. But the direction of the normal force and the angle λ between Q and R remains the same, whatever be the direction of Q.
• Hence, taking into consideration the different directions of the applied force in the same plane, it is seen that the resultant Q makes an imaginary cone, with the normal force R as its axis. This imaginary cone is called the cone of friction.
• The vertex of the cone is the contact point O of the body and the plane, its axis is along the normal force R and the semi-vertical angle is the angle of friction λ.

Vector Long Answer Type Questions

Question 1. Under which condition will the magnitude of the scalar sum be equal to the magnitude of the vector sum?
Answer:

A scalar has only magnitude and therefore, the addition of scalars is the addition of magnitudes. When some vec¬tors have the same direction, then only their magnitudes are added in vector addition. In such a case, the magnitude of the vector sum is equal to the sum of the scalars.

Question 2. If \(|\vec{A}| \neq|\vec{B}|\), then is it possible that \(\vec{A}\) + \(\vec{B}\) =0? Explain.
Solution:

If \(\vec{A}\) + \(\vec{B}\) = 0 or, \(\vec{A}\) = –\(\vec{B}\); it means that \(\vec{A}\) and \(\vec{B}\) are equal in magnitude, but opposite in direction.

But it is given that \(|\vec{A}| \neq|\vec{B}|\), i.e., the magnitudes are not equal. So, \(\vec{A}\) + \(\vec{B}\) +0.

Understanding Vector Concepts Questions

Question 3. Can the sum of three vectors, i.e., their resultant, be equal to zero? Explain.
Answer:

Yes, if the resultant of any two of the vectors is equal and opposite to the third one, the resultant of the three vectors will be zero. Let a, b, c be three vectors related as \(\vec{a}, \vec{b}, \vec{c}\)

∴ \(\vec{a}+\vec{b}=-\vec{c} \text {. Then } \vec{a}+\vec{b}+\vec{c}=0\)

Question 4. Two stones are dropped at the same time from a height. The first is dropped from rest while the other is released with a small horizontal velocity. Which stone will reach the ground first?
Answer:

The initial velocity of the first stone is zero and that of the second stone is in the horizontal direction. Hence, the vertical component of the initial velocity of the second stone is also zero.

Consequently, the two stones will touch the ground at the same instant, but not at the same point. Due to an initial horizontal velocity, the second stone will have some horizontal displacement.

Question 5. A boy throws a ball vertically upward from a vehicle moving with a constant acceleration. Where would the ball land?
Answer:

The ball will fall behind the boy if he is facing the direction of acceleration.

• When the boy releases the ball, the horizontal velocities of both the ball and the vehicle are the same. The distance the ball covers after being released has nothing to do with its vertical velocity.
• However, the ball’s horizontal velocity remains the same, whereas that of the vehicle increases as it is moving with a constant acceleration. Therefore, the ball will travel a shorter distance than the vehicle will. Thus, it will land behind the boy.

WBBSE Class 11 Vector Q&A

Question 6. Can the value of a component of a vector be greater than the value of the vector Itself? Discuss the case of rectangular components in this context
Answer:

By the law of parallelogram of vectors, two components of a vector form two adjacent sides of a parallelogram with the vector as the diagonal. Since both the adjacent sides of a parallelogram can be longer than the diagonal, the value of the components can be greater than that of the vector.

But the rectangular components of a vector, being two adjacent sides of a rectangle, cannot be greater than the value of the vector, because the diagonal of a rectangle is always longer than any of the sides.

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Question 7. At zero wind speed, rainwater falls vertically with velocity V cm • s^-1 and is collected in a pot at a fixed rate. How will the rate of collection of rainwater change when the wind is blowing with a velocity of W cm s^-1 perpendicular to V?
Answer:

There will be no change in the rate of collection of rainwater due to wind velocity.

Rainwater falls vertically in the pot when there is no wind. If A cm² is the area of the cross-section of the pot, the rate of collection of rainwater is, x = AV cm³ · s^-1 …..(1)

Because of the wind, water will be filled in the pot obliquely. The component of the cross-sectional area of the pot perpendicular to the velocity of rain, is A₁ = A cosθ.

Here, the effective velocity of rain is \(V_1=\frac{V}{\cos \theta}\)

Hence, the rate of collection of water, \(x_1=A_1 \cdot V_1\)

= \(A \cos \theta \cdot \frac{V}{\cos \theta}=A V \mathrm{~cm}^3 \cdot \mathrm{s}^{-1}\)………(2)

Thus, from (1) and (2), it is inferred that the rate of collection of rainwater remains the same.

Question 8. Are the magnitudes of the two vectors (\(\vec{A}-\vec{B}\)) and (\(\vec{B}-\vec{A}\)) the same?
Answer:

The magnitudes of these two vectors are the same. But since \(\vec{B}-\vec{A}\) = –\(\vec{A}-\vec{B}\), their directions are opposite.

Question 9. Show that, if three forces acting on a particle can be taken sequentially to form the three sides of a trian¬gle, their resultant Is zero.
Answer:

Let \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) be the three forces acting at a point O. Now, taking these forces sequentially and keeping their magnitudes and directions the same, we obtain the three sides of a triangle, \(\overrightarrow{P Q}, \overrightarrow{Q R} \text { and } \overrightarrow{R P}\)

We have to show that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)

According to the triangle law of vector addition, \(\vec{a}+\vec{b}=\overrightarrow{P R}=-\overrightarrow{R P} \text { or, } \vec{a}+\vec{b}=-\vec{c}\)

or, \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\)

∴ The resultant of the three forces \(\vec{a}, \vec{b}, and \vec{c}\) is zero.

Applications of Vectors in Physics Q&A

Question 10. If the position coordinates of the points A and B are (x₁, y₁, z₁) and (x₂, y₂, z₂) respectively, determine the magnitude and direction of the vector \(\overrightarrow{A B}\).
Answer:

The position coordinates of A and B are (x₁, y₁, z₁) and (x₂, y₂, z₂) respectively.

∴ \(\overrightarrow{O A}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\) and \(\overrightarrow{O B}=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\)

From ΔOAB, we get, \(\overrightarrow{O A}+\overrightarrow{A B}=\overrightarrow{O B}\)

or \(\overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}\)

= \(\left(x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\right)-\left(x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\right)=\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}\)

The magnitude of the vector \(\overrightarrow{A B}\) is, AB = \(\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

The direction of \(\overrightarrow{A B}\) is obtained from its direction cosines \(\frac{x_2-x_1}{A B}, \frac{y_2-y_1}{A B} \text { and } \frac{z_2-z_1}{A B} \text {. }\)

Question 11. A person leans out of a train moving with uniform velocity and drops a coin. How does the path of motion of the coin appear to a co-passenger and a person standing outside the train near the rail tracks?
Answer:

When the coin is dropped from the train, due to inertia of motion it has a horizontal component of velocity equal to that of the train. Simultaneously, it experiences a vertical downward acceleration due to the force of gravity. If the train’s velocity is v, the velocity of the co-passenger is also v. Therefore, with respect to this person, the horizontal component of the velocity of the coin is, v- v = 0.

So, the passenger will see the coin drop vertically downwards due to the action of gravity. With respect to any person standing outside the train near the rail tracks, the coin will have both horizontal and vertical movements. Thus, it is essentially a projectile motion, and the coin follows a parabolic path.

Vector Addition and Subtraction Q&A

Question 12. State whether any physical quantity having magni¬tude and direction is a vector.
Answer:

All physical quantities with magnitude and direction are not vectors. To be a vector, the physical quantity must obey the rules of vector addition. For example, the flow of current is not a vector quantity though it has both magni¬tude and direction.

Question 13. Can the magnitude of the resultant of two vectors be less than either of them? Explain.
Answer:

\(\vec{a}\) = \(\overrightarrow{A B}\) and \(\vec{b}\) = \(\overrightarrow{B C}\). If the resultant of these two vectors is \(\vec{c}\), then according to the triangle law of vectors, \(\vec{c}\) = \(\overrightarrow{A C}\).

Obviously, the side AC of the triangle ABC may be smaller than AB or BC, or both. Thus, the magnitude of the resultant vector \(\vec{c}\) may be smaller than that of \(\vec{a}\) or \(\vec{b}\) or both.

Question 14. By adding three unit vectors is it possible to get a unit vector?
Answer:

If any two unit vectors out of the three are equal in magnitude but opposite in direction, then adding all the three vectors we obtain a resultant equal to the third unit vector. For instance, on adding \(\vec{a}\), –\(\vec{a}\) and \(\vec{b}\), we get, \(\vec{a}\)+(\(\vec{a}\)) + \(\vec{b}\) = \(\vec{b}\).

Question 15. Resultant of two vectors \(\vec{F}_1\) and \(\vec{F}_2\) is \(\vec{P}\) When \(\vec{F}_2\) is reversed, the resultant is \(\vec{Q}\). Show that \(\left(P^2+Q^2\right)=2\left(F_1^2+F_2^2\right)\)
Answer:

When \(\vec{F}_2\) is reversed it becomes –\(\vec{F}_2\).

From given conditions, \(\vec{F}_1+\vec{F}_2=\vec{P}, \vec{F}_1+\left(-\vec{F}_2\right)=\vec{F}_1-\vec{F}_2=\vec{Q}\)

∴ \(P^2+Q^2=\left(\vec{F}_1+\vec{F}_2\right)^2+\left(\vec{F}_1-\vec{F}_2\right)^2\)

= \(F_1^2+F_2^2+2 \vec{F}_1 \cdot \vec{F}_2+F_1^2+F_2^2-2 \vec{F}_1 \cdot \vec{F}_2\)

= \(2\left(F_1^2+F_2^2\right)\)

Question 16. How does the change of acceleration due to gravity affect the path of a projectile?
Answer:

The locus of a projectile is denoted by y = \(x \tan \alpha-\frac{g}{2 u^2 \cos ^2 \alpha} x^2\)…..(1)

Let u and α be constants.

Now from equation (1) we can say that both the maximum height and the horizontal range of a projectile decrease with increasing g. In this case, the locus of the projectile is shown by line a.

On the other hand, both quantities increase with decreasing g. In this case, the locus of the projectile is shown by line b.

Real-Life Examples of Vector Applications

Question 17. Can four non-coplanar vectors produce equilibrium? Give reasons.
Answer:

Four non-coplanar vectors can produce equilib¬rium when they obey the following condition: The resultant of any two vectors must be equal in magnitude but opposite in direction to the resultant of the other two vectors.

Question 18. Show that a stretched wire cannot remain horizontal when weight is suspended from its mid-point.
Answer:

Let the two ends A and B of the wire be rigidly fixed and a weight W be suspended from the mid-point O. In this condition, the two parts of the string OA and OB make equal angles θ with the horizontal and the tension on each part is T.

For equilibrium, the vertical components of T on each wire together balance the weight W, and the horizontal components balance each other.

∴ 2T sinθ = W or, sinθ = \(\frac{W}{2T}\)

Since, T cannot be infinitely large and W ≠ 0,

sinθ ≠ 0 i.e., θ ≠ 0°

Hence, the wire cannot remain horizontal when a weight is suspended from its mid-point (or from any other point along its length).

Question 9. Two wooden blocks are falling from the same height. One is falling down an inclined plane and the other is In a free fall. Out of the two

1. Which one will reach the ground first and 
2. Which one will have a higher velocity when it touches the ground?

Answer:

1. Let the length, height, and angle of inclination of the inclined plane be l, h, and θ respectively.

The acceleration of the block falling along the inclined Plane = g sinθ. Let the time taken by the first and the second blocks respectively to touch the ground be t₁ and t₂

For the first block, \(l=\frac{1}{2} g \sin \theta \cdot t_1^2\)

or, \(t_1^2=\frac{2 l}{g \sin \theta}=\frac{2 h}{g \sin ^2 \theta}\) (because \(\sin \theta=\frac{h}{l}\))

or, \(t_1=\frac{1}{\sin \theta} \sqrt{\frac{2 h}{g}}\)…..(1)

For the second block, h = \(\frac{1}{2} g t_2^2 \quad \text { or, } t_2=\sqrt{\frac{2 h}{g}}\)……(2)

As \(\sin \theta<1\)(because \(\theta<90^{\circ}\)), from equations (1) and (2) it is seen that \(t_1>t_2\), i.e., the block falling freely touches the ground first.

2. Let the velocities acquired by the first and the second blocks respectively be \(v_1\) and \(v_2\).

For the first block, \(v_1^2=2 g \sin \theta \cdot l=2 g \cdot \frac{h}{l} \cdot l=2 g h \text { or, } v_1=\sqrt{2 g h}\)

For the second block, \(v_2^2=2 g h \text { or, } v_2=\sqrt{2 g h}\)

∴ \(v_1=v_2\)

Hence, both the blocks touch the ground with the same velocity.

Step-by-Step Solutions to Vector Problems

Question 20. In a circus, a joker stands on a highly elevated plank with a ball in his hand. Another joker also stands with a rifle in his hand pointing it directly at the ball. If the rifle is fired precisely at the moment when the ball is released, will the bullet hit the bail? Air resis¬tance is negligible.
Answer:

The gravitational acceleration acting on the bullet and the ball are equal in magnitude and act vertically downwards. As the bullet is fired at the very moment the ball is released, the vertical displacements of both the bullet and the ball are the same. Therefore, the bullet will hit the ball.

Alternative method: The bullet from the rifle is directed straight to the ball. Let v₀ be the velocity with which the bullet leaves the rifle at an angle θ with the horizontal. Let us consider, the time taken by the bullet to pass across the vertical line MB at A is t.

Horizontal distance travelled by the bullet = OB = x

x = \(v_0 \cos \theta \times t \quad \text { or, } t=\frac{x}{v_0 \cos \theta}\)

∴ \(t^2=\frac{x^2}{v_0^2 \cos ^2 \theta}\)

For the vertical motion of the bullet, \(A B =v_0 \sin \theta \times t-\frac{1}{2} g t^2\)

= \(v_0 \sin \theta \times \frac{x}{v_0 \cos \theta}-\frac{1}{2} g \times \frac{x^2}{v_0^2 \cos ^2 \theta}\)

= \(x \tan \theta-\frac{1}{2} \frac{g x^2}{v_0^2 \cos ^2 \theta}\)

From the ΔOMB, \(\tan \theta=\frac{M B}{O B}=\frac{M B}{x}\) M B = \(x \tan \theta\)

Now, MA = \(M B-A B=x \tan \theta-\left[x \tan \theta-\frac{1}{2} \cdot \frac{g x^2}{v_0^2 \cos ^2 \theta}\right]\)

= \(\frac{1}{2} \frac{g x^2}{v_0^2 \cos ^2 \theta}=\frac{1}{2} g t^2\)

Thus in a time t, the bullet falls through a vertical distance \(\frac{1}{2} g t^2\) below M.

The vertical distance fallen by the ball is

h = \(u t+\frac{1}{2} g t^2=\frac{1}{2} g t^2\) [because u=0]

Thus the bullet and the ball will always reach point A at the same time. Hence the bullet will always hit the ball whatever be the velocity of the bullet.

Question 21. Under which condition will the magnitude of the resultant of two vectors be equal to that of any one of the constituent vectors?
Answer:

Suppose, a and b are the magnitudes of two vectors and the angle between them is θ. According to the problem let the magnitude of the resultant of \(\vec{a}\) and \(\vec{b}\) be equal to a.

Therefore, \(a^2=a^2+b^2+2 a b \cos \theta \text { or, } b(b+2 a \cos \theta)=0\)

∴ \(b+2 a \cos \theta\) =0(b≠0]

or, \(\theta=\cos ^{-1}\left(-\frac{b}{2 a}\right)\)

that is if the angle between \(\vec{a}\) and \(\vec{b}\) is = \(\cos ^{-1}\left(-\frac{b}{2 a}\right)\) or, \(\cos ^{-1}\left(-\frac{a}{2 b}\right)\), the magnitude of the resultant vector is equal to a or b respectively.

Vector Components and Resolution Questions

Question 22. If the angle between two vectors is slowly increased from 0 then what changes will be found in the resultant?
Answer:

Let \(\vec{a}\) and \(\vec{b}\) be the two vectors whose resultant is \(\vec{R}\). \(\vec{a}\) is directed along the positive x-axis and is constant, while \(\vec{b}\) moves from the positive x-axis in the anticlockwise direction.

Let at any moment, the angle between \(\vec{a}\) and \(\vec{b}\) be θ

∴ R = (a² + b² + 2abcosθ)½

From this equation we get,

when, θ = 0, R = a+ b

when, θ = 90°, R = (a²+b²)  when, 9 = 180°, R = |a-b|

when, θ = 270°, R = (a² + b²)½

when, θ = 360°, R = a+b

So, when the angle increases from 0° to 180° gradually, the magnitude of R decreases from (a+b) to |a-b|. Again, when θ increases from 180° to 360°, the magnitude of R increases from |a-b| to (a+b).

Question 23. If \(|\vec{a} \times \vec{b}|=\vec{a} \cdot \vec{b}\), them what us tghe angle between \(\vec{a}\) and \(\vec{b}\)?
Answer:

According to the question, \(|\vec{a} \times \vec{b}|=\vec{a} \cdot \vec{b}\)

∴ absinθ = abcosθ

[Let the angle between \(\vec{a}\) and \(\vec{b}\) be θ]

or, tan θ = 1 [a and b≠0]

∴ θ = 45°.

Question 24. What is the vector product of two equal vectors?
Answer:

The vector product of two equal vectors leads to a zero vector or null vector, i.e., the resultant vector has a magnitude equal to zero without any fixed direction. For any vector \(\vec{A},|\vec{A} \times \vec{A}|=(A)(A) \sin 0^{\circ}=0\).

∴\(\vec{A}\) x \(\vec{A}\)= \(\vec{0}\)

Question 25. Show that the projection or component of a vector \(\vec{R}\) on another vector \(\vec{A}\) is \(\vec{R}\) \(\hat{a}\), where a is a unit vector along \(\vec{A}\).
Answer:

In general, the component of a vector \(\vec{R}\) along a direction of another vector making an angle θ with it, is Rcosθ which is the projection of \(\vec{R}\) along the direction of that vector.

Consider two planes passing through the initial and terminal points of \(\vec{R}\), which are perpendicular to \(\vec{A}\) at F and H respectively.

Then, projection of \(\vec{R}\) on \(\vec{A}\) is FH = EG = Rcosθ = \(\vec{R}\) – \(\hat{a}\)

Question 26. If \(\vec{A}\) is a constant vector, then show that \(\frac{d \vec{A}}{d t}\) is perpendicular to \(\vec{A}\).
Answer:

⇒\(|\vec{A}|=\) constant

∴ \(\vec{A} \cdot \vec{A}=A^2=\) constant

∴ \(\frac{d}{d t}(\vec{A} \cdot \vec{A})=0\)

or, \(\vec{A} \cdot \frac{d \vec{A}}{d t}+\frac{d \vec{A}}{d t} \cdot \vec{A}=0 \text { or, } 2 \vec{A} \cdot \frac{d \vec{A}}{d t}=0 \text { or, } \vec{A} \cdot \frac{d \vec{A}}{d t}=0\)

So, \(\frac{d \vec{A}}{d t}\) is perpendicular to \(\vec{A}\).

Question 27. A water fountain on the ground sprinkles water all around it. If the speed of water coming out of the fountain is v, then what will be the total area around the fountain that gets wet?
Answer:

Let the angle with the horizontal be 6 and the hor¬izontal range of water coming out of the fountain be R.

∴ R = \(\frac{v^2 \sin 2 \theta}{g}\)

The value of R will be maximum when sin 2θ = 1

i.e., 2θ = 90° or, θ = 45°

∴ \(R_{\max }=\frac{v^2}{g}\)

So a circular area of radius R_max around the fountain gets wet.

∴ Total area, \(A=\pi R_{\max }^2=\pi\left(\frac{\nu^2}{g}\right)^2=\frac{\pi v^4}{g^2}\)

WBCHSE Class 11 Physics Vector Short Answer Questions

Vector Short Answer Type Questions And Answers

Question 1. The angle subtended by the vector \(\vec{A}=\sqrt{3} \hat{i}-\hat{j}\) with the y-axis is

1. \(60^{\circ}\)
2. \(240^{\circ}\)
3. \(120^{\circ}\)
4. \(45^{\circ}\)

Answer: 3. \(120^{\circ}\)

The option 3 is correct

Question 2. Two non-collinear unit vector \(\hat{a}\) and \(\hat{b}\) are such that \(|\hat{a}+\hat{b}|=\sqrt{3}\). Find the angle between the two unit vectors.
Answer:

⇒ \(|\vec{a}+\vec{b}|=\sqrt{3}\)

or, \(a^2+b^2+2 a b \cos \theta=(\sqrt{3})^2=3\)

or, \(\cos \theta=\frac{3-a^2-b^2}{2 a b}=\frac{3-\left(1^2-1\right)}{2 \cdot 1 \cdot 1}=\frac{1}{2}=\cos 60^{\circ}\)

∴ \(\theta=60^{\circ}\)

Question 3. The ratio between the values of the cross product and the dot product of two vectors is \(\frac{1}{\sqrt{3}}\). The angle between them is

1. 30
2. 45
3. 60
4. 120

Answer:

For \(\vec{A}\) and \(\vec{B}\), \(\frac{\text { value of cross product }}{\text { value of dot product }}=\frac{A B \sin \theta}{A B \cos \theta}=\tan \theta\)

∴ \(\tan \theta=\frac{1}{\sqrt{3}}=\tan 30^{\circ}\)

WBCHSE Class 11 Physics Vector Short Answer Questions

Question 4. \((2 \hat{i}+\hat{j}-\hat{k}) \mathrm{N}\) force is acting on a body of \(10 \mathrm{~kg}\) mass. If the body starts from rest, then after \(20 \mathrm{sec}\) what will be its velocity?
Answer:

Acceleration, \(\vec{a}=\frac{\text { force }}{\text { mass }}=\frac{2 \hat{i}+\hat{j}-\hat{k}}{10} \mathrm{~m} \cdot \mathrm{s}^{-2}=\text { constant }\)

So, the particle is moving with uniform acceleration.

After 20s, velocity, \(\vec{v}=\vec{u}+\vec{a} t=0+\frac{2 \hat{i}+\hat{j}-\hat{k}}{10} \cdot 20=(4 \hat{i}+2 \hat{j}-2 \hat{k}) \mathrm{m} \cdot \mathrm{s}^{-1}\)

Magnitude of the velocity, \(v=\sqrt{4^2+2^2+(-2)^2}=\sqrt{24}=2 \sqrt{6} \mathrm{~m} \cdot \mathrm{s}^{-1}\)

Question 5. If \(\vec{A}+\vec{B}+\vec{C}=0\) then show that \(\vec{A} \times \vec{B}=\vec{B} \times \vec{C}=\vec{C} \times \vec{A}\)
Answer:

⇒ \(\vec{B} \times \vec{C}=\vec{B} \times(-\vec{A}-\vec{B})=-\vec{B} \times \vec{A}-\vec{B} \times \vec{B}=+\vec{A} \times \vec{B}-0\)

= \(\vec{A} \times \vec{B}\)

Now, \(\vec{C} \times \vec{A}=(-\vec{A} \times-\vec{B}) \times \vec{A}=-\vec{A} \times \vec{A}-\vec{B} \times \vec{A}\)

= \(-0+\vec{A} \times \vec{B}=\vec{A} \times \vec{B}\)

∴ \(\vec{A} \times \vec{B}=\vec{B} \times \vec{C}=\vec{C} \times \vec{A}\)

WBCHSE Class 11 Physics Vector Short Answer Questions

WBBSE Class 11 Vector Short Answer Questions

Question 6. If \(\vec{A} \cdot \vec{B}=|\vec{A} \times \vec{B}|\), then the angle between \(\vec{A}\) and \(\vec{B}\) is

1. \(\pi\)
2. \(\frac{\pi}{2}\)
3. 0
4. \(\frac{\pi}{4}\)

Answer:

⇒ \(\vec{A} \cdot \vec{B}=|\vec{A} \times \vec{B}|\)

or, \(A B \cos \theta=A B \sin \theta\) or, \(\tan \theta=1\)

∴ \(\theta=\frac{\pi}{4}\)

The option 4 is correct

Class 11 Physics	Class 12 Maths	Class 11 Chemistry
NEET Foundation	Class 12 Physics	NEET Physics

Question 7. \(\vec{a} \times \vec{b}\) is not equal to \(\vec{b} \times \vec{a}\). Why?
Answer:

∴ \(\vec{a} \times \vec{b}\) and \(\vec{b} \times \vec{a}\) both are equal in magnitude but opposite in direction.

West Bengal Class 11 Physics Vector SAQs 

Question 8. Find a unit vector that is perpendicular to both \(\vec{A}=3 \hat{i}+\hat{j}+2 \hat{k}\) and \(\vec{B}=2 \hat{i}-2 \hat{j}+4 \hat{k}\).
Answer:

The unit vector perpendicular to both \(\vec{A}\) and \(\vec{B}\),

∴ \(\hat{n}= \pm \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}\)

Now, \(\vec{A} \times \vec{B}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 2 \\ 2 & -2 & 4\end{array}\right|=8 \hat{i}-8 \hat{j}-8 \hat{k}\)

∴ \(|\vec{A} \times \vec{B}|=8 \sqrt{3}\)

∴ \(\hat{n}= \pm \frac{1}{\sqrt{3}}(\hat{i}-\hat{j}-\hat{k})\)

Question 9.

1. Determine the relation between the kinetic energy of a projectile at maximum height and at initial position for maximum range.
2. What is a null vector?

Answer:

Let the mass of the projectile = m and the velocity of projection = v.

Kinetic energy at the time of projection, E = 1/2 mv².

For maximum horizontal range, the angle of projection, = 45

At highest point, kinetic energy, \(E^{\prime}=\frac{1}{2} m\left(v \cos 45^{\circ}\right)^2=\frac{1}{2} m v^2 \times \frac{1}{2}=\frac{E}{2}\)

A null vector is a vector having a magnitude zero and no fixed direction.

Question 10. Which quantity remains unchanged in the case of a projectile?

1. Momentum
2. Kinetic energy
3. Vertical component of velocity
4. Horizontal component of velocity

Answer:

There is no horizontal acceleration of a projectile.

The option 4 is correct.

West Bengal Class 11 Physics Vector SAQs 

Vector Addition and Subtraction Short Answers

Question 11. Determine the unit vector along the vector \(\vec{A}=\hat{i}+3 \hat{j}+4 \hat{k} \text {. }\)
Answer:

The unit vector, \(\hat{A}=\frac{\vec{A}}{|\vec{A}|}=\frac{\hat{i}+3 \hat{j}+4 \hat{k}}{\sqrt{1^2+3^2+4^2}}=\frac{1}{\sqrt{26}} \hat{i}+\frac{3}{\sqrt{26}} \hat{j}+\frac{4}{\sqrt{26}} \hat{k}\)

Question 12. At what angle should the two forces \((\vec{A}+\vec{B})\) and \((\vec{A}-\vec{B})\) act so that their resultant be \(\sqrt{3 A^2+B^2} \text {. }\)
Answer:

If \(\vec{a}=(\vec{A}+\vec{B})\), then \(a^2=\vec{a} \cdot \vec{a}=(\vec{A}+\vec{B}) \cdot(\vec{A}+\vec{B})=A^2+B^2+2 \vec{A} \cdot \vec{B}\)

If \(\vec{b}=(\vec{A}-\vec{B})\), similarly \(b^2=\vec{b} \cdot \vec{b}=A^2+B^2-2 \vec{A} \cdot \vec{B}\)

∴ \(\vec{a} \cdot \vec{b}=(\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})\)

= \(A^2-\vec{A} \cdot \vec{B}+\vec{A} \cdot \vec{B}-B^2=A^2-B^2\)

If the angle between \(\vec{a}\) and \(\vec{b}\) is \(\alpha, \vec{a} \cdot \vec{b}=a b \cos \alpha\)

∴ \(a b \cos \alpha=A^2-B^2\)

If \(\vec{c}\) be the resultant of \(\vec{a}\) and \(\vec{b}\),

\(c^2=a^2+b^2+2 a b \cos \alpha or, \quad 3 A^2+B^2=\left(A^2+B^2+2 \vec{A} \cdot \vec{B}\right)\)

+ \(\left(A^2+B^2-2 \vec{A} \cdot \vec{B}\right)+2\left(A^2-B^2\right)\)

or, \(3 A^2+B^2=4 A^2\) or, \(A^2-B^2=0\)

Hence, \(a b \cos \alpha=A^2-B^2=0\)

or, \(\cos \alpha=0\) or, \(\alpha=90^{\circ}\)

If the two forces act at an angle \(90^{\circ}\), their resultant will be \(\sqrt{3 A^2+B^2}\).

Class 11 Physics Vector Short Questions WBCHSE 

Question 13. If \(\vec{A}=0.4 \hat{i}+0.3 \hat{j}+c \hat{k}\) be a unit vector, then what is the value of c?
Answer:

∴ \(|\vec{A}|=\sqrt{(0.4)^2+(0.3)^2+c^2}=1\)

or, \(0.16+0.09+c^2=1\) or, \(c=\frac{\sqrt{3}}{2}\)

Question 14. Find the angle between the two vectors \(\vec{A}=\hat{i}-2 \hat{j}+3 \hat{k}\) and \(\vec{B}=2 \hat{i}+\hat{j}+3 \hat{k}\).
Answer:

⇒ \(\vec{A}= \hat{i}-2 \hat{j}+3 \hat{k}, \vec{B}=2 \hat{i}+\hat{j}+4 \hat{k}\)

⇒ \(\vec{A} \cdot \vec{B}=1 \times 2+(-2) \times 1+3 \times 4=12\)

⇒ \(|\vec{A}|=A=\sqrt{1^2+(-2)^2+3^2}=\sqrt{14}\)

⇒ \(|\vec{B}|=B=\sqrt{2^2+1^2+4^2}=\sqrt{21}\)

Now, \(\vec{A} \cdot \vec{B}=A B \cos \theta[\theta\) is the angle between the vectors \(\vec{A}\) and \(\vec{B}\).

or, \(\cos \theta=\frac{\vec{A} \cdot \vec{B}}{A B}=\frac{12}{7 \sqrt{6}}\)=\(\frac{2 \sqrt{6}}{7}\) or, \(\theta=\cos ^{-1}\left(\frac{2 \sqrt{6}}{7}\right)\)

Question 15. What are the quantities that remain constant during the motion of the particle?
Answer:

The horizontal component of the velocity of the particle and its downward acceleration (acceleration due to gravity) remain constant

Question 16. Consider three vectors \(\vec{A}=\hat{i}+\hat{j}-2 \hat{k}, \vec{B}=\hat{i}-\hat{j}+\hat{k}\) and \(\vec{C}=2 \hat{i}-3 \hat{j}+4 \hat{k}\). A vector \(\vec{X}\) of the form \(\alpha \vec{A}+\beta \vec{B}\)(α and β are numbers) is perpendicular to \(\vec{C}\). The ratio of α and β is

1. 1: 1
2. 2: 1
3. -1: 1
4. 3: 1

Answer:

∴ \((\alpha \vec{A}+\beta \vec{B}) \cdot \vec{C}=0\)

or, \(2(\alpha+\beta)-3(\alpha-\beta)+4(\beta-2 \alpha)=0\)

or, \(-9 \alpha+9 \beta=0 or, \alpha: \beta=1: 1\)

The option 1 is correct.

Real-Life Examples of Vector Applications

Question 17. A cricket ball thrown across a field is at heights h₁ and h₂ from the point of projection at times t₁ and t₂ respectively after the throw. The ball is caught by a fielder at the same height as that of projection. The time of flight of the ball in this journey is

1. \(\frac{h_1 t_2^2-h_2 t_1^2}{h_1 t_2-h_2 t_1}\)
2. \(\frac{h_1 t_2^2+h_2 t_2^2}{h_2 t_1+h_1 t_2}\)
3. \(\frac{h_1 t_2^2+h_2 t_1^2}{h_1 t_2+h_2 t_1}\)
4. \(\frac{h_1 t_1^2-h_2 t_2^2}{h_1 t_1-h_2 t_2}\)

Answer:

⇒ \(h_1=(u \sin \theta) t_1-\frac{1}{2} g t_1^2 ; h_2=(u \sin \theta) t_2-\frac{1}{2} g t_2^2\)

∴ \(\frac{h_1+\frac{1}{2} g t_1^2}{h_2+\frac{1}{2} g t_2^2}=\frac{t_1}{t_2} \quad \text { or, } h_1 t_2-h_2 t_1=\frac{g}{2}\left(t_1 t_2^2-t_1^2 t_2\right)\)

So, time of flight is given by T = \(\frac{2 u \sin \theta}{g}=\frac{2}{g}\left[\frac{h_1+\frac{1}{2} g t_1^2}{t_1}\right]=\frac{2}{t_1}\left[\frac{h_1}{g}+\frac{t_1^2}{2}\right]\)

= \(\frac{h_1}{t_1} \times\left(\frac{t_1 t_2^2-t_1^2 t_2}{h_1 t_2-h_2 t_1}\right)+t_1=\frac{h_1 t_2^2-h_2 t_1^2}{h_1 t_2-h_2 t_1}\)

Class 11 Physics Vector Short Questions WBCHSE 

Question 18. Particle A moves along the x-axis with a uniform velocity of magnitude 10 m/s. Particle B moves with a uniform velocity of 20 m/s along a direction making an angle of 60° with the positive direction of the x-axis as shown The relative velocity of B with respect of that of A is

1. 10 m/s along x-axis
2. 10√3 m/s along the y-axis (perpendicular to the x-axis)
3. 10√5 m/s along the bisection of the velocities of A and B
4. 30 m/s along negative x-axis

Answer:

cos60° = \(\frac{1}{2}\) = \(\frac{10}{20}\)

So, the third side of the triangle will be parallel to the y-axis

Length of the third side = \(\sqrt{20^2-10^2}=10 \sqrt{3}\)

Therefore, the relative velocity  of B with respect to that of A = 10√3 m/s along the y-axis

The option 2 is correct.

Question 19. The vectors \(\vec{A}\) and  \(\vec{B}\) are such that \(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\).

1. 0°
2. 60°
3. 90°
4. 45°

Answer:

Here, \(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\)

or, \(A^2+B^2+2 A B \cos \theta=A^2+B^2-2 A B \cos \theta\)

(θ is the angle between \(\vec{A}\) and \(\vec{B}\))

or, \(4 A B \cos \theta=0\) or, \(\cos \theta=0\) or, \(\theta=90^{\circ}\)

The option 3 is correct.

Class 11 Physics Vector Short Questions WBCHSE 

Question 20. Three vectors \(\vec{A}=a \hat{i}+\hat{j}+\hat{k}; \quad \vec{B}=\hat{i}+b \hat{j}+\hat{k}\) and \(\vec{C}=\hat{i}+\hat{j}+c \hat{k}\) are mutually perpendicular (\(\hat{i}, \hat{j} \text { and } \hat{k}\) are unit vectors along X Y and Z axis respectively). The respective values of a, b, and c are

1. 0,0,0
2. \(-\frac{1}{2}, \frac{-1}{2}, \frac{-1}{2}\)
3. 1,-1,1
4. \(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}\)

Answer:

As \(\vec{A}\), \(\vec{B}\) and \(\vec{C}\) are perpendicular to each other, \(\vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{C}=\vec{C} \cdot \vec{A}=0\)

a + b + 1 = 1 + b + c = a + 1 + c = 0

a+ b + 1= 0……(1)

1 + b + c = 0…..(2)

a + 1 + c = 0………(3)

Solving equations (1), (2), and (3) we get a = b = c = -1/2

The option 4 is correct.

Step-by-Step Solutions to Vector Problems

Question 21. In a triangle ABC, the sides AB and AC are represented by the vectors \(3 \hat{i}+\hat{j}+\hat{k} \text { and } \hat{i}+2 \hat{j}+\hat{k}\) respectively. Calculate the angle ∠ABC.

1. \(\cos ^{-1} \sqrt{\frac{5}{11}}\)
2. \(\cos ^{-1} \sqrt{\frac{6}{11}}\)
3. \(\left(90^{\circ}-\cos ^{-1} \sqrt{\frac{5}{11}}\right)\)
4. \(\left(180^{\circ}-\cos ^{-1} \sqrt{\frac{5}{11}}\right)\)

Answer:

Given, \(\overrightarrow{A B}=(3 \hat{i}+\hat{j}+\hat{k}) and \overrightarrow{A C}=(\hat{i}+2 \hat{j}+\hat{k})\)

From the triangle law of vector, \(\overrightarrow{A B}+\overrightarrow{B C}=\overrightarrow{A C}\)

or, \(\overrightarrow{B C}=\overrightarrow{A C}-\overrightarrow{A B}=(\hat{i}+2 \hat{j}+\hat{k})-(3 \hat{i}+\hat{j}+\hat{k})=-2 \hat{i}+\hat{j}\)

∠ABC is the angle between \(\overrightarrow{B A}\) and $\(\overrightarrow{B C}\).

∴ \(\overrightarrow{B A} \cdot \overrightarrow{B C}=|\overrightarrow{B A}||\overrightarrow{B C}| \cos \theta$\)

or, \((6-1)=\sqrt{3^2+1^2+1^2} \times \sqrt{(-2)^2+1^2} \times \cos \theta\)

or, \(\frac{5}{\sqrt{55}}=\cos \theta\) or, \(\theta=\cos ^{-1}\left(\sqrt{\frac{5}{11}}\right)\)

The option 1 is correct

WBCHSE Physics Vector Chapter Short Answer Questions 

Question 22. A projectile is fired from the surface of the earth with a velocity of 5 m · s^-1 and angle θ with the horizontal. Another projectile fired from another planet with a velocity of 3m · s^-1 at the same angle follows a trajectory that is identical to the trajectory of the projectile fired from the Earth. The value of the acceleration due to gravity on the planet is (in m · s^-2) (given g = 9.8m· s^-2)

1. 3.5
2. 5.9
3. 16.3
4. 110.8

Answer:

Since trajectories of both cases are identical, \(R_{\max }=R_{\max }^{\prime}\) (Here \(R_{\max }^{\prime}\) is the horizontal range in the second case)

∴ \(R_{\max }=\frac{u_1^2}{g} \sin 2 \theta \text { and } R_{\max }^{\prime}=\frac{u_2^2}{g^{\prime}} \sin 2 \theta\)

So, \(\frac{u_1^2}{g} \sin 2 \theta=\frac{u_2^2}{g^{\prime}} \sin 2 \theta\)

∴ \(g^{\prime}=\frac{u_2^2 g}{u_1^2}=\frac{3^2 \times 9.8}{5^2}=3.5\)

The option 1 is correct.

Question 23. A particle is moving such that its position coordinates (x, y) are (2 m, 3 m) at time t = 0, (6 m, 7 m) at time t = 2s and (13m, 14m) at time t = 5s. Average velocity vector \(\left(\vec{V}_{\mathrm{av}}\right)\) from t = 0 and  t = 5s.

1. \(\frac{1}{5}(13 \hat{i}+14 \hat{j})\)
2. \(\frac{7}{3}(\hat{i}+\hat{j})\)
3. \(2(\hat{i}+\hat{j})\)
4. \(\frac{11}{5}(\hat{i}+\hat{j})\)

Answer:

At t=0, \(\vec{r}_1=2 \hat{i}+3 \hat{j}\) and at \(t=5 \mathrm{~s}, \overrightarrow{r_3}=13 \hat{i}+14 \hat{j}\)

∴ \(V_{\mathrm{av}}=\frac{\left(x_2 \hat{i}-x_1 \hat{i}\right)+\left(y_2 \hat{j}-y_1 \hat{j}\right)}{5}\)

= \(\frac{(13-2) \hat{i}+(14-3) \hat{j}}{5}\)

= \(\frac{11 \hat{i}+11 \hat{j}}{5}=\frac{11}{5}(\hat{i}+\hat{j})\)

Option 4 is correct.

Question 24. A ship A is moving westwards with a speed of 10 km · h^-1 and ship B, initially 100 km south of A, is moving northwards with a speed of 10 km · h^-1. The time after which the distance between them becomes the shortest is

1. 0 h
2. 5 h
3. 5√2 h
4. 10√2 h

Answer:

Let the required time be t hours.

If we suppose the x-axis along the eastward direction and the y-axis along the northward direction, then the position of ship A after time t,  \(\vec{r}_A=(-10 \hat{i}) t\)

The position of ship B after time t, \(\vec{r}_B=-100 \hat{j}+(10 \hat{j}) t\)

Therefore, the position of B with respect to A, \(\vec{r}_B-\vec{r}_A=(10 t) \hat{i}+(10 t-100) \hat{j}\)

So, \(\left|\vec{r}_B-\vec{r}_A\right|=\sqrt{(10 t)^2+(10 t-100)^2}\)

= \(\sqrt{100 t^2+100 t^2-2000 t+10000}\)

= \(10 \sqrt{2} \sqrt{t^2-10 t+50}\)

This distance becomes the shortest when \(\left(t^2-10 t+50\right)\) becomes minimum, i.e., \(\frac{d}{d t}\left(t^2-10 t+50\right)=0 \quad \text { or, } 2 t-10=0\)

∴ t = 5 hours

The option 2 is correct

Question 25. If the magnitude of the sum of two vectors is equal to the magnitude of the difference of the two vectors, the angle between these vectors is

1. 90°
2. 45°
3. 180°
4. 0°

Answer: 1. 90°

The option 1 is correct.

WBCHSE Physics Vector Chapter Short Answer Questions 

Question 26. A ball of mass 1 kg is thrown vertically upwards and returns to the ground after 3 seconds. Another ball, thrown at 60° with vertical also stays in air for the same time before it touches the ground. The ratio of the two heights are

1. 1:3
2. 1:2
3. 1:1
4. 2:1

Answer:

Both of the balls stay in the air for the same time before touching the ground. Hence, the vertical components of the velocities of the balls along with the heights are equal. So, the ratio of the two heights is 1:1.

The option 3 is correct.

Vector Components and Resolution Questions

Question 27. The angle between \(\vec{A}-\vec{B} \text { and } \vec{A} \times \vec{B} \text { is }(\vec{A} \neq \vec{B})\)

1. 60°
2. 90°
3. 120°
4. 45°

Answer:

⇒ \(\vec{A}-\vec{B}\) lies on the same plane of \(\vec{A}\) and \(\vec{B}\). Again, the direction of vector \(\vec{A}\) x \(\vec{B}\) is along the perpendicular to \(\vec{A}\) – \(\vec{B}\).

∴ The angle between \(\vec{A}\)–\(\vec{B}\) and \(\vec{A}\) x \(\vec{B}\) =90°

The option 2 is correct.

Question 28. The moment of the force, \(\vec{F}=4 \hat{i}+5 \hat{j}-6 \hat{k}\) at (2, 0, -3), about the point (2, -2, -2), is given by

1. \(-7 \hat{i}-8 \hat{j}-4 \hat{k}\)
2. \(-4 \hat{i}-\hat{j}-8 \hat{k}\)
3. \(-8 \hat{i}-4 \hat{j}-7 \hat{k}\)
4. \(-7 \hat{i}-4 \hat{j}-8 \hat{k}\)

Answer:

⇒ \(\vec{F}=4 \hat{i}+5 \hat{j}-6 \hat{k}\)

⇒ \(\vec{r}=(2 \hat{i}+0 \hat{j}-3 \hat{k})-(2 \hat{i}-2 \hat{j}-2 \hat{k})=0 \hat{i}+2 \hat{j}-\hat{k}\)

⇒ \(\vec{\tau}=\vec{r} \times \vec{F}=(2 \hat{j}-\hat{k}) \times(4 \hat{i}+5 \hat{j}-6 \hat{k})\)

= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
0 & 2 & -1 \\
4 & 5 & -6
\end{array}\right|=-7 \hat{i}-4 \hat{j}-8 \hat{k}\)

The option 4 is correct.

WBCHSE Physics Vector Chapter Short Answer Questions 

Question 29. A body is projected horizontally from the top of a building of height h. The velocity of projection is u. Find

1. The time it will take to reach the ground,
2. The horizontal distance between the foot of the building and the ground where it will strike,
3. The velocity with which the body will reach the ground.

Answer:

u = initial velocity at the point A

So, u = horizontal component of initial velocity,

0 = vertical component of the initial velocity.

The horizontal component, u = constant, since there is no acceleration in that direction. But the vertical motion is under a constant acceleration g, the acceler¬ation due to gravity.

Let B be the point where the body strikes the ground.

The corresponding vertical motion is through the distance AO = h.

1. For the vertical motion AO, h = \(0 \cdot t+\frac{1}{2} g t^2 \quad \text { or, } t=\sqrt{\frac{2 h}{g}}\)

As it corresponds to the actual of flight is, \(t=\sqrt{\frac{2 h}{g}}\)

2. Horizontal range, OB = uniform horizontal velocity x time of flight.

or, \(R=u \sqrt{\frac{2 h}{g}}\)

3. After a time t, the horizontal component of velocity = u, and the vertical component of velocity = 0 + gt = gt.

∴ The resultant velocity at B = \(\sqrt{u^2+(g t)^2}=\sqrt{u^2+g^2 \frac{2 h}{g}}=\sqrt{u^2+2 g h}\)

If it makes an angle θ with the horizontal, then \(\theta=\tan ^{-1}\left(\frac{g t}{u}\right)\)

Common Vector Questions for Class 11

Question 30. Explain why it is easier to pull a lawn mower than to push it.
Answer:

The force F is usually applied obliquely relative to the ground, during either a push or a pull of the lawn mower. In both cases, force F is resolved into two components, one is horizontal and the other is vertical.

The horizontal component H of the force F is responsible for the horizontal motion. The difference is, during the push the vertical component V increases the effective weight of the mower, whereas during the pull the effective weight decreases due to V. So it is easier to pull the mower than to push it.

Question 31. On a two-lane road, car A is traveling at a speed of 36 km · h^-1. Two cars B and C approach car A from opposite directions with speeds of 54 km · h^-1 each. At a certain instant, when both cars B and C are at a distance of 1 km from A, B decides to overtake car A before C does. What minimum acceleration is required of B to avert an accident?
Answer:

Velocity of car A, v_A = 36km · h^-1 = 10m · s^-1

Let v_B and v_C be the velocities of cars B and C.

∴ v_B = v_C = 54 km · h^-1 = 15 m · s^-1

Velocity of car B relative to A, V_BA = V_B – V_A = 15-10 = 5 m · s^-1

Velocity of car C relative to A, V_CA = v_C + (v_A) = V_C+V_A= 15 + 10 = 25 m · s^-1

Also, the distance of B and C from A is 1 km = 100 m. Let t = time taken by car C to travel a distance of 1 km towards A.

∴ S = v_CAt or, 1000 = 25 for, t = 40 s

Suppose a = minimum acceleration required for car B to avoid the accident.

With this acceleration, car B overtakes car A in 40 s.

So, \(S=v_{B A} t+\frac{1}{2} a t^2\) or, \(1000=5 \times 40+\frac{1}{2} a \times(40)^2\)

∴ a = \(\frac{(1000-200) \times 2}{(40)^2}=1 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

Vector SAQs for Class 11 Physics WBCHSE 

Question 32. Find the condition for which \(|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}| ?\)
Answer:

⇒ \(|\vec{A}+\vec{B}|^2=|\vec{A}-\vec{B}|^2\)

or, \(A^2+B^2+2 A B \cos \theta=A^2+B^2-2 A B \cos \theta\)

or, \(4 A B \cos \theta=0\) or, \(\cos \theta=0=\cos 90^{\circ}\) or, \(\theta=90^{\circ}\)

∴ The condition is, \(\vec{A} \perp \vec{B}\)

Key Terms in Vectors Explained

Question 33. Two parallel rail tracks run north-south. Train A moves north at a speed of 54 km/h, and Train B moves south at a speed of 90 km/h. What is the

1. The velocity of B with respect to A
2. The velocity of ground with respect to B
3. Velocity of a monkey, running on the roof of train A against its motion with a velocity of 18 km/h with respect to the train, as observed by a man standing on the ground.

Answer:

Let us consider the south-to-north direction to be positive.

Here, V_A = +54 km/h = 15 m/s

v_B = -90 km/h = -25 m/s

1. Velocity of B with respect to A, V_BA = V_B – V_A = -25-15 = -40 m/s towards south

2. Velocity of ground with respect to B = 0-v_B =0-(-25) =25 m/s towards north

3. Let the velocity of the monkey with respect to the ground be v_m.

So, the velocity of monkey with respect to A = v_m– v_A =-18 km/h =-5 m/s

∴ v_m = V_A -5 = 15-5 =10 m/s towards north.

Question 34. Is it possible to have a constant rate of change of velocity when velocity changes both in magnitude and direction?
Answer:

Yes, for example—a body moving upwards or downwards where acceleration is constant while magnitude and direction change.

Question 35. What are two angles of projection of a projectile projected with a velocity 30 m/s, so that the horizontal range is 45m? Take g = 10 m/s².
Answer:

Horizontal range, R = \(\frac{u^2 \sin 2 \theta}{g}\)

or, \(\sin 2 \theta=R \cdot g / u^2\)

or, \(\sin 2 \theta=\frac{R \cdot g}{30^2}=\frac{45 \times 10}{30 \times 30}=\frac{1}{2}=\sin 30^{\circ}\)

∴ \(\theta=15^{\circ}\) and \(\left(90^{\circ}-15^{\circ}\right)=75^{\circ}\)

Vector SAQs for Class 11 Physics WBCHSE 

Question 36. If a projectile has a constant initial speed and angle of projection, find the relation between the changes in the horizontal range due to a change in acceleration due to gravity.
Answer:

R = \(\frac{u^2 \sin 2 \theta}{g}\)

⇒ \(R_1=\frac{u^2 \sin 2 \theta}{g_1} ; R_2=\frac{u^2 \sin 2 \theta}{g_2}\)

∴ \(\frac{R_1}{R_2}=\frac{g_2}{g_1}\)

Hence, the horizontal range is inversely proportional to the acceleration due to gravity.

Question 37. An old man walks 10 m due east from his house and then turns to his left at an angle of 60° east. He then walked 10 m in that direction fell down on the ground and got injured. His grandson observing him moves straight towards him from the initial position of his grandfather, helps him to stand, and takes him safely home,

1. Should the boy follow the same path followed by the old man? If not, why?
2. What are the values you suggest for the boy’s reply?

Answer:

Boy should take the shortest path, i.e., direct from A to C.

Caring, by holding and supporting his grandfa¬ther from his shoulders,

1. Call nearby people for help,
2. Gives first aid to his grandfather at home,
3. Taking grandfather to hospital.

 

Vector Very Short Answer Type Questions

Question 1. We usually say that ‘time moves in a forward direction’, but time is not a vector quantity. Why?
Answer: Does not obey vector algebra

Question 2. What change takes place in the value of the resultant of two vectors when the angle between them is increased from 0 to 90°?
Answer: Decreases

Read And Learn More WBCHSE Class 11 Physics Very Short Question And Answers

Question 3. Is any physical quantity having a magnitude and a direction a vector quantity?
Answer: No

Question 4. Can the resultant of three coplanar vectors be zero?
Answer: Yes

Question 5. “If the magnitudes and directions of three forces acting on a particle are represented by three sides of a triangle taken in order, the particle remains in equilibrium”— state whether the statement is true or false.
Answer: True

Question 6. What is a free vector?
Answer: Whose initial and final points are not fixed

Question 7. What are orthogonal unit vectors?
Answer: \(\hat{i}, \hat{j} \text { and } \hat{k}\) which are mutually perpendicular to each other

Question 8. What is the position vector of the origin of a coordinate system?
Answer: Zero

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Question 9. Magnitude of the resultant of two vectors is minimal when they are
Answer: In the opposite direction

Question 10. The resultant of two vectors of magnitudes 3 units and 4 units 5 units. What is the angle between the vectors?
Answer: \(\frac{\pi}{2}\)

Question 11. If an acceleration acts on a moving object along the direction of motion, the velocity of the object
Answer: Increases

Question 12. Value of the resultant of \((\vec{A}+\vec{B}) \text { and }(\vec{A}-\vec{B})\) is
Answer: 2  \(\vec{A}\)

Question 13. If \(\left|\vec{v}_1+\vec{v}_2\right|=\left|\vec{v}_1-\vec{v}_2\right| \text { and } \vec{v}_1 \text { and } \vec{v}_2\) have finite values then \(\vec{v}_1 \text { and } \vec{v}_2\) are
Answer: Mutually perpendicular

Question 14. Can commutative law be applied to vector subtraction?
Answer: No

Question 15. Can we apply associative law to vector subtraction?
Answer: Yes

WBBSE Class 11 Vector Very Short Answer Questions

Question 16. How many components can a vector be resolved into?
Answer: Infinite

Question 17. Is a rocket in flight an illustration of a projectile?
Answer: No

Question 18. What is the angle of projection for attaining maximum vertical height?
Answer: 90°

Question 19. What is the angle between two vectors whose vector product is zero?
Answer: 0°

Question 20. What is the scalar product of two vectors perpendicular to each other?
Answer: 0

Question 21. What is the angle between \((\vec{A}+\vec{B}) \text { and }(\vec{A} \times \vec{B}) \text { ? }\)
Answer: 90°

Question 22. Can the value of  \(\vec{A} \times \vec{A}\) be 0?
Answer: Yes

Question 23. If \(\hat{i}\) and \(\hat{j}\) are unit vectors along x and y axes respectively then the angle made by (\(\hat{i}\) + \(\hat{j}\)) vector with the x-axis is
Answer: 45°

Question 24. What is the angle between the vectors \(\vec{A} \text { and } \vec{A} \times \vec{B} \text {? }\)
Answer: 90°

Question 25. What is the angle between vector \(\vec{A}\) and the resultant of \((\vec{A}+\vec{B}) \text { and }(\vec{A}-\vec{B}) ?\)
Answer: Zero

Question 26. Skating on a circular ice slab of a radius of 200 m, three girls travel between diametrically opposite points P and Q along three different paths. Find out the magnitude of the displacement vector for each of them. For which of the girls, this magnitude is equal to the actual distance traveled by her?
Answer: 400 m for each B

Real-Life Examples of Vector Applications

Vector Assertion Reason Type Questions And Answers

These questions have statement 1 and statement 2. Of the four choices given below, choose the one that best describes the two statements.

1. Statement 1 is true, statement 2 is true; statement 2 Is a correct explanation for statement 1.
2. Statement 1 is true, and statement 2 is true; statement 2 is not a correct explanation for statement 1.
3. Statement 1 is true, statement 2 is false.
4. Statement 1 is false, and statement 2 is true.

Question 1. Statement 1: The minimum number of unequal vectors on a plane required to give zero resultant is three.

Statement 2: If \(\vec{B}+\vec{A}+\vec{C}=0\), then they must lie on the same plane.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 2. Statement 1: If <p is the angle between \(\vec{P}\) and \(\vec{Q}\), then \(\tan \phi=\frac{|\vec{P} \times \vec{Q}|}{\vec{P} \cdot \vec{Q}}\)

Statement 2. Statement 1 is true, statement 2 is false.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 3. Statement 1: If two vectors \(\vec{a}\), and \(\vec{b}\), are such that \(|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|\), then the angle between \(\vec{a}\) and \(\vec{b}\) is 90°.

Statement 2: \(\vec{a}+\vec{b}=\vec{b}+\vec{a}\).

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 4. Statement 1: A physical quantity cannot be called a vector if its magnitude is zero.

Statement 2: A vector has both magnitude and direction.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 5. Statement 1: Finite angular displacement is not a vector quantity.

Statement 2: A vector must obey the proper law of addition.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 Is a correct explanation for statement 1.

Question 6. Statement 1: The vector sum of two vectors is always greater than their vector difference.

Statement 2: If \(\vec{A}\) and \(\vec{B}\) are perpendicular to each other, the magnitudes of \(\vec{A}\) + \(\vec{B}\) and \(\vec{A}\)–\(\vec{B}\) are the same.

Answer: 4. Statement 1 is false, statement 2 is true.

Question 7. Statement 1: (Ax B)- (BxA) is -A²B²sin²θ. Here d is the angle between A and B.

Statement 2: (A x B) and (B x A) are two antiparallel vectors provided A and B are neither parallel nor antiparallel.

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 Is a correct explanation for statement 1.

Question 8. Statement 1: The horizontal range of a projectile is the same for angles 30° and 60° of projection.

Statement 2: The horizontal range of the projectile is independent of the angle of projection.

Answer: 3. Statement 1 is true, statement 2 is false.

Question 9. Statement 1: During the flight of a projectile, the horizontal component of its velocity remains uniform.

Statement 2: The vertical component of the velocity of a projectile becomes zero at the highest point of its path.

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Unit Vectors and Their Importance Short Answers

Question 10. Statement 1: Two orthogonal components of a force of magnitude 25 N may be 24 N and 7N.

Statement 2: If \(\quad|\vec{A}|=|\vec{B}|=1, |\vec{A} \times \vec{B}|^2+|\vec{A} \cdot \vec{B}|^2=1\)

Answer: 2. Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1.

Question 11. Statement 1: The angle between the vectors \(\vec{A}=\hat{i}+\hat{j}\) and \(\vec{B}=\hat{j}+\hat{k} \text { is } \frac{\pi}{3}\)

Statement 2: The angle between vector \(\vec{A}\) and \(\vec{B}\) is, \(\theta=\cos ^{-1}\left(\frac{\vec{A} \cdot \vec{B}}{A B}\right)\)

Answer: 1. Statement 1 is true, statement 2 is true; statement 2 Is a correct explanation for statement 1.

Question 12. Statement 1: The initial velocity of projectile = (a\(\hat{i}\)+b\(\vec{j}\)). The horizontal range becomes maximum for a = b.

Statement 2: for the same magnitude of initial velocity, the horizontal range of a projectile becomes maximum for the angle 45° of projection.

Answer: 4. Statement 1 is false, statement 2 is true.

Vector Multiple Choice Questions And Answers

Question 1. What is the condition for \(\vec{A}+\vec{B}=\vec{A}-\vec{B}\) to be valid?

1. \(\vec{A}=0\)
2. \(\vec{B}=0\)
3. \(\vec{A}=\vec{B}\)
4. \(\vec{A}=-\vec{B}\)

Answer: 2. \(\vec{B}=0\)

Question 2. The magnitudes of the sum and difference of the two vectors are equal. The angle between the vectors is

1. 0°
2. 90°
3. 120°
4. 60°

Answer: 2. 90°

Read And Learn More WBCHSE Class 11 Physics MCQs

Question 3. The magnitude of each of the two vectors is P. The Magnitude of the resultant of the two is also P. The Angle between the vectors is

1. 0°
2. 60°
3. 120°
4. 90°

Answer: 3. 120°

Question 4.Two forces of equal magnitude act simultaneously on a particle. If the resultant of the forces is equal to the magnitude of each of them then the angle between the forces is

1. An acute angle
2. An obtuse angle
3. A right angle
4. Of any value

Answer: 2. An obtuse angle

Question 5. A man travels 30 m to the north, then 20 m to the east, and after that 30√2 m to the south-west. His displacement from the starting point is

1. 15 m to the east
2. 28 m to the south
3. 10 m to the west
4. 15 m to the southwest

Answer: 3. 10 m to the west

WBBSE Class 11 Vector MCQs

Question 6. \(0.2 \hat{i}+0.6 \hat{j}+a \hat{k}\) is a unit vector. The value of a should be

1. 0.3
2. 0.4
3. 0.6
4. 0.8

Answer: 3. 0.6

Question 7. \(\vec{R}\) is the resultant of the vectors \(\vec{A}\) and \(\vec{B}\). If R = \(\frac{B}{\sqrt{2}}\), then the angle θ is

1. 30°
2. 45°
3. 60°
4. 75°

Answer: 2. 45°

Question 8. The position vector of a particle is related to time t as \(\vec{r}=\left(t^2-1\right) \hat{i}+2 t \hat{j}\). The locus of the particle on the x-y plane is

1. Parabolic
2. Circular
3. Straight line
4. Elliptical

Answer: 1. Parabolic

Question 9. Two forces of the same magnitude F are at right angles to each other. The magnitude of the net force (total force) acting on the object is

1. F
2. 2 F
3. Between F and 2F
4. More than 2 F

Answer: 3. Between F and 2 F

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Question 10. If a and b are two unit vectors inclined at an angle of 60° to each other, then

1. \(|a+b|>1\)
2. \(|a+b|<1\)
3. \(|a-b|>1\)
4. \(|a-b|<1\)

Answer: 1. \(|a+b|>1\)

Question 11. The condition (a-b)² = a²b² is satisfied when

1. a is parallel to b
2. a ≠ b
3. a-b = 1
4. a ⊥ b

Answer: 3. a-b = 1

Vector Operations and Their Applications MCQs

Question 12. If \(\vec{P}+\vec{Q}=\vec{R} \text { and }|\vec{P}|=|\vec{Q}|=\sqrt{3} \text { and }|\vec{R}|=3\), then the angle between \(\vec{P}\) and \(\vec{Q}\) is

1. \(\frac{\pi}{4}\)
2. \(\frac{\pi}{6}\)
3. \(\frac{\pi}{3}\)
4. \(\frac{\pi}{2}\)

Answer: 3. \(\frac{\pi}{3}\)

Question 13. The angles which the vector \(\vec{A}=3 \hat{i}+6 \hat{j}+2 \hat{k}\) makes with the coordinate axes are

1. \(\cos ^{-1} \frac{3}{7}, \cos ^{-1} \frac{6}{7}\) and \(\cos ^{-1} \frac{2}{7}\)
2. \(\cos ^{-1} \frac{4}{7}, \cos ^{-1} \frac{5}{7} \)and \(\cos ^{-1} \frac{3}{7}\)
3. \(\cos ^{-1} \frac{3}{7}, \cos ^{-1} \frac{4}{7}\) and \(\cos ^{-1} \frac{1}{7}\)
4. None of the above

Answer: 1. \(\cos ^{-1} \frac{3}{7}, \cos ^{-1} \frac{6}{7}\) and \(\cos ^{-1} \frac{2}{7}\)

Question 14. If  \(\vec{a}+\vec{b}=\vec{c}\), then the angle between \(\vec{a}\) and \(\vec{b}\) is

1. 90°
2. 180°
3. 120°
4. Zero

Answer: 4. Zero

Question 15. If at and are two non-collinear unit vectors and if \(\left|a_1+a_2\right|=\sqrt{3}\) then the value of \(\left(a_1-a_2\right) \cdot\left(2 a_1+a_2\right)\) is

1. 2
2. 3/2
3. 1/2
4. 1

Answer: 3. 1/2

Question 16. Which of the following is a vector quantity?

1. Temperature
2. Impulse
3. Gravitational potential
4. Power

Answer: 2. Impulse

Question 17. The resultant of two forces of magnitude (x+y) and (x-y) is \(\sqrt{x^2+y^2}\). The angle between them is

1. \(\cos ^{-1}\left[-\frac{\left(x^2+y^2\right)}{2\left(x^2-y^2\right)}\right]\)
2. \(\cos ^{-1}\left[-\frac{2\left(x^2-y^2\right)}{x^2+y^2}\right]\)
3. \(\cos ^{-1}\left[-\frac{x^2+y^2}{x^2-y^2}\right]\)
4. \(\cos ^{-1}\left[-\frac{x^2-y^2}{x^2+y^2}\right]\)

Answer: 1. \(\cos ^{-1}\left[-\frac{\left(x^2+y^2\right)}{2\left(x^2-y^2\right)}\right]\)

Question 18. ABC is an equilateral triangle and O is its centre of mass. Find n if \(\overrightarrow{A B}+\overrightarrow{A C}=n \overrightarrow{A O}\)

1. 1
2. 2
3. 3
4. 4

Hint: \(\overrightarrow{A B}+\overrightarrow{A C}=(\overrightarrow{A O}+\overrightarrow{O B})+(\overrightarrow{A O}+\overrightarrow{O C})\)

= \(2 \overrightarrow{A O}+(\overrightarrow{O B}+\overrightarrow{O C})=2 \overrightarrow{A O}+\overrightarrow{A O}\)

= \(3 \overrightarrow{A O}\)

Answer: 3. 3

Short Notes on Vector Properties with MCQs

Question 19. The sum of the magnitudes of two forces acting at a point is 16 N. The resultant has a magnitude of 8N and is perpendicular to the force of lower magnitude. The two forces are

1. 6N and 10N
2. 8N and 8N
3. 4N and 12N
4. 2N and 14N

Answer: 1. 6N and 10N

Question 20. Two men have velocities of 4 m/s towards the east and 3 m/s towards west, respectively. The velocity of the first man relative to the second is

1. \((4 \hat{i}+3 \hat{j})\)
2. \((3 \hat{i}+4 \hat{j})\)
3. \((4 \hat{i}-3 \hat{j})\)
4. \((3 \hat{i}-4 \hat{j})\)

Answer: 1. \((4 \hat{i}+3 \hat{j})\)

Question 21. Two forces in the ratio 1:2 act simultaneously on a particle. The result of these forces is three times the first force. The angle between them is

1. 0°
2. 60°
3. 90°
4. 45°

Answer: 1. 0°

Question 22. Given \(\vec{A}=2 \hat{i}+3 \hat{j} \text { and } \vec{B}=\hat{i}+\hat{j}\). The component of vector \(\vec{A}\) along vector \(\vec{B}\) is

1. \(\frac{1}{\sqrt{2}}\)
2. \(\frac{3}{\sqrt{2}}\)
3. \(\frac{5}{\sqrt{2}}\)
4. \(\frac{7}{\sqrt{2}}\)

Answer: 3. \(\frac{5}{\sqrt{2}}\)

Question 23. One of the components of a velocity vector of magnitude 50 m · s^-1 Is 30 m · s^-1, Its other orthogonal component Is

1. 15 m · s^-1
2. 20 m · s^-1
3. 25 m · s^-1
4. 40 m · s^-1

Answer: 4. 40 m · s^-1

Question 24. The times of flight of the two projectiles are t₁ and t₂ If R is the horizontal range of each of them, then

1. \(t_1 t_2 \propto R\)
2. \(t_1 t_2 \propto R^2\)
3. \(t_1 t_2 \propto R^3\)
4. \(t_1 t_2 \propto R^{\frac{1}{2}}\)

Hint: The angles of projection are \(\theta\) and \(\left(90^{\circ}-\theta\right)\).

∴ \(t_1 t_2=\frac{2 u \sin \theta}{g} \cdot \frac{2 u \sin \left(90^{\circ}-\theta\right)}{g}\)

= \(\frac{4 u^2}{g^2} \sin \theta \cos \theta=\frac{2}{g} \cdot \frac{u^2 \sin 2 \theta}{g}=\frac{2}{g} R\)

∴ \(t_1 t_2\propto R\)

Answer: 1. \(t_1 t_2 \propto R\)

Question 25. For angles θ and (90° – θ) of projection, a projectile has the same horizontal range R. The maximum heights attained are H₁ and H₂ respectively. Then the relation among R, H_1, and H₂ is

1. \(R=\sqrt{H_1 H_2}\)
2. \(R=\sqrt{H_1^2+H_2^2}\)
3. \(R=H_1+H_2\)
4. \(R=4 \sqrt{H_1 H_2}\)

Answer: 4. \(R=4 \sqrt{H_1 H_2}\)

Question 26. An airplane is flying with a velocity of 216 km/h at an altitude of 1960 m relative to the ground. It drops a bomb when it is just above point A on the ground. The bomb hits the ground at B. The distance AB is

1. 1.2 km
2. 0.33 km
3. 3.33 km
4. 33 km

Answer: 1. 1.2 km

Real-Life Examples of Vectors in Motion

Question 27. The initial velocity and the acceleration of a particle are \(\vec{u}\) = 3\(\hat{i}\)+4\(\hat{j}\) and \(\vec{a}\) = 0.3 \(\hat{i}\)+0.4\(\hat{j}\). The magnitude of its velocity after 10 s is

1. 10 units
2. 8.5 units
3. 77√2 units
4. 7 units

Answer: 1. 10 units

Question 28. The equations of motion of a projectile are x = 36t and 2y = 96t- 9.8 t². The angle of projection is

1. \(\sin ^{-1}\left(\frac{4}{5}\right)\)
2. \(\sin ^{-1}\left(\frac{3}{5}\right)\)
3. \(\sin ^{-1}\left(\frac{3}{4}\right)\)
4. \(\sin ^{-1}\left(\frac{4}{3}\right)\)

Hint: x= 36t and y = 48t – \(\frac{1}{2}\) x 9.8t²—they suggest that the horizontal and vertical components of initial velocity are 36 and 48 units, respectively.

∴ \(\tan \theta=\frac{48}{36}=\frac{4}{3} \text { or, } \sin \theta=\frac{4}{5}\)

Answer: 1. \(\sin ^{-1}\left(\frac{4}{5}\right)\)

Question 29. Two projectiles, projected with angles (45°-θ) and (45° +θ) respectively, have their horizontal ranges in the ratio

1. 2:1
2. 1:1
3. 2:3
4. 1:2

Answer: 2. 1:1

Question 30. For a projectile, (horizontal range)² = 48 x (maximum height)². The angle of projection is

1. 45°
2. 60°
3. 75°
4. 30°

Answer: 4. 30°

Question 31. Two railway tracks are parallel to the west-east direction. Along one track train A moves with a speed of 30 m · s^-1 from west to east, while along the second track, train B moves with a speed of 48 m · s^-1 from east to west. The relative speed of B with respect to A is

1. 48 m · s^-1
2. -78 m · s^-1
3. 30 m · s^-1
4. Zero

Answer: 2. 48 m · s^-1

Question 32. Angle between the vectors \(\hat{i}+\hat{j} \text { and } \hat{i}-\hat{k}\) is

1. 60°
2. 30°
3. 45°
4. 90°

Answer: 1. 60°

Question 33. A vector is multiplied by (-2). As a result

1. The magnitude of the vector is doubled and the direction is unaltered
2. The magnitude of the vector remains the same and the direction is reversed
3. The magnitude of the vector is doubled and the direction is reversed
4. No change in the magnitude or direction of the vector

Answer: 3. The Magnitude of the vector is doubled and the direction is reversed

Question 34. In a clockwise system

1. \(\hat{j} \times \hat{j}=1\)
2. \(\hat{k} \cdot \hat{i}=1\)
3. \(\hat{j} \times \hat{k}=\hat{i}\)
4. \(\hat{i} \cdot \hat{i}=0\)

Answer: 3. \(\hat{j} \times \hat{k}=\hat{i}\)

Question 35. A vector \(\vec{P}=3 \hat{i}-2 \hat{j}+a \hat{k}\) is perpendicular to the vector \(\vec{Q}=2 \hat{i}+\hat{j}-\hat{k}\). The value of a is

1. 2
2. 1
3. 4
4. 3

Answer: 3. 4

Step-by-Step Solutions to Vector MCQs

Question 36. For any two vectors \(\vec{A} \text { and } \vec{B} \text {, if } \vec{A} \cdot \vec{B}=|\vec{A} \times \vec{B}|\), the magnitude of \(\vec{C}=\vec{A}+\vec{B}\) is equal to

1. \(\sqrt{A^2+B^2}\)
2. \(A+B\)
3. \(\sqrt{A^2+B^2+\frac{A B}{\sqrt{2}}}\)
4. \(\sqrt{A^2+B^2+\sqrt{2} A B}\)

Answer: 4. \(\sqrt{A^2+B^2+\sqrt{2} A B}\)

Question 38. A vector perpendicular to both of \((3 \hat{i}+\hat{j}+2 \hat{k})\) and \((2 \hat{i}- 2 \hat{j}+4 \hat{k})\) is

1. \(\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}-\hat{k})\)
2. \(\hat{i}-\hat{j}-\hat{k}\)
3. \(\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k})\)
4. \((\sqrt{3} \hat{i}-\hat{j}-\hat{k})\)

Answer: 2. \(\hat{i}-\hat{j}-\hat{k}\)

Question 39. A vector normal to \(a \cos \theta \hat{i}+b \sin \theta \hat{j}\) is

1. \(b \sin \theta \hat{i}-a \cos \theta \hat{j}\)
2. \(\frac{1}{a} \sin \theta \hat{i}-\frac{1}{b} \cos \theta \hat{j}\)
3. \(5 \hat{k}\)
4. All of the above

Answer: 4. All of the above

Question 40. A vector \(\vec{A}\) of magnitude 2 units is inclined at angles 30° and 60° with positive x- and y-axes, respectively. Another vector of magnitude 5 units is aligned along the positive x-axis. Then \(\vec{A}\) • \(\vec{B}\) is

1. 5√3
2. 3√5
3. 2√3
4. 3√2

Answer: 1. 5√3

Question 41. For what values of a and b, the vector \((a \hat{i}+b \hat{j})\) will be a unit vector perpendicular to the vector \((\hat{i}+\hat{j})\)?

1. 1,0
2. 0,1
3. \(\frac{1}{\sqrt{3}},-\frac{2}{\sqrt{3}}\)
4. \(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\)

Answer: 4. \(\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\)

Question 42. Two billiard balls, starting from the same point, have velocities \((\hat{i}+\sqrt{3} \hat{j}) \text { and }(2 \hat{i}+2 \hat{j})\), respectively. The angle between them is

1. 60°
2. 15°
3. 45°
4. 30°

Answer: 2. 15°

Question 43. If \(|\vec{A} \times \vec{B}|=\sqrt{3} \vec{A} \cdot \vec{B}\), then \(|\vec{A}+\vec{B}|=\)?

1. \(\left(A^2+B^2+A B\right)^{\frac{1}{2}}\)
2. (\(\left(A^2+B^2+\frac{A B}{\sqrt{3}}\right)^{\frac{1}{2}}\)
3. A+B
4. \(\left(A^2+B^2+\sqrt{3} A B\right)^{\frac{1}{2}}\)

Answer: 1. \(\left(A^2+B^2+A B\right)^{\frac{1}{2}}\)

Question 44. The vector product of \(\vec{A}\) and \(\vec{B}\) is zero. The scalar product of \(\vec{A}\) and \((\vec{A}+\vec{B})\) is

1. 0
2. \(A^2\)
3. \(A B\)
4. \(A^2+A B\)

Answer: 4. \(A^2+A B\)

Question 45. If \(\vec{A} \cdot \vec{B}=\vec{A} \cdot \vec{C}=0\), then the vector parallel to \(\vec{A}\) would be

1. \(\vec{C}\)
2. \(\vec{B}\)
3. \(\vec{B} \times \vec{C}\)
4. \(\vec{A} \times(\vec{B} \times \vec{C})\)

Answer: 3. \(\vec{B} \times \vec{C}\)

In this type of question, more than one option is correct.

Question 46. In the following figure which of the statements is correct?

1. The sign of x -component of \(\vec{l}_1\) and \(\vec{l}_2\) is negative
2. The signs of the y -component of \(\vec{l}_1 \text { and } \vec{l}_2\) are positive and negative respectively
3. The signs of x and y-components of \(\vec{l}_1+\vec{l}_2\) are both positive
4. None of these

Answer:

1. The sign of x -component of \(\vec{l}_1\) and \(\vec{l}_2\) is negative

3. The signs of x and y-components of \(\vec{l}_1+\vec{l}_2\) are both positive

Common Vector MCQs for Class 11 –

Question 47. Two particles are projected in the air with speed v₀ at angles θ₁ and θ₂ (both acute) to the horizontal, respectively. If the height reached by the first particle is greater than that of the second, then which are the correct choices?

1. The angle of projection:  θ₁ > θ₂
2. Time of flight: T₁ > T₂
3. Horizontal range: R₁ > R₂
4. Total energy: U₁ > U₂

Answer:

Question 48. For two vectors \(\vec{A} \text { and } \vec{B},|\vec{A}+\vec{B}|=|\vec{A}-\vec{B}|\) is always true when

1. \(|\vec{A}|=|\vec{B}| \neq 0\)
2. \(\vec{A} \perp \vec{B}\)
3. \(|\vec{A}|=|\vec{B}| \neq 0\) and \(\vec{A}\) and \(\vec{B}\) are parallel or antiparallel
4. When either \(|\vec{A}|\) or \(|\vec{B}|\) is zero

Answer:

2. \(\vec{A} \perp \vec{B}\)

4. When either \(|\vec{A}|\) or \(|\vec{B}|\) is zero

Question 49. Given \(\vec{a}+\vec{b}+\vec{c}+\vec{d}=\overrightarrow{0}\). Which of the following statements is correct?

1. \(\vec{a}, \vec{b}, \vec{c}\) and \(\vec{d}\) must each be a null vector
2. The magnitude of \((\vec{a}+\vec{c})\) equals the magnitude of \((\vec{b}+\vec{d})\)
3. The magnitude of \(\vec{a}\) can never be greater than the sum of the magnitudes of \(\vec{b}, \vec{c}\), and \(\vec{d}\)
4. \((\vec{b}+\vec{c})\) must lie on the plane of \(\vec{a}\) and \(\vec{d}\) if \(\vec{a}\) and \(\vec{d}\) are not collinear and in the line of \(\vec{a}\) and \(\vec{d}\), if they are collinear

Answer:

2. The magnitude of \((\vec{a}+\vec{c})\) equals the magnitude of \((\vec{b}+\vec{d})\)

3. The magnitude of \(\vec{a}\) can never be greater than the sum of the magnitudes of \(\vec{b}, \vec{c}\) and \(\vec{d}\)

4. \((\vec{b}+\vec{c})\) must lie on the plane of \(\vec{a}\) and \(\vec{d}\) if \(\vec{a}\) and \(\vec{d}\) are not collinear and in the line of \(\vec{a}\) and \(\vec{d}\), if they are collinear

Question 50. State which of the following statements are false. A scalar quantity is one that

1. Is conserved in a process
2. Can never take negative values
3. Must be dimensionless
4. Has the same value for observers with different orientations of axes

Answer:

1. Is conserved in a process
2. Can never take negative values
3. Must be dimensionless

Question 51. The resultant of \(\vec{P}\) and \(\vec{Q}\) is \(\vec{R}\). The angle between \(\vec{P}\) and \(\vec{Q}\) is

1. 90°, if R² = P² + Q²
2. Less than 90°, if R2 > (P² + Q²)
3. Greater than 90°, if R² < (P2 + Q²)
4. Greater than 90°, if R² > (P² + Q²)

Answer:

1. 90°, if R² = P² + Q²
2. Less than 90°, if R2 > (P² + Q²)
3. Greater than 90°, if R² < (P2 + Q²)

Question 52. The magnitude of the vector product of A and B may be

1. Greater than AB
2. Equal to AB
3. Less than AB
4. Zero

Answer:

2. Equal to AB

3. Less than AB

4. Zero

Question 53. If \(\vec{A}=5 \hat{i}+6 \hat{j}+3 \hat{k} \text { and } \vec{B}=6 \hat{i}-2 \hat{j}-6 \hat{k}\), then

1. \(\vec{A}\) and \(\vec{B}\) are perpendicular
2. \(\vec{A} \times \vec{B}=\vec{B} \times \vec{A}\)
3. \(\vec{A}\) and \(\vec{B}\) have the same magnitude
4. \(\vec{A} \cdot \vec{B}=0\)

Answer:

1. \(\vec{A}\) and \(\vec{B}\) are perpendicular

4. \(\vec{A} \cdot \vec{B}=0\)

Question 54. Two projectiles, thrown from the same point with the same magnitude of velocity, have angles 60° and 30° of their projection. Then their

1. Maximum heights attained are the same
2. Horizontal ranges are the same
3. Magnitudes of velocity at the instants of hitting the ground are the same
4. Times of flight are the same

Answer:

2. Horizontal ranges are the same

3. Magnitudes of velocity at the instants of hitting the ground are the same

Question 55. A body is projected with a velocity u at an angle θ with the horizontal. At t = 2s, the body makes an angle of 30° with the horizontal. 1 s later, it attains its maximum height. Then

1. u = 20√3 m/s
2. θ = 60°
3. θ = 45
4. u = \(\frac{20}{\sqrt{3}} \mathrm{~m} / \mathrm{s}\)

Answer:

1. u = 20√3 m/s
2. θ = 60°

Question 56. \(\vec{A} \perp \vec{B}\); \(\vec{C}\) is coplanar with \(\vec{A}\) and \(\vec{B}\). Therefore,

1. \(\vec{A}=x \vec{B}+y \vec{C}\), where x and y are scalars
2. \(\vec{A} \cdot(\vec{B} \times \vec{C})=0\)
3. \(|(\vec{A} \times \vec{B}) \times \vec{C}|=A B C\)
4. \(\vec{A} \cdot \vec{B}=0\)

Answer:

1. \(\vec{A}=x \vec{B}+y \vec{C}\), where x and y are scalars
2. \(\vec{A} \cdot(\vec{B} \times \vec{C})=0\)
3. \(|(\vec{A} \times \vec{B}) \times \vec{C}|=A B C\)
4. \(\vec{A} \cdot \vec{B}=0\)

Product Of A Scalar And A Vector

WBBSE Class 11 Scalar and Vector Product Notes

A scalar has magnitude but no direction. On the other hand, a vector has both magnitude and direction. Therefore, when a scalar is multiplied by a vector, the product has both magnitude and direction. Thus, the product is a vector.

1. When a vector is multiplied by a scalar, the product is also a vector. If the scalar is positive then the direction of the vector remains unaltered but, if the scalar is negative, the direction of the vector becomes opposite to that of the original vector.

Example: 10 eastwards x 5 = 50 eastwards but, 10 eastwards x (-5) =(-50) eastwards = 50 westwards

2. The product of a scalar quantity with a vector quantity produces, in general, some other vector quantity.

3. If the vector is a unit vector, then the product’s magnitude is the same as the scalar’s. For example, if \(\vec{n}\) is a unit vector due east, then \(\vec{n}\) x 5 =5 due east or, \(\vec{n}\) x 5 m · s^-1 speed = 5 m · s^-1 velocity eastwards.

Product Of Two Vectors

Two vectors, when multiplied, may produce either a scalar or a vector. Accordingly, they are called scalar or dot product and vector or cross product.

Scalar Or Dot Product

Two vectors may give a scalar product like: force • displacement = work

Here, although force and displacement are both vector quantities, their product, work, is a scalar quantity. Such type of product of two vectors is called a scalar product or dot product. A dot (⋅) symbol is used between the two vectors to represent such a product mathematically.

Scalar Or Dot Product Definition: The scalar product of two vectors \(\vec{A}\) and \(\vec{B}\) is defined as \(\vec{A}\)–\(\vec{B}\) = AB cosθ, where θ is the angle between A and B.

Some Properties Of Scalar Products:

1. \(\vec{A}\) · \(\vec{A}\) = A², i.e., the scalar product of a vector with itself is the square of its magnitude.

If the same vector is taken twice, then the angle between them =0°.

∴\(\vec{A}\) ⋅ \(\vec{A}\) = AAcosθ = A²

Hence, A = \(\sqrt{\vec{A} \cdot \vec{A}}\)

This rule is often used for determining the magnitude of any vector.

Understanding Scalar Multiplication of Vectors

2. \(\vec{A}\) ⋅ \(\vec{B}\)= \(\vec{B}\) ⋅ \(\vec{A}\), i.e., scalar products are commutative.

If the angle from \(\vec{A}\) to \(\vec{B}\) is θ, the angle from \(\vec{B}\) to \(\vec{A}\) is -θ.

∴ \(\vec{A}\) ⋅ \(\vec{B}\) = ABcosθ

and \(\vec{B}\)–\(\vec{A}\) = BAcos(-θ) = ABcosθ

Hence, \(\vec{A}\) ⋅ \(\vec{B}\) = \(\vec{B}\) ⋅ \(\vec{A}\)

3. The scalar product of two orthogonal vectors, i.e., the vectors that are mutually at right angles, is zero.

If \(\vec{A}\) ⊥\(\vec{B}\), then the angle between them, θ = 90°. Hence, cosθ = 0°

∴ \(\vec{A}\) ⋅ \(\vec{B}\) = ABcosθ = 0

Alternatively, if \(\vec{A}\) ⋅ \(\vec{B}\) = 0 for two non-zero vectors \(\vec{A}\) and \(\vec{B}\), then \(\vec{A}\) ⊥ \(\vec{B}\).

4. If θ is the angle between two vectors \(\vec{A}\) and \(\vec{B}\), then \(\vec{A}\) ⋅ \(\vec{B}\) = ABcosθ or, \(\frac{\vec{A} \cdot \vec{B}}{A B}=\cos \theta\)

or, \(\theta=\cos ^{-1}\left(\frac{\vec{A} \cdot \vec{B}}{A B}\right)\)

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5. Scalar Products Of Mutually Perpendicular Unit Vectors: This rule is often used to determine the angle between two vectors, \(\hat{i}, \hat{j}, \hat{k}\) are three unit vectors along the directions of positive x, y, z axes respectively. Each has a value of 1 and they are mutually perpendicular.

If we take the same vector twice, the angle between them will be 0°; for example, the angle between \(\hat{i} \text { and } \hat{i}\) is 0°. If we take two different vectors, then the angle between them will be 90°; for example, the angle between \(\hat{i} \text { and } \hat{J}\)is 90°.

Therefore, \(\hat{i} \cdot \hat{i}=(1)(1) \cos 0^{\circ}=1\) and \(\hat{i} \cdot \hat{j}=(1)(1) \cos 90^{\circ}=0\)

So, the general rule for the scalar product of these unit vectors \(\hat{i} \cdot \hat{i}=1, \quad \hat{j} \cdot \hat{j}=1, \quad \hat{k} \cdot \hat{k}=1\)

and \(\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{i}=\hat{j} \cdot \hat{k}=\hat{k} \cdot \hat{j}=\hat{k} \cdot \hat{i}=\hat{i} \cdot \hat{k}=0\)

Effects of Scalar Multiplication on Vector Magnitude

6. Scalar Product Of Two Vectors Using The Position Coordinates: Let \(\vec{A}=A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\) and \(\vec{B}=B_x \hat{i}+B_y \hat{j}+B_z \hat{k}\) i.e., components of \(\vec{A}\) and \(\vec{B}\) along x, y and z axes are \(A_x\), \(A_y, A_z\) and \(B_x, B_y, B_z\) respectively.

Now, \(\vec{A} \cdot \vec{B}=\left(A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\right) \cdot\left(B_x \hat{i}+B_y \hat{j}+B_z \hat{k}\right)\)

Following the rules of simple multiplication, and omitting the 6 terms containing \(\hat{i} \cdot \hat{j}, \hat{i} \cdot \hat{k}\), etc. that are equal to zero, \(vec{A} \cdot \vec{B}=A_x B_x+A_y B_y+A_z B_z\)

6. Dot product of a vector with \(\hat{i}, \hat{j}, \hat{\boldsymbol{k}}\) Let a vector be \(\vec{A}=A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\).

So, \(\hat{i} \cdot \vec{A}=\hat{i} \cdot\left(A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\right)\)

= \(A_x \text { [using, } \hat{i} \cdot \hat{i}=\cdots=1 \text {, and } \hat{i} \cdot \hat{j}=\cdots=0 \text { ] }\)

Similarly, \(\hat{j} \cdot \vec{A}=A_y\), and \(\hat{k} \cdot \vec{A}=A_z\).

Thus, the dot product of any vector with \(\hat{i}, \hat{j}, \hat{k}\) denotes the magnitudes of the x, y, and z components respectively of the vector itself. This property stands for any unit vector \(\hat{n}\):

∴ \(\hat{n} \cdot \vec{A}=\) magnitude of the component of \(\vec{A}\) along the direction of the unit vector \(\hat{n}\). i.e., the projection of \(\vec{A}\) along the direction of \(\hat{n}\).

Product Of Two Vectors Numerical Examples

Short Answer Questions on Scalar Multiplication

Example 1. Find the magnitude of the vectors \(3 \hat{i}-4 \hat{j}+12 \hat{k}\).
Solution:

⇒ \(\vec{A}=(3 \hat{i}-4 \hat{j}+12 \hat{k})\)

∴ \(\vec{A} \cdot \vec{A}=(3 \hat{i}-4 \hat{j}+12 \hat{k}) \cdot(3 \hat{i}-4 \hat{j}+12 \hat{k})\)

= \(3^2+(-4)^2+12^2=169\)

Now \(\vec{A} \cdot \vec{A}=A^2\); therefore \(A^2=169\)

∴ A = \(\sqrt{169}=13 \text { units. }\)

Example 2. Find the angle between the vectors \(\hat{i}+\hat{j} \text { and } \hat{i}-\hat{k}\)
Solution:

Let us consider, \(\vec{A}=(\hat{i}+\hat{j})\) and \(\vec{B}=(\hat{i}-\hat{k})\)

∴ \(\vec{A} \cdot \vec{A}=(\hat{i}+\hat{j}) \cdot(\hat{i}+\hat{j})=1^2+1^2=2=A^2\)

∴ A = \(sqrt{2} \text { units }\)

and \(\vec{B} \cdot \vec{B}=(\hat{i}-\hat{k}) \cdot(\hat{i}-\hat{k})=1^2+(-1)^2=2=B^2\)

∴ B = \(\sqrt{2} \text { units }\)

⇒ \(\vec{A} \cdot \vec{B}=(\hat{i}+\hat{j}) \cdot(\hat{i}-\hat{k})=\hat{i} \cdot \hat{i}-\hat{i} \cdot \hat{k}+\hat{j} \cdot \hat{i}-\hat{j} \cdot \hat{k}\)

= 1-0+0-0 = 1

If the angle between the two vectors is θ, \(\vec{A} \cdot \vec{B}=A B \cos \theta\)

or, \(\cos \theta =\frac{\vec{A} \cdot \vec{B}}{A B}=\frac{1}{\sqrt{2} \cdot \sqrt{2}}=\frac{1}{2}=\cos 60^{\circ}\)

∴ θ = 60°

Example 3. Find the angle between the vectors \(\vec{A}=2 \hat{i}+3 \hat{j}\) and \(\vec{B}=-3 \hat{i}+2 \hat{j}\)
Solution:

Let the angle between the two vectors \(\vec{A}\) and \(\vec{B}\) be θ.

Now, \(\vec{A} \cdot \vec{B}=(2 \hat{i}+3 \hat{j}) \cdot(-3 \hat{i}+2 \hat{j})=-6+6=0\)

As this dot product is zero, \(\vec{A}\) and \(\vec{B}\) are perpendicular to each other. So, the angle between them = 90°.

Example 4. Prove that die diagonals of a rhombus are perpendicular to each other.
Solution:

ABCD is a rhombus. \(\overrightarrow{A B}=\overrightarrow{D C}=\vec{a}\)

and \(\overrightarrow{A D}=\overrightarrow{B C}=\vec{b}\) the values of \(\vec{a}\) and \(\vec{b}\) are the same for a rhombus.

Let a = b = m

⇒ \(\overrightarrow{A C}=\overrightarrow{A B}+\overrightarrow{B C}=\vec{a}+\vec{b}\)

and \(\overrightarrow{B D}=\overrightarrow{B A}+\overrightarrow{A D}=-\vec{a}+\vec{b}\)

∴ \(\overrightarrow{A C} \cdot \overrightarrow{B D}=(\vec{a}+\vec{b}) \cdot(-\vec{a}+\vec{b})\)

= \(-\vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}-\vec{b} \cdot \vec{a}+\vec{b} \cdot \vec{b}\)

= \(-a^2+\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{b}+b^2=-m^2+m^2=0\)

As the dot product is 0, the angle between the vectors = 90°

i.e., \(\overrightarrow{A C}=\overrightarrow{B D}\) are perpendicular to each other.

Example 5. \(\vec{a}, \vec{b}\) and \(\vec{c}\) are three unit vectors. Show that, \(|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2\) ≤ 9
Solution:

⇒ \(|\vec{a}+\vec{b}+\vec{c}|^2=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c})\)

= \(\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{c}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\)

= \(3+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\)

Now, \(|\vec{a}+\vec{b}+\vec{c}|^2\) ≥ 0

∴ \(2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\) ≥ -3……(1)

Also, \(|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2\)

= \(2[\vec{a} \cdot \vec{a}+\vec{b} \cdot \vec{b}+\vec{c} \cdot \vec{c}-(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})]\)

or, \(2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})\)

= \(6-\left[|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2\right]\)

From (1) and (2), we get, \(6-\left[|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2\right]\) ≥-3

or, \(\left[|\vec{a}-\vec{b}|^2+|\vec{b}-\vec{c}|^2+|\vec{c}-\vec{a}|^2\right]\) ≤ 9

Example 6. Find the pojetion of vector \(\vec{P}=2 \hat{i}-3 \hat{j}+6 \hat{k}\) on the vector = \(\vec{Q}=\hat{i}+2 \hat{j}+2 \hat{k}\)
Solution:

The projection of \(\vec{P}\) on \(\vec{Q}\) is

P cosθ = \(\frac{\vec{P} \cdot \vec{Q}}{|\vec{Q}|}=\frac{P_x Q_x+P_y Q_y+P_z Q_z}{\sqrt{Q_x^2+Q_y^2+Q_z^2}}\)

(where θ is the angle between \(\vec{P}\) and \(\vec{Q}\))

= \(\frac{(2 \cdot 1)+[(-3) \cdot 2]+(6 \cdot 2)}{\sqrt{1^2+2^2+2^2}}=\frac{2-6+12}{\sqrt{9}}=\frac{8}{3}\)

Example 7. A particle is moving in a curvilinear path defined by the equations x = 2t², y = t²-4t, and z = 3t-5. Find out the magnitudes of the components of velocity and acceleration along (\(\hat{i}-3 \hat{j}+2 \hat{k})\)) at time t = 1
Solution:

Velocity, \(\vec{v}=\frac{d \vec{r}}{d t}=\frac{d}{d t}(x \hat{i}+y \hat{j}+z \hat{k})\)

= \(\frac{d x}{d t} \hat{i}+\frac{d y}{d t} \hat{j}+\frac{d z}{d t} \hat{k}\)

= \(4 t \hat{i}+(2 t-4) \hat{j}+3 \hat{k}\)

∴ At t=1, \(\vec{v}=4 \hat{i}-2 \hat{j}+3 \hat{k}\)……(1)

Acceleration, \(\vec{a}=\frac{d \vec{v}}{d t}=4 \hat{i}+2 \hat{j}=\) constant ….(2)

Unit vector along \(\vec{A}=\hat{i}-3 \hat{j}+2 \hat{k}\), \(\hat{n}=\frac{\vec{A}}{A}=\frac{\hat{i}-3 \hat{j}+2 \hat{k}}{\sqrt{1^2+(-3)^2+2^2}}=\frac{1}{\sqrt{14}}(\hat{i}-3 \hat{j}+2 \hat{k})\)….(3)

∴ At t=1, the component of \(\vec{v}\) along \(\vec{A}\) is, from (1) and (3),

∴ \(\nu_A =\hat{n} \cdot \vec{v}=\frac{1}{\sqrt{14}}(\hat{i}-3 \hat{j}+2 \hat{k}) \cdot(4 \hat{i}-2 \hat{j}+3 \hat{k})\)

= \(\frac{1}{\sqrt{14}} \times 16=\frac{8 \sqrt{14}}{7}\)

The component of \(\vec{a}\) along \(\vec{A}\) is, from (2) and (3), \(a_A=\hat{n} \cdot \vec{a}=\frac{1}{\sqrt{14}}(\hat{i}-3 \hat{j}+2 \hat{k}) \cdot(4 \hat{i}+2 \hat{j})\)

= \(\frac{1}{\sqrt{14}} \times(-2)=-\frac{\sqrt{14}}{7}\)

∴ \(\left|a_A\right|=\frac{\sqrt{14}}{7}\)

Example 8. Prove that a right-angled triangle can be formed using the vectors \(\vec{A}=\hat{i}-3 \hat{j}+5 \hat{k}, \quad \vec{B}=2 \hat{i}+\hat{j}-4 \hat{k}\) and \(\vec{C}=3 \hat{i}-2 \hat{j}+\hat{k}\)
Solution:

∴ \(\vec{A}+\vec{B}=(\hat{i}-3 \hat{j}+5 \hat{k})+(2 \hat{i}+\hat{j}-4 \hat{k})\)

= \(3 \hat{i}-2 \hat{j}+\hat{k}=\vec{C}\)

i.e., \(\vec{A}+\vec{B}=\vec{C}\)

So, \(\vec{A}, \vec{B} and \vec{C}\) can form a triangle.

Again, \(\vec{B} \cdot \vec{C}=(2 \hat{i}+\hat{j}-4 \hat{k}) \cdot(3 \hat{i}-2 \hat{j}+\hat{k})=6-2-4=0\)

∴ In the triangle formed by \(\vec{A}, \vec{B}\) and \(\vec{C}\), the angle between \(\vec{B}\) and \(\vec{C}\) is \(90^{\circ}\). Thus, it is a right-angled triangle.

[The dot product \(\vec{B} \cdot \vec{C}\) has been calculated, because \(\vec{A}\) forms the largest side of the triangle. An angle of \(90^{\circ}\) is always opposite to the largest side. In this example, \(A=\sqrt{35}, B=\sqrt{21}\) and \(C=\sqrt{14}\), i.e., A>B, C]

Applications of Scalar and Vector Multiplication

Vector Or Cross Product: The product of two vectors can be a vector. For example, angular velocity x position vector = linear velocity

Here, both angular velocity and position vector are vector quantities, and their product, linear velocity, is also a vector quantity. Such type of product of two vectors is called a vector product or cross product and is represented by putting a cross (x) sign between the two vectors.

Vector Or Cross Product Definition: The vector product or cross product of two vectors \(\vec{a}\) and \(\vec{b}\) is defined as \(\vec{a}\) x \(\vec{b}\)= ab sin\(\theta \hat{n}\), where d is the angle from \(\vec{a}\) to \(\vec{b}\), and \(\hat{n}\) is a unit vector. The direction of \(\hat{n}\) is the direction of advance of a right-handed screw when it is rotated from \(\vec{a}\) and \(\vec{b}\).

Here, \(\hat{n} \perp \vec{a} \text { and } \hat{n} \perp \vec{b}\). So the product \(\vec{a} \times \vec{b}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\). The unit vector \(\hat{n}\) has a magnitude, [latx]|\hat{n}|[/latex] = 1.

So the magnitude of the vector product is \(|\vec{a} \times \vec{b}|=a b \sin \theta\)

Some Properties Of Vector Products:

1. \(\vec{A}\) x \(\vec{A}\) = \(\vec{0}\), i.e., vector product of a vector with itself is a null vector.

The angle between the same vector taken twice is zero.

∴ The magnitude of \(\vec{A}\)x \(\vec{A}\) = \(|\vec{A} \times \vec{A}|=A A \sin 0^{\circ}=0\)

Hence, \(\vec{A}\) x \(\vec{A}\) = \(\vec{0}\)

2. \(\vec{A}\) x \(\vec{B}\) = –\(\vec{B}\) x \(\vec{A}\), i.e., vector product is not commutative.

If the angle from \(\vec{A}\) to \(\vec{B}\) is θ, the angle from \(\vec{B}\) to \(\vec{A}\) is -θ. As sin(-θ) = -sinθ, we get, \(\vec{A}\) x \(\vec{B}\) = –\(\vec{B}\) x \(\vec{A}\).

3. Vector Product Of Two Mutually Perpendicular Vectors: If \(\vec{A}\) ⊥ \(\vec{B}\), then, the magnitude of \(\vec{A}\) x \(\vec{B}\) = \(|\vec{A} \times \vec{B}|\) = ABsin90° = AB, and the direction of \(\vec{A}\) x \(\vec{B}\) is perpendicular to both of \(\vec{A}\) and \(\vec{B}\).

Hence, \(\vec{A}\), \(\vec{B}\) and (\(\vec{A} \times \vec{B}\))—all these three vectors are mutually perpendicular, i.e., if \(\vec{A}\) and \(\vec{B}\) are along x and y axes respectively, then their product \(\vec{A} \times \vec{B}\) will be along z -axis.

4. Vector Products Of Mutually Perpendicular Unit Vectors: \(\hat{i}, \hat{j}, \hat{k}\) are the unit vectors along positive x, y, and z axes respectively. The magnitude of each is 1.

As \(\vec{A}\) x \(\vec{A}\) = 0, we get, \(\hat{i} \times \hat{i}=\hat{j} \times \hat{j}=\hat{k} \times \hat{k}=0\)

Again, the magnitude of \(\hat{i} \times \hat{j}=|\hat{i} \times \hat{j}|=(1)(1) \sin 90^{\circ}=1\)

According to the right-handed corkscrew rule, the direction of \(\hat{i}\) x \(\hat{j}\) is along the z-axis. Again, the unit vector along the z-axis is \(\hat{k}\). Hence, \(\hat{i}\) x \(\hat{j}\) = \(\hat{k}\).

Similarly, \(\hat{i} \times \hat{j}=\hat{k}, \hat{j} \times \hat{k}=\hat{i}, \hat{k} \times \hat{i}=\hat{j}\)

Again, since \(\vec{A} \times \vec{B}=-\vec{B} \times \vec{A}, \hat{j} \times \hat{i}=-\hat{k}, \hat{k} \times \hat{j}=-\hat{i}\), \(\hat{i} \times \hat{k}=-\hat{j}\)

Here, the cyclic order of \(\hat{i}, \hat{j}, \hat{k}\) is important. If this cyclic order is maintained, the product of the first two vectors leads to the third with a positive sign, whereas if the cyclic order is not maintained, the product becomes the third vector with a negative sign.

Example: \(\hat{i} \times \hat{j}=\hat{k}, \hat{j} \times \hat{k}=\hat{i}, \hat{k} \times \hat{i}=\hat{j}, \hat{i} \times \hat{k}=-\hat{j}, \hat{k} \times \hat{j}=-\hat{i}, \hat{j} \times \hat{i}=-\hat{k}\)

5. Vector Product In Terms Of Positional Coordinates: Let \(\vec{A}=A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\) and \(\vec{B}=B_x \hat{i}+B_y \hat{j}+B_z \hat{k}\).

∴ \(\vec{A} \times \vec{B}=\left(A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\right) \times\left(B_x \hat{i}+B_y \hat{j}+B_z \hat{k}\right)\)

Out of the 9 terms of this product, the terms containing \(\hat{i} \times \hat{i}, \hat{j} \times \hat{j}\) and \(\hat{k} \times \hat{k}\) are zero. Writing the remaining 6 terms using the relations like \(\hat{i} \times \hat{j}=\hat{k}\),

we get, \(\vec{A} \times \vec{B}=\hat{i}\left(A_y B_z-A_z B_y\right)+\hat{j}\left(A_z B_x-A_x B_z\right)\) + \(\hat{k}\left(A_x B_y-A_y B_x\right)\)

It is convenient to write this product in a determinant form:

⇒ \(\vec{A} \times \vec{B}=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
A_x & A_y & A_z \\
B_x & B_y & B_z
\end{array}\right|\)

Area Of Triangle: If two vectors \(\vec{a}\) and \(\vec{b}\) represent the two sides of a triangle then half of the magnitude of their cross-product will give the area of the triangle.

In ΔABC \(\overrightarrow{C B}=\vec{a}, \overrightarrow{C A}=\vec{b}\) and the angle between them = C.

Height of the triangle, AD = \(A C \cdot \frac{A D}{A C}=b \sin C\)

∴ Area of \(\triangle A B C=\frac{1}{2} \times(\text { base }) \times(\text { height })\)

= \(\frac{1}{2} \times(B C) \times(A D)=\frac{1}{2} a b \sin C\)

= \(\frac{1}{2}|\vec{a} \times \vec{b}|\)

Area Of Parallelogram: ABCD is a parallelogram. Let us denote the two adjacent sides BC and BA by two vectors, \(\vec{a}=\overrightarrow{B C}\) and \(\vec{b}=\overrightarrow{B A}\); θ = angle between the vectors \(\vec{a}\) and \(\vec{b}\).

Now, we draw AD ⊥ BC.

From AD = AB \(\frac{A D }{A B}\) = b sinθ. So, the area of the parallelogram, S = base x height = (BC)(AD) = \(a b \sin \theta=|\vec{a} \times \vec{b}|\)

This means that the magnitude of the cross product of two vectors is geometrically represented by the area of a parallelogram whose adjacent sides stand for the two given vectors.

Surace Of A Vector: The preceding relation S = \(|\vec{a} \times \vec{b}|\) hints at the possibility of writing S as a vector product, \(\vec{S}\) = \(\vec{a}\) x \(\vec{b}\)

• But it would mean that the area S is a vector quantity. Now, we have to check whether it can be true. For this, let us consider a plane mirror in a room. To describe its effect precisely, we have to specify not only its surface area but also its orientation, i.e., how it is placed in the room.
• A statement like a mirror of area 600 cm² is not sufficient; it should be stated like ‘a mirror of area 600 cm” facing north’.
• This confirms that a plane surface should indeed be treated as a vector quantity; its magnitude equals the area of the surface, and its direction is perpendicular to the surface. A curved surface does not have a definite direction, and cannot be treated as a vector.
• However, an infinitesimally small area ds on this surface is effectively plane, and may be treated as a vector \(\overrightarrow{d s}\). This concept is widely used in different branches of physics.
• In this context, the relation \(\vec{S}\) = \(\vec{a}\) x \(\vec{b}\) is valid for a parallelogram. Consequently, the cross product of two vectors is geometrically represented by the area vector of a parallelogram whose adjacent sides stand for the two given vectors.

Vector Or Cross Product Numerical Examples

Example 1. \(\vec{A}=2 \hat{i}+3 \hat{j}+4 \hat{k}\) and \(\vec{B}=\hat{i}-\hat{j}+\hat{k}\) are two vectors. Find \(\vec{A} \times \vec{B}\).
Solution:

⇒ \(\vec{A} \times \vec{B}=\hat{i}\left(A_y B_z-A_z B_y\right)+\hat{j}\left(A_z B_x-A_x B_z\right)\)

+ \(\hat{k}\left(A_x B_y-A_y B_x\right)\)

= \(\hat{i}\{3 \cdot 1-4 \cdot(-1)\}+\hat{j}(4 \cdot 1-2 \cdot 1)\) + \(\hat{k} \hat{i}+2 \hat{j}-5 \hat{k}\)

Alternative method:

⇒ \(\vec{A} \times \vec{B} =\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
1 & -1 & 1
\end{array}\right|=\hat{i}(3+4)-\hat{j}(2-4)+\hat{k}(-2-3)\)

= \(7 \hat{i}+2 \hat{j}-5 \hat{k}\)

Example 2. Using the vector method in a triangle, prove that,

1. \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\) and
2. \(\cos A=\frac{b^2+c^2-a^2}{2 b c}\).

Solution: According to \(\vec{a}+\vec{c}=\vec{b}\)

1. \(\vec{a} \times \vec{b}=\vec{a} \times(\vec{a}+\vec{c})=\vec{a} \times \vec{c}\) (because \(\vec{a} \times \vec{a}=0\))

∴ \(|\vec{a} \times \vec{b}|=|\vec{a} \times \vec{c}|\)

or, \(a b \sin C=a c \sin \theta\)

(where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{c}\))

or, \(b \sin C=c \sin \left(180^{\circ}-B\right)\)

or, \(b \sin C=c \sin B or, \frac{b}{\sin B}=\frac{c}{\sin C}\)

Proceeding in the same way, we get \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\)

2. \(b^2=\vec{b} \cdot \vec{b}=(\vec{a}+\vec{c}) \cdot(\vec{a}+\vec{c})=\vec{a} \cdot \vec{a}+\vec{c} \cdot \vec{c}+2 \vec{a} \cdot \vec{c}\)

= \(a^2+c^2+2 a c \cos \theta\)

= \(a^2+c^2+2 a c \cos \left(180^{\circ}-B\right)\)

= \(a^2+c^2-2 a c \cos B\)

Similarly, \(a^2=b^2+c^2-2 b c \cos A\)

or, \(\cos A=\frac{b^2+c^2-a^2}{2 b c}\).

Example 3. If the two diagonals of a parallelogram are given by \(\vec{R}_1=3 \hat{i}-2 \hat{j}+7 \hat{k}\) and \(\vec{R}_2=5 \hat{i}+6 \hat{j}-3 \hat{k}\), find out the area of this parallelogram.
Solution:

Let, the two adjacent sides of the parallelogram \(\vec{A}\) and \(\vec{B}\).

Then, \(\vec{R}_1=\vec{A}+\vec{B} \text { and } \overrightarrow{R_2}=\vec{A}-\vec{B}\)

Now, \((\vec{A}+\vec{B}) \times(\vec{A}-\vec{B})=\vec{A} \times \vec{A}-\vec{A} \times \vec{B}+\vec{B} \times \vec{A}-\vec{B} \times \vec{B}\)

= \(0-\vec{A} \times \vec{B}-\vec{A} \times \vec{B}-0=-2(\vec{A} \times \vec{B})\)

From this given data, \((\vec{A}+\vec{B}) \times(\vec{A}-\vec{B})=\vec{R}_1 \times \vec{R}_2\)

= \((3 \hat{i}-2 \hat{j}+7 \hat{k}) \times(5 \hat{i}+6 \hat{j}-3 \hat{k})\)

= \(\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
3 & -2 & 7 \\
5 & 6 & -3
\end{array}\right|=\hat{i}[(-2)(-3)-7 \times 6]+\hat{j}[7 \times 5-3(-3)]\) + \(\hat{k}[3 \times 6-5(-2)]\)

= \(-36 \hat{i}+44 \hat{j}+28 \hat{k}\)

On comparison, \(-2(\vec{A} \times \vec{B})=-36 \hat{i}+44 \hat{j}+28 \hat{k}\)

or, \(\vec{A} \times \vec{B}=18 \hat{i}-22 \hat{j}-14 \hat{k}\)

Here, \(\vec{A} \times \vec{B}=\vec{S}=\) area vector for the parallelogram.

∴ Its area \(|\vec{S}|=|\vec{A} \times \vec{B}|=\sqrt{18^2+(-22)^2+(-14)^2}\)

=31.7 square units.

Example 4. Find the angle betwen force \(\vec{F}=(3 \hat{i}+4 \hat{j}-5 \hat{k})\) and displacement \(\vec{d}=(5 \hat{i}+4 \hat{j}-3 \hat{k})\). Also find the projection of \(\vec{F}\) on \(\vec{d}\).
Solution:

Let \(\theta\) be the angle between the vectors \(\vec{F}\) and \(\vec{d}\). Then by the definition of the scalar product of two vectors, we nave \(\vec{F} \cdot \vec{d}=F d \cos \theta \quad \text { or, } \cos \theta=\frac{\vec{F} \cdot \vec{d}}{F d}\)

Now, \(|\vec{F}|=\sqrt{3^2+4^2+(-5)^2}=\sqrt{50}\)

and \(\vec{d}=\sqrt{5^2+4^2+(-3)^2}=\sqrt{50}\)

∴ \(\vec{F} \cdot \vec{d}=(3 \hat{i}+4 \hat{j}-5 \hat{k}) \cdot(5 \hat{i}+4 \hat{j}-3 \hat{k})\)

= \((3 \cdot 5)+(4 \cdot 4)+(-5) \cdot(-3)\)

= \(15+16+15=46\)

∴ \(\cos \theta=\frac{46}{\sqrt{50} \times \sqrt{50}}=\frac{46}{50}=0.92\)

or, \(\theta=\cos ^{-1} 0.92 \approx 23^{\circ}\)

Hence, the projection of \(\vec{F}\) on \(\vec{d}\) is \(F \cos \theta=\sqrt{50} \times(0.92)=7.07 \times 0.92=6.50 \text { units. }\)

Example 5. Prove that \(|\vec{P} \times \vec{Q}|^2=|\vec{P}|^2|\vec{Q}|^2-|\vec{P} \cdot \vec{Q}|^2\).
Solution:

∴ \(|\vec{P} \times \vec{Q}|^2+|\vec{P} \cdot \vec{Q}|^2=(P Q \sin \theta)^2+(P Q \cos \theta)^2\)

(θ= angle from \(\vec{P} \text { to } \vec{Q}\))

= \(P^2 Q^2\left(\sin ^2 \theta+\cos ^2 \theta\right)=P^2 Q^2\)

= \(|\vec{P}|^2|\vec{Q}|^2\)

∴ \(|\vec{P} \times \vec{Q}|^2=|\vec{P}|^2|\vec{Q}|^2-|\vec{P} \cdot \vec{Q}|^2\).

Example 6. The resultant of two vectors A and B acting through a point O Is R. A certain straight line Intersects the lines representing the vectors \(\vec{A}\), \(\vec{B}\) and \(\vec{R}\) at points P, Q, and S, respectively. Prove that \(\frac{A}{O P}+\frac{B}{O Q}=\frac{R}{O S}\)
Solution:

Let, the unit vectors along \(\vec{A}, \vec{B}\) and \(\vec{R}\) be \(\hat{a}\), and \(\hat{c}\) respectively.

Then \(\vec{A}=A \hat{a}, \vec{B}=B \hat{b}, \vec{R}=R \hat{c}\)

and \(A \hat{a}+B \hat{b}=R \hat{c}\)

or, \(A \frac{\overrightarrow{O P}}{O P}+B \frac{\overrightarrow{O Q}}{O Q}=R \frac{\overrightarrow{O S}}{O S}\)

Graphical Representation of Scalar Multiplication

If \(\hat{n}\) is the unit vector along the line C D, \(\overrightarrow{O P} \times \hat{n}=O P \sin \alpha \hat{m}=O N \hat{m}\): \(\overrightarrow{O Q} \times \hat{n}=O N \hat{m} ; \overrightarrow{O S} \times \hat{n}=O N \hat{m}\)

where \(\hat{m}\) is a unit vector perpendicular to the plane of the vectors.

Then, \(\frac{A}{O P} \overrightarrow{O P} \times \hat{n}+\frac{B}{O Q} \overrightarrow{O Q} \times \hat{n}=\frac{R}{O S} \overrightarrow{O S} \times \hat{n}\)

or, \(\frac{A}{O P} O N \hat{m}+\frac{B}{O Q} O N \hat{m}=\frac{R}{O S} O N \hat{m}\)

or, \(\frac{A}{O P}+\frac{B}{O Q}=\frac{R}{O S}\).

Relative Velocity Of Rain With Respect To A Moving Observer

WBBSE Class 11 Relative Velocity of Rain Notes

In the absence of wind, rain falls vertically. But when a man moves forward during a rainfall, he has to slant his umbrella in front of him. This is due to the fact that the relative velocity of the falling rain with respect to the man makes an angle with the vertical.

Let the actual downward velocity of rain = \(\vec{v}\); the horizontal velocity of the man = \(\vec{u}\).

Angle of Rain with Respect to a Moving Observer

Hence, the relative velocity of rain with respect to the man, \(\vec{w}\) = \(\vec{v}\) – \(\vec{u}\)

∴ w = \(\sqrt{v^2+u^2}\)….(1)

If the relative velocity makes an angle θ with the vertical, then, tanθ = 2……(2)

From equation (1) the magnitude of the relative velocity of rain and from equation (2) its direction can be determined. As the relative velocity of rain makes an angle θ with the vertical, it seems that rain comes down at an angle when we move. This is why, while riding a cycle, one has to hold an umbrella along CO to save himself from getting wet.

For the same reason, a speeding vehicle receives more rain on the front windscreen than on the rear one. Similarly, a person sitting inside a moving train feels that the raindrops follow a slanted path.

Read and Learn More: Class 11 Physics Notes

Relative Velocity Of Rain With Respect To A Moving Observer Numerical Examples

Short Answer Questions on Relative Velocity of Rain

Example 1. A man is walking on a horizontal road at 3 km · h^-1 t while rain is falling vertically with a velocity of 4 km · h^-1. Find the magnitude and direction of the i velocity of rain with respect to the man.
Solution:

Velocity of the man, u = 3 km· h^-1; velocity of rainfall, v = 4 km · h^-1

Since, ∠AOB = 90°, the magnitude of the relative velocity of rain with respect to the man,

w = \(\sqrt{u^2+v^2}=\sqrt{3^2+4^2}=5 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

If \(\angle B O C=\theta, \tan \theta=\frac{u}{v}\) or, \(\theta=\tan ^{-1} \frac{u}{v}\)

∴ \(\theta=\tan ^{-1} \frac{3}{4}=36.9^{\circ}\)

Hence, the velocity of rain with respect to the man makes an angle of 36.9° with the vertical.

Example 2. To a man, walking on a horizontal path at 2 km · h^-1, rain appears to fall vertically at 2 km · h^-1. Find the magnitude and the direction of the actual velocity of rainfall.
Solution:

⇒ \(\vec{u}\) = velocity of the man, \(\vec{v}\)= actual velocity of the raindrops ; so, the relative velocity of the rain¬drops with respect to the man

⇒ \(\vec{w}=\vec{v}-\vec{u} \quad \text { or, } \vec{v}=\vec{u}+\vec{w}\)

Therefore, the resultant of the velocity of the man (\(\vec{u}\)) and the relative velocity of rain (\(\vec{w}\)) will give the real velocity of rain (\(\vec{v}\)).

From the figure, \(v=\sqrt{2^2+2^2}=2 \sqrt{2} \mathrm{~km} \cdot \mathrm{h}^{-1}\)

If the actual direction of rain is inclined at an angle θ with the vertical, then

tanθ = \(\frac{u}{w}\) = \(\frac{2}{2}\) = 1

∴ θ = 45°.

Example 3. To a car driver moving at 40 km · h^-1 towards the south, the wind appears to blow towards the east. When the speed of the car is reduced to 20km· h^-1 wind appears to blow from the north-west. Find the magnitude and direction of the actual velocity of the wind.
Solution:

Let us choose: the x-axis along the east and the y-axis along the north.

Initial velocity of the car, \(\vec{u}_1=-40 \hat{j} \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Final velocity of the car, \(\vec{u}_2=-20 \hat{j} \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Let, \(\vec{v}\) actual velocity of the wind.

Then, the relative velocity of the wind with respect to the car, \(\vec{w}=\vec{v}-\vec{u} \quad \text { or, } \vec{v}=\vec{u}+\vec{w}\)

Initially, \(\vec{w}_1=w_1 \hat{i}\); so \(\vec{v}=\vec{u}_1+\vec{w}_1=w_1 \hat{i}-40 \hat{j}\)…..(1)

Finally, \(\vec{w}_2=w_2 \cos 45^{\circ} \hat{i}-w_2 \sin 45^{\circ} \hat{j}\) (as it is towards south-east)

= \(\frac{w_2}{\sqrt{2}} \hat{i}-\frac{w_2}{\sqrt{2}} \hat{j}\)

∴ \(\vec{v}=\vec{u}_2+\vec{w}_2=\frac{w_2}{\sqrt{2}} \hat{i}-\left(\frac{w_2}{\sqrt{2}}+20\right) \hat{j}\)

Comparing the coefficients of \(\hat{j}\) in (1) and (2), \(\frac{w_2}{\sqrt{2}}+20=40 \text { or, } \frac{w_2}{\sqrt{2}}=20\)

Then from (2), \(\vec{v}=20 \hat{i}-(20+20) \hat{j}=(20 \hat{i}-40 \hat{j}) \mathrm{km} \cdot \mathrm{h}^{-1}\) (between east and south)

Magnitude of \(\vec{v}=v=\sqrt{(20)^2+(40)^2}=20 \sqrt{1+4}\)

= \(20 \sqrt{5} \mathrm{~km} \cdot \mathrm{h}^{-1}\)

If \(\vec{v}\) is inclined at an angle θ with east, then \(\tan \theta=\frac{-40}{20}=-2=\tan \left(-63.4^{\circ}\right) \quad \text { or, } \theta=-63.4^{\circ}\)

So, the wind velocity is at an angle of \(63.4^{\circ}\) south of east.

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Application Of Relative Velocity In One An Two Dimensional Motion

Applications of Relative Velocity in Rain Problems

Motion Of A Boat On A River

Let \(\vec{v}\) be the speed of a boat in still water. If the river current has a velocity of u along its length, the boat is subjected to this velocity too. Therefore, it will move with a resultant velocity, \(\vec{w}\) = \(\vec{v}\) + \(\vec{u}\), and the motion of the boat will, in general, be oblique.

Shortest Distance Traversed In Crossing The River: To find the direction of \(\vec{v}\) such that the distance traveled by the boat is the least, let the angle made by \(\vec{v}\) with the shortest distance of crossing be θ. Resolving the components of \(\vec{v}\) in two mutually perpendicular directions

1. Against the current \(\vec{u}\) and
2. In the direction of the shortest distance AB, we get v sinθ and v cosθ respectively. The component v sinθ balances the effect of the current u.

Therefore, \(u=v \sin \theta \quad \text { or, } \theta=\sin ^{-1} \frac{u}{v}\)

Understanding Relative Velocity of Rain

Component v cosθ enables the boat to cross the river along AB and the time t required for that is

∴ t = \(\frac{l}{v \cos \theta}\)……(1)

The vector vcosθ is actually \(\vec{v}\) + \(\vec{u}\) = \(\vec{w}\) = the resultant velocity. So, to cross along the shortest distance, the boat should be steered in such a way that the resultant velocity is directly across the river.

w = \(\sqrt{v^2-u^2}\), and the angle of steering is, \(\theta=\sin ^{-1} \frac{u}{v}\). If l be the width of the river, the time taken to cross it is,

t = \(\frac{l}{w}=\frac{l}{\sqrt{v^2-u^2}}\)….(2)

Crossing The River In Minimum Time: In equation (1). t will be minimum when the value of cos# is maximum, i.e., cosθ = 1 or, θ = 0°. Hence, to cross the river in minimum time, the boat should be driven along the width of the river.

In that case, the situation is as shown. The boat crosses obliquely along AB with the resultant velocity w. The time required to cross the river is

t’ = \(\frac{A B}{w}=\frac{A C}{v}=\frac{l}{v}\)…..(3)

This is exactly the time required when no current is present. The longitudinal displacement of the boat due to oblique crossing is, \(C B=A C \tan \alpha=\frac{l u}{v}\)….(4)

If there is no current, u=0. Then from equations (2), (3), and (4), we get, t = \(t^{\prime}=\frac{l}{v} \quad \text { and } C B=0\)

Motion Of A Boat On A River Numerical Examples

Example 1. The velocity of a boat in still water is 5 km · h^-1. It takes 15 min to cross a river along the width. The river is 1 km wide. Find the velocity of the current.
Solution:

The boat crosses a river of width 1 km in 15 min or in \(\frac{1}{4}\) h.

The resultant of the velocity of the boat and that of the current is \(\vec{w}\)(say) and

∴ w = \(\frac{1}{1 / 4}=4 \mathrm{~km} \cdot \mathrm{h}^{-1} \text {. }\)

Let the velocity of the boat in still water be \(\vec{v}\) and the velocity of the current be \(\vec{u}\).

Frome figure, \(v^2=u^2+w^2\)

or, \(u^2=v^2-w^2\)

or, \(u=\sqrt{v^2-w^2}=\sqrt{5^2-4^2}=3 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Example 2. A man can reach the point directly opposite on the other bank of a river by swimming across the river in time t₁ and crossing the same distance in time t₂ while swimming along the current. If the velocity of the man in still water is v and the velocity of the water current is u, find the ratio between t₁ and t₂.
Solution:

Let the resultant velocity with which the man swims across the river be w. Hence, w = \(\sqrt{v^2-u^2}\) and therefore, time required to cross the river,

∴ \(t_1=\frac{l}{\sqrt{v^2-u^2}}\)…..(1)

Where l is the width of the river

When the man swims in the direction of the current, the resultant velocity, w’ = v+u, and time required to cross the same distance,

∴ \(t_2=\frac{l}{v+u}\)…….(2)

From (1) and (2), \(\frac{t_1}{t_2}=\frac{v+u}{\sqrt{v^2-u^2}}=\frac{v+u}{\sqrt{v-u} \cdot \sqrt{v+u}}=\sqrt{\frac{v+u}{v-u}}\)

Real-Life Examples of Rain and Relative Velocity

Example 3. Two boats, each with a velocity 8 km · h^-1, attempt to cross a river of width 800 m. The velocity of river current is 5 km · h^-1. One of the boats crosses the river following the shortest path and the other follows the route in which the time taken is minimum. If they start simultaneously, what would be the time difference between their arrivals at the other bank?
Solution:

In order to follow the shortest path, the boat should set itself at a particular angle with the current such that the boat’s resultant velocity is perpendicular to the direction of the current. Hence, the magnitude of the resultant velocity of the boat,

w = \(\sqrt{v^2-u^2}\)

and the time required to cross the river along the shortest path, \(t_1=\frac{x}{w}=\frac{x}{\sqrt{v^2-u^2}}=\frac{0.8}{\sqrt{8^2-5^2}}=0.128 \mathrm{~h}=7.69 \mathrm{~min}\)

[x = 800 m = 0.8 km, v = 8 km · h^-1 and u = 5 km · h^-1]

To cross in minimum time, the boat should travel at right angles to the current and the time required is

Required time difference, t₁ – t₂ = 7.69 – 6 = 1.69 min

Example 4. A person can swim at 4 km · h^-1 in still water. At what angle should he set himself to cross the river in a direction perpendicular to the river current of velocity 2 km · h^-1?
Solution:

Let the velocity of the person be \(\vec{v}\), velocity of the current be \(\vec{u}\) and the resultant of \(\vec{u}\) and \(\vec{v}\) be \(\vec{w}\). The person crosses the river along its width from one bank to the other.

Hence, \(\vec{w}\) and \(\vec{u}\) are perpendicular to each other. If the angle between \(\vec{v}\) and \(\vec{w}\) is θ, then \(\sin \theta=\frac{u}{v}\)= \(\frac{2}{4}=\frac{1}{2}\) or, \(\theta=30^{\circ} \text {. }\)

Hence, the person has to swim at an inclination of 30° with the width of the river, or at an angle of (90° + 30°) or, 120° with the direction of the current as shown.

Example 5. The width of a river is D. A man can cross the river in time t₁ in the absence of any river current. But in the presence of a certain river current, the man takes a time t₂ to cross the river directly. Show that the velocity of the current is v = \(D \sqrt{\frac{1}{t_1^2}}-{\frac{1}{t_2^2}}\)
Solution:

Let u = velocity of the man on the river. \(t_1=\frac{D}{u} \quad \text { or, } u=\frac{D}{t_1}\)….(1)

In the presence of a river current of velocity v, the effective velocity of the man will be the resultant w of u and v. \(\vec{w} = \vec{u}+\vec{v}\)

To cross the river directly, the man has to swim in such a direction that the resultant w is across the width of the river.

Clearly, w = \(\sqrt{u^2-v^2}\)

and the time taken to cross the river is \(t_2=\frac{D}{w}=\frac{D}{\sqrt{u^2-v^2}}\).

∴ \(t_2^2=\frac{D^2}{u^2-v^2}\) or, \(u^2=v^2+\frac{D^2}{t_2^2}\)……(2)

From (1) and (2), \(\frac{D^2}{t_1^2}=v^2+\frac{D^2}{t_2^2}\) or, \(v^2=D^2\left(\frac{1}{t_1^2}-\frac{1}{t_2^2}\right)\)

∴ v = \(D \sqrt{\frac{1}{t_1^2-\frac{1}{t_2}}}\)

Relative Velocity And Relative Acceleration

WBBSE Class 11 Relative Velocity Notes

Relative Velocity: When we consider the motion of a body either on the earth’s surface or close to it, it is assumed that the earth is a stationary frame of reference. The velocity of a body with respect to a stationary observer on the earth’s surface is considered as the actual velocity.

• Often we consider the motion of a body with respect to another one.
• A passenger in a moving train observes the surrounding objects in apparent motion even if they actually are stationary. When both of two trains are moving side by side, an observer in one of them would not be able to perceive the actual motion of the other—only an apparent motion relative to himself would be observed.
• In such cases, it is necessary to consider the idea of relative motion or relative velocity between two bodies where the reference point may not be at rest.
• We already know that there is nothing like absolute rest or absolute motion. So there should be nothing like absolute velocity—all observed velocities are effectively relative.

Relative Velocity Definition: Relative velocity is defined as the apparent velocity of a body with respect to another body that may be in motion.

Calculation Of Relative Velocity:

Case 1: The observer and the object move in the same direction: Let us assume that the velocity of the observer is u and that of the object is v. If they move in the same direction, the object would seem to be slower to the observer.

The relative velocity of the object with respect to the observer is, w = actual velocity of the object – actual velocity of the observer

= V— u…..(1)

Case 2: The observer and the object move in opposite directions: Let the velocity of the observer be u and that of the object moving in the opposite direction be -v.

Read and Learn More: Class 11 Physics Notes

Here again, the relative velocity of the object with respect to the observer is,

w = actual velocity of the object – actual velocity of the observer

= – v- u = -(v+ u)….(2)

The magnitude of the relative velocity of the object, (u + v) will always be greater than the actual velocities (u or v).

Case 3: The observer and the object move in different directions: If the object and the observer are moving in different directions that are not in the same line or plane, then again equation (1) is used to find the relative velocity. However, the subtraction has to be done using vector algebra.

In general, if \(\vec{u}\) = velocity of an object,

and \(\vec{v}\) = velocity of an observer, then, the relative velocity of the object with respect to the observer is, \(\vec{w}=\vec{v}-\vec{u}\)….(3)

We may rewrite equation (3) as \(\vec{w}=\vec{v}-\vec{u}=\vec{v}+(-\vec{u})\) = the resultant of the vectors \(\vec{v}\) and —\(\vec{u}\).

If we interchange the observer and the object keeping the velocities unchanged, then the relative velocity is given by, \(\vec{w}^{\prime}=\vec{u}-\vec{v}=-(\vec{v}-\vec{u})=-\vec{w}\)

As \(\vec{w}^{\prime}=-\vec{w}\), these two relative velocities Eire eqeal but opposite. Thus, out of two trains running on parallel tracks if the second train moves forward relative to the first one, then the first train will be observed to move backward relative to the second.

Since velocity is a vector quantity, the relative velocity of a body involves a vector subtraction and can be found out by the vector addition of \(\vec{v}\) and \(-\vec{u}\).

For example, a train moving due east with a velocity u sees another train moving due west with the same velocity. Then, relative velocity = \(\vec{u}-(-\vec{u})=2 \vec{u}\)and they will appear to approach each other at twice the speed.

If both the trains are moving in the same direction, relative velocity = \((\vec{u}-\vec{u})\) = 0. The trains will appear to be at rest with respect to each other.

Understanding Relative Velocity in Physics

Magnitude And Direction Of Relative Velocity: Let us consider that a body A (for example, an observer) is moving with velocity \(\vec{u}\) along OX and B, and another body is moving with velocity \(\vec{v}\) along OY as shown. Taking O as origin, \(\overrightarrow{O A}=\vec{u}\), \(\overrightarrow{O B}=\vec{v}\) and the angle between them is ∠AOB = α.

To compute the relative velocity of B with respect to A, OA’ is drawn so that AO = OA’. Then \(\overrightarrow{O A^{\prime}}\) is the opposite vector of \(\overrightarrow{O A}\), i.e., \(\overrightarrow{O A^{\prime}}\) = –\(\vec{u}\). Completing the parallelogram OA’CB, the diagonal \(\overrightarrow{O C}\) is drawn. OC represents the resultant of \(\vec{v}\) and –\(\vec{u}\), i.e., the relative velocity of B with respect to A.

∴ \(\overrightarrow{O A}\) = \(\vec{w}\)

And OC  = \(|\vec{w}|=\sqrt{u^2+v^2+2 u v \cos \left(180^{\circ}-\alpha\right)}\)

= \(\sqrt{u^2+v^2-2 u v \cos \alpha}\)…..(4)

If the relative velocity \(\overrightarrow{O C}\) is inclined to \(\overrightarrow{O B}\) at an angle θ then,

tanθ \(=\frac{u \sin \left(180^{\circ}-\alpha\right)}{v+u \cos \left(180^{\circ}-\alpha\right)}=\frac{u \sin \alpha}{\nu-u \cos \alpha}\)….(5)

Also, since, \(\vec{w}^{\prime}=-\vec{w}\), the relative velocity of A with respect to B is represented by the vector \(\overrightarrow{C P}\).

Relative Acceleration Explained

Relative Acceleration: By similar reasoning, relative acceleration is defined as the apparent acceleration of a body with respect to another body that may be in an accelerated motion.

Let A and B be two moving objects. The rate of change of velocity of A with respect to B is called the relative acceler¬ation of A with respect to B. Actual acceleration of A is obtained when B is at rest or in uniform motion.

If the actual accelerations of A and B are \(\vec{a}_A\) and \(\vec{a}_B\) respectively, then the relative acceleration of B with respect to A

∴ \(\vec{a}_{B A}=\vec{a}_B-\vec{a}_A\)…..(1)

and relative acceleration of A with respect to B

∴ \(\vec{a}_{A B}=\vec{a}_A-\vec{a}_B\)….(2)

Relative Acceleration Example:

1. The actual acceleration of a freely falling body is the downward acceleration due to gravity (\(\vec{g}\)). Hence, two bodies, falling freely, will have a relative acceleration \(\vec{g}\)–\(\vec{g}\) = 0. They are, therefore, stationary or moving with uniform velocity with respect to each other.
2. Let a body A fall vertically downwards with an acceleration \(\vec{g}\) due to gravity. Another body B is moving horizontally with a uniform acceleration \(\vec{a}\).

Relative Velocity Formula and Examples

Then the relative acceleration of A with respect to B is, \(\vec{g}^{\prime}=\vec{g}-\vec{a}=\vec{g}+(-\vec{a})\) = resultant of \(\vec{g}\) and –\(\vec{a}\).

This resultant \(\vec{g}^{\prime}\) is shown.

Its magnitude is, g’ = \(\sqrt{g^2+a^2} \text { (obviously, } g^{\prime}>g \text { ) }\)

g’ is inclined with the vertical at an angle, \(\theta=\tan ^{-1} \frac{a}{g}\)

For example, when a car or a train accelerates suddenly, any object hanging from its roof is tilted away from the vertical due to the inclination of the relative acceleration \(\vec{g}^{\prime}\). For any random motion of the car or of the train, the relative acceleration also changes randomly, and angle θ does not have a steady value. That is why hanging objects are often observed to be oscillating.

Relative Acceleration Numerical Examples

Short Answer Questions on Relative Motion

Example 1. A car is moving at 80 km · h^-1 towards the north. Another car is moving at 80^2 km · h^-1 towards the northwest. Find the relative velocity of the second car with respect to the first.
Solution:

The velocity of the first car, \(\vec{u}\) = 80 km • h^-1 towards north = \(\overrightarrow{A B}\) the velocity of the 2nd car, \(\vec{v}\) = 80√2 km · h^-1 towards north-west = \(\overrightarrow{A C}\).

Hence, relative velocity of the second car with respect to the \(\vec{w}=\vec{v}-\vec{u}=\vec{v}+(-\vec{u})=\overrightarrow{A C}+\overrightarrow{B A}=\overrightarrow{B C}\) (from triangle law of vector addition)

As ∠BAC = \(45^{\circ}\), we get from \(\triangle A B C\), \(w^2=u^2+v^2-2 u v \cos 45^{\circ}\)

= \((80)^2+(80 \sqrt{2})^2-2.80 \cdot 80 \sqrt{2} \cdot \frac{1}{\sqrt{2}}\)

∴ \(w^2=80^2\left[1+(\sqrt{2})^2-2\right]=80^2\)

∴ w = 80 km · h^-1

Since, w = u, from ΔABC,

∠ACB = ∠BAC = 45°

∴ ∠ABC = 90°

Hence, the vector \(\overrightarrow{B C}=\vec{w}\) is directed towards the west.

The relative velocity of 1st car with respect to the second will also have the same magnitude but will be directed towards the east.

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Example 2. Two bodies are moving such that the velocity of one is twice that of the other and they make an angle of 60° with each other. Find the relative velocity of one with respect to the other.
Solution:

Representing the velocities in the vector diagram, we get, \(\overrightarrow{A B}=\vec{u}\) = velocity of one body, \(\overrightarrow{A B}=\vec{2 u}\) = velocity of the other.

The relative velocity of the second with respect to the first, \(\vec{w}=\overrightarrow{A C}-\overrightarrow{A B}=\overrightarrow{A C}+\overrightarrow{B A}=\overrightarrow{B C}\) [from triangle law of vector addition]

From ΔABC, \(w^2=(-u)^2+(2 u)^2+2 \cdot(-u) \cdot 2 u \cos 60^{\circ}\)

= \(5 u^2-2 u^2=3 u^2\)

∴ \(w=\sqrt{3} u\)

Also from the trigonometric rule, \(\frac{2 u}{\sin \angle A B C}=\frac{w}{\sin 60^{\circ}}=\frac{\sqrt{3} u \times 2}{\sqrt{3}}\)

or, \(\sin \angle A B C=1=\sin 90^{\circ}\)

∴ \(\angle A B C=90^{\circ} \text {. }\)

Hence \(\vec{w}\), i.e., \(\overrightarrow{B C}\) is perpendicular to \(\vec{u}\), i.e., \(\overrightarrow{A B}\).

Similarly, the relative velocity of the first with respect to the second can be found. In this case, the relative velocity will be –\(\vec{w}\) as the direction will be opposite.

Example 3. At any instant of time, two ships A and B are 70 km apart along a line AB which is directed from north to south. A starts moving towards west at 25 km · h^-1 and at the same time B starts moving towards the north at 25 km · h^-1. Find the distance of closest approach between the two ships and the time required for this.
Solution:

Solution: Let us choose the origin at point A; x-axis along east; y-axis along north.

Initial position of ship A=0 and that of ship B=-70\(\hat{j}\) km

Velocity of ship A = \(-25 \hat{i} \mathrm{~km} \cdot \mathrm{h}^{-1}\) and that of ship \(B=25 \hat{j} \mathrm{~km} \cdot \mathrm{h}^{-1}\).

After a time of t hours, the positions of the ships are,

⇒ \(\vec{r}_A=0+(-25 \hat{i}) t=-25 t \hat{i} \mathrm{~km}\)

⇒ \(\vec{r}_B=-70 \hat{j}+(25 \hat{j}) t=(25 t-70) \hat{j} \mathrm{~km}\)

Position of ship B relative to ship A, \(\vec{r}=\vec{r}_B-\vec{r}_A=25 t \hat{i}+(25 t-70) \hat{j}\)

The distance between them is r = \(|\vec{r}|\).

So, \(r^2 =(25 t)^2+(25 t-70)^2\)

= \(625 t^2+625 t^2-3500 t+4900\)

= \(50\left(25 t^2-70 t+98\right)\)

= \(50\left[(5 t-7)^2-7^2+98\right]\)

= \(50\left[(5 t-7)^2+49\right]\)

When the distance r between the ships is minimum, \(r^2\) is also minimum.

This happens when \((5 t-7)^2\), a squared quantity, is minimum, i.e., zero.

∴ 5t – 7 = 0

or, t = \(\frac{7}{5} \mathrm{~h}=84 \mathrm{~min}=1 \mathrm{~h} 24 \mathrm{~min}\)

At this instant of time, \(r^2=50[0+49]=50 \times 49=25 \times 49 \times 2\)

∴ The distance of the closest approach between the two ships, \(r_{\min }=\sqrt{25 \times 49 \times 2}=35 \sqrt{2} \mathrm{~km}\)

Example 4. A ship is moving towards the east at 10 km · h^-1. A boat is moving north of east making an angle of 30° with the north. What should be the velocity of the boat so that the boat always appears, from the ship, to move towards the north?
Solution:

Let the velocity of the ship be \(\vec{u}=\overrightarrow{O A}\) and velocity of the boat be \(\vec{v}=\overrightarrow{O B}\)

Hence, the velocity of the boat with respect to the ship, \(\vec{w}=\vec{v}-\vec{u}=\overrightarrow{A B}\)

From the figure, \(\sin 30^{\circ}=\frac{u}{v} \text { or, } \frac{1}{2}=\frac{u}{v}\)

or, \(v =2 u=2 \times 10 \mathrm{~km} \cdot \mathrm{h}^{-1} \text { [Given, } u=10 \mathrm{~km} \cdot \mathrm{h}^{-1} \text { ] }\)

= \(20 \mathrm{~km} \cdot \mathrm{h}^{-1}\)

Graphical Representation of Relative Velocity

Example 5. A man is in a car moving with an acceleration of 5 m · s^-2. Find the apparent value of the acceleration due to gravity and the direction of pull of the earth with respect to him.
Solution:

Let the acceleration of the man in the car be \(\vec{a}\) and acceleration due to gravity be \(\vec{g}\). Hence acceleration due to gravity’s relation to the man is, \(\overrightarrow{g^{\prime}}=\vec{g}-\vec{a}=\vec{g}+(-\vec{a})\)

g’ = \(\sqrt{g^2+a^2}=\sqrt{9.8^2+5^2}\)

= \(\sqrt{121.04}=11 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

The angle θ, which \(\vec{g}^{\prime}\) makes with \(\vec{g}\) is given by \(\tan \theta=\frac{a}{g}=\frac{5}{9.8}=0.51=\tan 27^{\circ}\)

∴ \(\theta=27^{\circ}\)

Hence, with respect to the man the earth’s pull is acting downward at an angle 27° with the vertical in the side opposite to his direction of motion.

Example 6. A lift is moving up with a constant acceleration a. A man standing on the lift, throws a ball vertically upwards with a velocity v, which returns to the thrower after a time t. Show that v = (a + g)\(\frac{t}{2}\) where g is the acceleration due to gravity.
Solution:

The downward acceleration of the ball with respect to the lift = g-(-a) = g+ a.

The initial velocity of the ball is v upward. As the ball returns to the thrower, its relative displacement is zero.

Hence, from the equation h = ut- \(\frac{1}{2}\)gt², we get,

0 = \(v t-\frac{1}{2}(g+a) t^2 \text { or, } 0=t\left\{v-\frac{1}{2}(g+a) t\right\}\)

Since, t ≠ 0,

∴ \(v-\frac{1}{2}(g+a) t=0 \quad \text { or, } v=(a+g) \frac{t}{2}\)

Example 7. A lift is moving up with an acceleration of 2 m · s^-2. A nail gets dislodged from the roof of the lift when its speed reaches 8 m · s^-1. If the height of the lift cage is 3 m, find the time taken by the nail to touch the floor of the lift.
Solution:

Relative to the lift, the initial velocity of the nail is u = 0 and it falls through a height of h = 3 m. Inside the lift, the relative acceleration due to gravity acting on the nail downwards is

g’ = \(g-(-a)=9.8-(-2)=11.8 \mathrm{~m} \cdot \mathrm{s}^{-2}\)

∴ h = \(u t+\frac{1}{2} g^{\prime} t^2 \text { or, } 3=0 \cdot t+\frac{1}{2} \times 11.8 \times t^2\)

or, \(t^2=\frac{2 \times 3}{11.8} \text { or, } t=\sqrt{\frac{6}{11.8}}=0.713 \mathrm{~s}\)

Example 8. A simple pendulum is suspended from the roof of a car moving horizontally with an acceleration of 10 m · s^-2. What will be the angle made by the pendulum in its equilibrium position with the vertical? [g = 10 m · s^-2]
Solution:

The acceleration of the car, a = 10 m · s^-2, and acceleration due to gravity g = 10 m · s^-2. The pendulum will be inclined along the direction of the apparent pull of the earth.

Let the angle made by the pendulum with the vertical be θ.

∴ tanθ = \(\frac{a}{g}\) = \(\frac{10}{10}\) = 1 or, θ = 45

Example 9. Two parallel rail lines are directed as north-south. A train X runs towards north with a speed of 15 m · s^-1 and another train Y runs towards south with a speed of 25 m · s^-1. Find

1. The velocity of Y relative to X,
2. The velocity of ground with respect to Y,
3. The velocity of a monkey running on the roof of X against its motion with a velocity of 5 m · s^-1 relative to X, as observed by a man standing on the ground.

Solution:

Let us assume the velocity from south to north is positive.

Then, the velocity of X, v_x = +15 m · s^-1

and velocity of Y, u_y = -25 m · s^-1 m

The velocity of train Y relative to train X is, v_YX = v_Y – v_X=-25-15 = -40 m · s^-1

The relative velocity of ground with respect to train Y is, v_GY = v_G – V_Y = o – (-25) = 25 m · s^-1

The relative velocity of the monkey with respect to training X is, VMX= v_M-u_X=-5m · s^-1

(where is the velocity of the monkey with respect to the ground)

∴ v_M = v_X + v_MX = 15 + (-5) = 10 m · s^-1

Example 10. A steamer is moving towards the east with a velocity u. A second steamer is moving with a velocity 2u at angle θ north of east The motion of the second steamer relative to the first is along the northeast. Show that, cosθ-sinθ = \(\frac{1}{2}\)
Solution:

Let us choose, the x-axis along the east and the y-axis along the north. Then the velocities are of the first steamer, \(\vec{v}_1=u \hat{i}\); of the second steamer, \(\overrightarrow{v_2}=2 u \cos \theta \hat{i}+2 u \sin \theta \hat{j}\)

∴ Velocity of the second steamer relative to the first is, \(\vec{w}=\vec{v}_2-\vec{v}_1=u(2 \cos \theta-1) \hat{i}+2 u \sin \theta \hat{j}\)

As \(\vec{w}\) is along the north-east, its x- and y-components are equal.

∴ u(2cosθ-1) = 2usinθ or, 2cosθ-2sinθ = 1

or, cosθ – sinθ = \(\frac{1}{2}\)

Example 11. A stone is dropped from a tower 400 m high. Simultaneously, another stone is thrown upwards from the earth’s surface with a velocity of 100 m/s. When and where would these two stones meet? (g = 9.8 m/s²)
Solution:

Both the stones move with a uniform acceleration g, the acceleration due to gravity, acting downwards.

∴ The acceleration of one relative to the other = g- g = 0.

At time t = 0, the downward velocities of the two stones are, respectively, 0 and -100m/s. So, relative velocity = 0 – (-100) s 100 m/s. The stones move with this relative velocity, which is uniform in the absence of any relative acceleration.

The initial distance between the stones = 400 m.

Thus, they will meet after a time t = \(\frac{400}{100}\) = 4 s.

The height through which the first stone falls in time t = 4 s is, h = \(\frac{1}{2}\)gt² = \(\frac{1}{2}\)x 9.8 x 42 = 78.4 m

∴ The stones meet after 4 s at a height of (400 – 78.4), or 321.6 m.

Example 12. A rubber bail Is thrown downwards from the top of a lower with a velocity of 14 m/s. A second ball Is dropped from the same place 1 s later. The first ball reaches the ground in 2 seconds and rebounds with the same magnitude of velocity. How much later would the two balls collide with each other?
Solution:

Height of the tower, h = Distance travelled by the first ball in 2 s (t = 0 to t = 2s)

= 14 x 2 + \(\frac{1}{2}\) x 9.8 x 2² = 47.6 m

The second ball starts at t = 1 s. From t = Is to t = 2s, this ball comes down through a height, h’ = \(\frac{1}{2}\) x 9.8 x (2 – 1)² = 4.9 m

∴ At t = 2 s , the distance between the two balls, H = h-h’ = 47.6-4.9 = 42.7 m

For the first ball, velocity at t = 2 s is, v₁ = 14 + 9.8 x 2 = 33.6 m/s

As it rebounds with the same magnitude of velocity, its upward velocity at the same instant, i.e., at t = 2 s is v’₁= -33.6 m/s.

For the second belli, at t = 1 s, the velocity u = 0.

So, at t = 2 s, the velocity, v₂ = 0 + 9.8 x (2 – 1) = 9.8 m/s

∴ Relative velocity of the second ball with respect to the first at t = 2 s is,

V = v₂-v’₁ = 9.8-(-33.6) = 43.4 m/s

Relative acceleration =g-g = 0. So, for relative motion, the velocity V is uniform. Thus, time required to travel the height of H = 42.7 m is,

t = \(\frac{H}{V}\) = \(\frac{42.7}{43.4}\) = 0.98 s-1

Therefore, the balls will collide at T = 2 + t = 2.98 s.

Example 13. An object falling freely from a height H hits an inclined plane at a height h in its trajectory. At the instant of collision, the velocity of the object changes to become horizontal. What is the value of \(\frac{h}{H}\) for which it spends maximum time to reach the ground?

Solution:

The trajectory of the object is ABC

Let AB is a free fall through a height (H-h) in time t₁.

∴ (H- h) = \(0+\frac{1}{2} g t_1^2 \quad \text { or, } t_1=\sqrt{\frac{2(H-h)}{g}}\)

BC is a projectile motion for which the vertical component of velocity initially (at B) is zero. Also for the path BC, the vertical drop is through a height h. If the time taken is t₂

h = \(0+\frac{1}{2} g t_2^2 \quad \text { or, } t_2=\sqrt{\frac{2 h}{g}}\)

Total time, t = \(t_1+t_2=\sqrt{\frac{2(H-h)}{g}}+\sqrt{\frac{2 h}{g}}=\sqrt{\frac{2}{g}}(\sqrt{H-h}+\sqrt{h}]\)

∴ \(t^2=\frac{2}{g}[(H-h)+h+2 \sqrt{h(H-h)}]\)

= \(\frac{2}{g}[H+\sqrt{4 h(H-h)}]=\frac{2}{g}\left[H+\sqrt{H^2-(2 h-H)^2}\right]\)

For maximum t, i.e., for maximum t², the term (2h- H)² should be minimum. This minimum value is zero, as it is a square.

∴ (2h— H)² = 0 or, 2h = H or, \(\frac{h}{H}\) = \(\frac{1}{2}\)

Composition Of Several Vectors By Resolution

WBBSE Class 11 Vector Resolution Notes

Three or more vectors can be added by the law of polygon of vectors. This method is geometrically convenient but is inconvenient mathematically. However, when vectors lie on a plane, an elegant method is to resolve each of the vectors into two mutually perpendicular components while performing addition.

Composition Of Several Vectors By Resolution Process:

The vectors are drawn on a plane from a common initial point. Two mutually perpendicular axes X and Y are drawn from that point and each vector is resolved into components along these two axes.

Next, algebraic sums of the x-components and of the y -y-components of the vectors with proper sign (+ or -) are determined separately. Now, vector addition of the resultant x and y components gives the resultant of all the vectors.

Composition Of Several Vectors By Resolution Calculation:

Let \(\vec{P}, \vec{Q}, \vec{R}\) be three coplanar vectors having a common initial point O. To find the resultant, axes OX and OY are drawn, coplanar with the vectors.

Let  \(\vec{P}, \vec{Q}, \vec{R}\) make angles α, β, γ respectively with the x-axis. Thus their respective components along the x-axis are \(P_x=P \cos \alpha, Q_x=Q \cos \beta, and R_x=R \cos \gamma\).

Similarly \(P_y=P \sin \alpha, Q_y=Q \sin \beta and R_y=R \sin \gamma\).

Composition of Vectors by Resolution Explained

Hence, the resultants (sums) F_x and F_y along x and y axes are \(F_x=P \cos \alpha+Q \cos \beta+R \cos \gamma\)….(1)

and \(F_y=P \sin \alpha+Q \sin \beta+R \sin \gamma\)….(2)

Read and Learn More: Class 11 Physics Notes

For performing the summation, positive or negative values of sine and cosine functions for the angles α, β, γ should be taken into consideration. \(\vec{F}\) represents the resultant of F_x and F_y, according to the triangle or the parallelogram law of vector addition.

F = \(\sqrt{F_x^2+F_y^2}\)…(3)

Key Concepts in Vector Addition and Resolution

and the inclination θ of \(\vec{F}\) with the x-axis is given by \(\tan \theta=\frac{F_y}{F_x}\)…(4)

Equations (3) and (4) give the magnitude and direction respectively of the resultant of the three vectors \(\vec{P}, \vec{Q}, \vec{R}\). The resultant of any number of coplanar vectors can be found using this method.

Composition Of Several Vectors By Resolution Numerical Examples

Examples of Vector Composition Using Resolution

Example 1. Compute the resultant of three coplanar vectors P, 2P and 3P inclined at 120° with one another.
Solution:

Computing The resultant of three coplanar vectors P, 2P and 3P inclined at 120° with one another

Let the x-axis be taken along vector P. Then the vectors 2P and 3P will make angles 120° and 240° respectively with the x-axis. Resolving the vectors into components along the x and y axes and adding, we get the stuns R_x and R_y respectively along the x and y axes.

⇒ \(R_x=P \cos 0^{\circ}+2 P \cos 120^{\circ}+3 P \cos 240^{\circ}\)

∴ \(R_x=P+2 P\left(-\frac{1}{2}\right)+3 P\left(-\frac{1}{2}\right)=P-P-\frac{3}{2} P=-\frac{3}{2} P\)

⇒ \(R_y=P \sin 0^{\circ}+2 P \sin 120^{\circ}+3 P \sin 240^{\circ}\)

= \(0+2 P \times \frac{\sqrt{3}}{2}+3 P\left(-\frac{\sqrt{3}}{2}\right)=(2-3) \frac{\sqrt{3}}{2} P=-\frac{\sqrt{3}}{2} P\)

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Hence, the value of the resultant \(\vec{R}\) is, R = \(\sqrt{R_x^2+R_y^2}=\sqrt{\frac{9}{4} P^2+\frac{3}{4} P^2}=\sqrt{3} P\)

∴ \(\vec{R}\) makes an angle θ with the x-axis such that,\(\tan \theta=\frac{R_y}{R_x}=\frac{-\frac{\sqrt{3}}{2} P}{-\frac{3}{2} P}=\frac{1}{\sqrt{3}}=\tan 30^{\circ} \text { or, } \tan 210^{\circ}\)

As \(R_x\) and \(R_y\) are both negative, R must be in the third

As R_x and R_y are both negative, R must be in the third quadrant, and θ = 210°.

Short Answer Questions on Vector Resolution

Example 2. Five coplanar forces, each of magnitude F, are acting on a particle. Each force is inclined at an angle of 30° with the previous one. Find out the magnitude and direction of the resultant force on the particle.
Solution:

Given

Five coplanar forces, each of magnitude F, are acting on a particle. Each force is inclined at an angle of 30° with the previous one.

Let us choose the x-axis along the first force, and y-axis along the perpendicular direction on the plane of the forces

The angles made by the five forces with the x-axis are 0°, 30°, 60°, 90° and 120°.

∴ Component of resultant force along x-axis, F_x = Fcos0° + Fcos30° + Fcos60° + Fcos90° + Fcos120°

= \(F\left[1+\frac{\sqrt{3}}{2}+\frac{1}{2}+0-\frac{1}{2}\right]\)

= \(F\left(1+\frac{\sqrt{3}}{2}\right)\)

and component of resultant force along y-axis, Fy = Fsin0° + Fsin30° + Fsin60° + Fsin90° + Fsinl20°

= \(F\left[0+\frac{1}{2}+\frac{\sqrt{3}}{2}+1+\frac{\sqrt{3}}{2}\right]=F\left(\frac{3}{2}+\sqrt{3}\right)\)

= \(\sqrt{3} F\left(1+\frac{\sqrt{3}}{2}\right)\)

So, the magnitude of the resultant force, \(F^{\prime}=\sqrt{F_x^2+F_y^2}=F\left(1+\frac{\sqrt{3}}{2}\right) \sqrt{1^2+(\sqrt{3})^2}\)

= \(2 F\left(1+\frac{\sqrt{3}}{2}\right)=(2+\sqrt{3}) F\)

The angle of inclination of F’ with the x-axis, \(\theta=\tan ^{-1} \frac{F_y}{F_x}=\tan ^{-1} \sqrt{3}=60^{\circ} \text { (along the third force) }\)

Three Dimensional Resolution Of Vectors Algebraic Representation

Let the magnitude and direction of a vector \(\vec{r}\), in space, be represented by \(\overrightarrow{O P}\). Our aim here is to find an elegant representation of \(\vec{r}\)

O is taken as the origin of the 3-dimensional space and three mutually perpendicular axes x, y, and z are drawn. A special rule is followed to indicate the directions of the three axes.

If we hold the first 3 fingers of our right hand at right angles to one another, then the forefinger points toward the x-axis, the middle finger toward the y-axis, and the thumb towards the z-axis. This is called the right-handed cartesian coordinate system.

Here, the origin O (0,0,0) is the initial point, and P (x, y, z) is the endpoint of the vector \(\overrightarrow{O P}\). Taking the line segment OP as a diagonal, the cuboid ADPEOBFC is drawn.

Understanding Three Dimensional Vector Components

The projections of the vector \(\vec{r}\) along the three axes are, from OA = x, OB = AD = y, and OC = DP = z. These x, y, and z are three mutually perpendicular components of the vector \(\vec{r}\).

When the line segments OA, AD, and DP are taken as vectors, the three-dimensional polygon OADP gives, as per the law of polygon of vector addition, \(\overrightarrow{O A}+\overrightarrow{A D}+\overrightarrow{D P}=\overrightarrow{O P}\)….(1)

If \(\hat{i}, \hat{j}, \hat{k}\) are the unit vectors along the x, y and z axes respectively in the positive direction,

∴ \(\overrightarrow{O A}=x \hat{i}, \overrightarrow{A D}=y \hat{j} \quad \text { and } \overrightarrow{D P}=z \hat{k}\)

Since, \(\vec{r}=\overrightarrow{O P}\), from equation (1)

∴ \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)………..(2)

⇒ \(\vec{r}\) is the position vector of the point P w.r.t origin O.

Equation (2) shows the algebraic representation of a position vector with initial point (0,0,0) and final point (x, y, Z).

Determination Of The Magnitude Of The Vector: From the geometrical property of a cuboid, (diagonal)² = (length)³ + (breadth)2 + (height)²

or, OP² = OA² + OB² + OC² or, r² = x² + y² + z²

or, \(r=\sqrt{x^2+y^2+z^2}\)….(3)

Hence, the magnitude or value of a vector with its initial point as the origin can be determined easily with the help of the coordinates of its terminal point.

Direction Cosine: Let \(\overrightarrow{O P}\) be inclined at an angle a with the x-axis, i.e., ∠POA = α. As per the property of the cuboid, ΔOAP is a right-angled triangle with ∠OAP = 90°.

Hence, \(\cos \alpha=\frac{O A}{O P}=\frac{x}{r}=l\) (say)….(4)

Similarly, if \(\overrightarrow{O P}\) makes angles β and γ with y and z axes respectively,

⇒ \(\left.\begin{array}{rl}\cos \beta =\frac{y}{r}=m \\ \text { and } \cos \gamma =\frac{z}{r}=n\end{array}\right\}\)

From (4) and (5), \(l^2+m^2+n^2=\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=\frac{x^2}{r^2}+\frac{y^2}{r^2}+\frac{z^2}{r^2}\)

= \(\frac{x^2+y^2+z^2}{r^2}=\frac{r^2}{r^2}=1\)…..(6) [with the help of equation (3)]

Hence, to know the direction of the vector, the angles α, β,γ can be determined using equations (4) and (5). cos α, cosβ, cosγ or l, m, n is called the direction cosines of vector r with respect to the axes x, y, and z respectively. Equation (6) indicates the relationship among the direction cosines.

 

WBBSE Class 11 Three Dimensional Vector Resolution Notes

Sum And Difference Of Two Vectors: Let \(\overrightarrow{r_1}=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k} \text { and } \vec{r}_2=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\) be two vectors.

Sum or resultant of these two vectors is \(\vec{r}_1+\vec{r}_2=\left(x_1+x_2\right) \hat{i}+\left(y_1+y_2\right) \hat{j}+\left(z_1+z_2\right) \hat{k}\)

and its magnitude is, \(\left|\vec{r}_1+\vec{r}_2\right|=\sqrt{\left(x_1+x_2\right)^2+\left(y_1+y_2\right)^2+\left(z_1+z_2\right)^2}\)

The direction of the resultant is specified by the direction cosines. They are \(\cos \alpha=\frac{x_1+x_2}{\left|\vec{r}_1+\vec{r}_2\right|}, \cos \beta=\frac{y_1+y_2}{\left|\vec{r}_1+\vec{r}_2\right|} \text { and } \cos \gamma=\frac{z_1+z_2}{\left|\vec{r}_1+\vec{r}_2\right|}\)

Similarly, the difference between the two vectors, \(\overrightarrow{r_1} \text { and } \overrightarrow{r_2}\) can also be found out. The resultant in this case will be \(\vec{r}_2-\vec{r}_1=\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}\)

The magnitude of the resultant will be, \(\left|\vec{r}_2-\vec{r}_1\right|=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

The corresponding direction cosines are \(\cos \alpha=\frac{x_2-x_1}{\left|\overrightarrow{r_2}-\vec{r}_1\right|}, \cos \beta=\frac{y_2-y_1}{\left|\vec{r}_2-\vec{r}_1\right|} \text { and } \cos \gamma=\frac{z_2-z_1}{\left|\vec{r}_2-\vec{r}_1\right|}\)

These relations are applied to determine the sum or difference of any number of vectors.

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Sum And Difference Of Two Vectors Discussions:

1. Equation (2) represents the 3-dimensional cartesian form of any vector \(\vec{r}\). However, the symbol \(\vec{r}\) is usually reserved for the position vector of a particle.

Any vector \(\vec{A}\) is represented in the system as \(\vec{A}=A_x \hat{i}+A_y \hat{j}+A_z \hat{k}\)….(7)

where A_x, A_y, and A_z stand for the x, y, and z components respectively of the vector \(\vec{A}\). Then, the magnitude of \(\vec{A}\) is

A= \(\sqrt{A_x^2+A_y^2+A_z^2}\)….(8)

The direction cosines, respectively, are l = \(\frac{A_x}{A}, m=\frac{A_y}{A} \quad \text { and } n=\frac{A_z}{A}\)…..(9)

with \(l^2+m^2+n^2=1\)….(10)

2. This is an example of the resolution of a vector \(\vec{A}\) into three components \(A_x \hat{i}, A_y \hat{j} \text { and } A_z \hat{k}\).

We dealt with coplanar vectors, and it was sufficient to resolve a vector into two components only. But, for a system of non-coplanar vectors, it is absolutely necessary to resolve each vector into three mutually perpendicular components.

3. Once every vector under consideration in a problem can be represented in the form of equation (7), vector geometry essentially transforms into vector algebra, and geometrical figures and rules are no longer necessary. Whereas vector geometry is practicable only for a few vectors, there is no limit on the number of vectors that can be handled with the help of vector algebra.

Algebraic Representation Numerical Examples

Short Answer Questions on 3D Vector Components

Example 1. The coordinates of the endpoint of a vector \(\overrightarrow{O P}\) is (4,3, -5). Express the vector in terms of its coordinates and find its absolute value.
Solution:

If the coordinates of P is (x, y, z) then, \(\overrightarrow{O P}=x \hat{i}+y \hat{j}+z \hat{k}\)

Here, x=4, y=3 and z=-5

∴ \(\overrightarrow{O P}=4 \hat{i}+3 \hat{j}-5 \hat{k}\)

and the absolute value of the vector,

OP = \(|\overrightarrow{O P}|=\sqrt{x^2+y^2+z^2}\)

= \(\sqrt{4^2+3^2+(-5)^2}\)

= \(\sqrt{16+9+25}=\sqrt{50}=5 \sqrt{2}\)

Example 2. Find the magnitude of the vector \(\vec{A}=\hat{i}-2 \hat{j}+3 \hat{k}\). Also, find the unit vector in the direction of \(\vec{A}\)
Solution:

Magnitude of \(\vec{A}\), A = \(\sqrt{1^2+(-2)^2+3^2}=\sqrt{14}\)

Unit vector in the direction of \(\vec{A}\) is \(\hat{n}=\frac{\vec{A}}{\vec{A}}=\frac{1}{\sqrt{14}} \hat{i}-\frac{2}{\sqrt{14}} \hat{j}+\frac{3}{\sqrt{14}} \hat{k}\)

Example 3. Vectors \(\vec{A}\) and \(\vec{B}\) can be expressed as \(\vec{A}=10 \hat{i}-12 \hat{j}+5 \hat{k} \text { and } \vec{B}=7 \hat{i}+8 \hat{j}-12 \hat{k}\), where \(\hat{i}\), \(\hat{j}\), k\(\hat{k}\) are unit vectors along x, y, z axes respectively. Find the resultant of the two vectors and its magnitude.
Solution:

The resultant vector \(\vec{C}=\vec{A}+\vec{B}=(10+7) \hat{i}+(-12+8) \hat{j}+(5-12) \hat{k}\)

or, \(\vec{C}=17 \hat{i}-4 \hat{j}-7 \hat{k}\)

Magnitude of the resultant, C = \(\sqrt{17^2+(-4)^2+(-7)^2}=\sqrt{354}=18.81 .\)

Example 4. Position coordinates of A and B are (-1,5,7) and (3,2,-5) respectively. Express \(\overrightarrow{A B}\) in terms of position coordinates.
Solution:

Let O be the origin, \(\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=\vec{b}\) and \(\overrightarrow{A B}=\vec{c}=\overrightarrow{O B}-\overrightarrow{O A}=\vec{b}-\vec{a}\)

Here, \(\vec{a}=-\hat{i}+5 \hat{j}+7 \hat{k} and \vec{b}=3 \hat{i}+2 \hat{j}+(-5) \hat{k}\)

∴ \(\overrightarrow{A B}=\vec{c}=\vec{b}-\vec{a}\)

= \((3 \hat{i}+2 \hat{j}-5 \hat{k})-(-\hat{i}+5 \hat{j}+7 \hat{k})\)

= \(4 \hat{i}-3 \hat{j}-12 \hat{k}\)

Example 5. \(3 \hat{i}+4 \hat{j}+12 \hat{k}\) is a vector. Find the magnitude of the vector and the angles it makes with the x, y, and z axes.
Solution:

If \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\), then the magnitude \(|\vec{r}|=\sqrt{x^2+y^2+z^2}\)

∴ Magnitude of the vector \(3 \hat{i}+4 \hat{j}+12 \hat{k}\) is,

a = \(\sqrt{3^2+4^2+12^2}=\sqrt{169}=13\)

Let the corresponding angles the vector makes with the x, y, and z axes be α, β, and γ respectively.

From definitions, \(\cos \alpha=\frac{x}{a}=\frac{3}{13} \text { or, } \alpha=\cos ^{-1} \frac{3}{13}\)

and \(\cos \beta=\frac{y}{a}=\frac{4}{13} \text { or, } \beta=\cos ^{-1} \frac{4}{13}\)

and \(\cos \gamma=\frac{z}{a}=\frac{12}{13} \text { or, } \gamma=\cos ^{-1} \frac{12}{13}\)

Mathematical Formulation for 3D Vector Resolution

Example 6. Two vectors \(\vec{A}\) and \(\vec{B}\) are \(\vec{A}=5 \hat{i}+3 \hat{j}-4 \hat{k}\) and \(\vec{B}=5 \hat{i}+2 \hat{j}+4 \hat{k}\). Find the unit vectors along \(\vec{A}+\vec{B}\) and \(\vec{A}-\vec{B}\).
Solution:

⇒ \(\vec{A}=5 \hat{i}+3 \hat{j}-4 \hat{k} and \vec{B}=5 \hat{i}+2 \hat{j}+4 \hat{k}\)

∴ \(\vec{A}+\vec{B}=(5+5) \hat{i}+(3+2) \hat{j}+(-4+4) \hat{k}=10 \hat{i}+5 \hat{j}\)

and \(|\vec{A}+\vec{B}|=\sqrt{10^2+5^2}=5 \sqrt{5}\)

Hence, unit vector along \(\vec{A}+\vec{B}=\frac{\vec{A}+\vec{B}}{|\vec{A}+\vec{B}|}=\frac{2}{\sqrt{5}} \hat{i}+\frac{1}{\sqrt{5}} \hat{j}\)

Similarly, \(\vec{A}-\vec{B}=\hat{j}-8 \hat{k}\)

and \(|\vec{A}-\vec{B}|=\sqrt{1^2+(-8)^2}=\sqrt{65}\)

and unit vector along \(\vec{A}-\vec{B}=\frac{\vec{A}-\vec{B}}{|\vec{A}-\vec{B}|}=\frac{1}{\sqrt{65}} \hat{j}-\frac{8}{\sqrt{65}} \hat{k}\).

Example 7. Two velocities \(\vec{v}_1\) and \(\vec{v}_2\) are 3 m/s towards north and 4 m/s towards east, respectively. Find \(\vec{v}_1-\vec{v}_2\).
Solution:

Let us choose, the x-axis along the east and the y-axis along the north.

∴ \(\vec{v}_1=3 \hat{j} \mathrm{~m} / \mathrm{s}\) and \(\vec{v}_2=4 \hat{i} \mathrm{~m} / \mathrm{s}\)

∴ \(\vec{v}_1-\vec{v}_2=-4 \hat{i}+3 \hat{j}\) (between west and north)

∴ \(\left|\vec{v}_1-\vec{v}_2\right|=\sqrt{(-4)^2+3^2}=5 \mathrm{~m} / \mathrm{s}\)

If \(\vec{\nu}_1-\vec{v}_2\) makes an angle θ with east, then \(\tan \theta=\frac{3}{-4}=\tan 143.1^{\circ}=\tan \left(180^{\circ}-36.9^{\circ}\right)\)

∴ \(\vec{v}_1-\vec{v}_2\) is inclined at an angle of 143.1^{\circ} north of east, i.e., \(36.9^{\circ}\) north of west.

Example 8. Find out the resultant of the following three displacement vectors: \(\vec{A}\) = 10 m, along north-west; \(\vec{B}\) = 20 m, 30° north of east; \(\vec{C}\) = 35 m, along the south.
Solution:

Let us choose, x-axis along the east and the y-axis along the north. Taking into account the x- and y-components of the given vectors, they can be written as (in meter),

⇒ \(\vec{A}=\hat{i}\left(-10 \cos 45^{\circ}\right)+\hat{j} 10 \sin 45^{\circ}=-5 \sqrt{2} \hat{i}+5 \sqrt{2} \hat{j}\)

⇒ \(\vec{B}=\hat{i} 20 \cos 30^{\circ}+\hat{j} 20 \sin 30^{\circ}=10 \sqrt{3} \hat{i}+10 \hat{j}\)

⇒ \(\vec{C}=-35 \hat{j}\)

∴ Resultant, \(\vec{R}=\vec{A}+\vec{B}+\vec{C}\)

= \(\hat{i}(-5 \sqrt{2}+10 \sqrt{3})+\hat{j}(5 \sqrt{2}+10-35)\)

= \(\hat{i} 5(2 \sqrt{3}-\sqrt{2})-\hat{j} 5(5-\sqrt{2})\) [between east and south]

∴ R = \(|\vec{R}|=\sqrt{\{5(2 \sqrt{3}-\sqrt{2})\}^2+\{-5(5-\sqrt{2})\}^2}\)

= \(5 \sqrt{(2 \sqrt{3}-\sqrt{2})^2+(5-\sqrt{2})^2}=20.65 \mathrm{~m}\)

If \(\vec{R}\) makes an angle θ with the x-axis, then \(\tan \theta=\frac{R_y}{R_x}=\frac{-5(5-\sqrt{2})}{5(2 \sqrt{3}-\sqrt{2})}=-1.75=\tan \left(-60.2^{\circ}\right)\)

So, \(\vec{R}\) is inclined at \(60.2^{\circ}\) south of east.

Example 9. The characteristic equations of a particle moving in a curved path are x = e^-t, y = 2 cos 3t, and z = 2sin3t, where t stands for time. Find out

1. velocity and acceleration at any instant,
2. velocity and acceleration at t = 0.

Solution:

Position vector of the particle at any instant, \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}=e^{-t} \hat{i}+2 \cos 3 \hat{t}+2 \sin 3 t \hat{k}\)

1. Velocity, \(\vec{v}=\frac{d \vec{r}}{d t}=-e^{-t} \hat{i}-6 \sin 3 t \hat{j}+6 \cos 3 t \hat{k}\)

= \(-x \hat{i}-3 z \hat{j}+3 y \hat{k}\)

Acceleration, \(\vec{a}=\frac{d \vec{v}}{d t}=e^{-t} \hat{i}-18 \cos 3 t \hat{j}-18 \sin 3 t \hat{k}\)

= \(x \hat{i}-9 y \hat{j}-9 z \hat{k}\)

2. At t = 0, \(\vec{v}=-e^0 \hat{i}-6 \sin (3 \times 0) \hat{j}+6 \cos (3 \times 0) \hat{k}\)

= \(-\hat{i}-6 \times 0 \hat{j}+6 \times 1 \hat{k}\)

= \(-\hat{i}+6 \hat{k}\)

v = \(|\vec{v}|=\sqrt{(-1)^2+6^2}=\sqrt{37}\)

At t = \(0, \vec{a}=e^0 \hat{i}-18 \cos (3 \times 0) \hat{j}-18 \sin (3 \times 0) \hat{k}\)

= \(\hat{i}-(18 \times 1) \hat{j}-(18 \times 0) \hat{k}\)

= \(\hat{i}-18 \hat{j}\)

a = \(|\vec{a}|=\sqrt{1^2+(-18)^2}=\sqrt{325}=5 \sqrt{13}\)

Real-Life Examples of Three Dimensional Vectors

Example 10. The position vector \(\vec{r}\) of a particle with respect to the origin changes with time t as \(\vec{r}=A t \hat{i}-B t^2 \hat{j}\), where A and B are positive constants. Determine

1. The locus of the particle,
2. The nature of variation with time of the velocity and acceleration vectors, and also the moduli of them.

Solution:

1. \(\vec{r}=A t \hat{i}-B t^2 \hat{j}=x i+y \hat{j}\)

x = \(A t \quad \text { and } y=-B t^2\)

Then, \(x^2=A^2 t^2 or, t^2=\frac{x^2}{A^2}\)

also, \(t^2=-\frac{y}{B}\)

∴ \(\frac{x^2}{A^2}=-\frac{y}{B}\) or, \(x^2=-\frac{A^2}{B} y\)

This is the locus of the particle, which is a parabola.

2. Velocity, \(\vec{v}=\frac{d \vec{r}}{d t}=A \hat{i}-2 B t \hat{j}\)

Its modulus, \(|\vec{v}|=\sqrt{A^2+(-2 B t)^2}=\sqrt{A^2+4 B^2 t^2}\)

Acceleration, \(\vec{a}=\frac{d \vec{v}}{d t}=-2 B \hat{j}\)

Its modulus, \(|\vec{a}|=\sqrt{(-2 B)^2}=2 B\)

Example 11. The position vector of a particle is, \(\vec{r}=3 t \hat{i}-2 t^2 \hat{j}+4 \hat{k}\). Find out

1. Its velocity \(\vec{v}\) and acceleration \(\vec{a}\),
2. The magnitude and direction of its velocity at t = 2 s.

Solution:

∴ \(\vec{r}=3 t \hat{i}-2 t^2 \hat{j}+4 \hat{k}\)

1. \(\vec{v}=\frac{d \vec{r}}{d t}=3 \hat{i}-4 t \hat{j}\)

∴ \(\vec{a}=\frac{d \vec{v}}{d t}=-4 \hat{j}\)

2. At t = \(2 \mathrm{~s} \quad \vec{v}=3 \hat{i}-(4 \times 2) \hat{j}=3 \hat{i}-8 \hat{j}\) (between the x-axis and negative y-axis)

∴ \(|\vec{v}|=\sqrt{3^2+(-8)^2}=\sqrt{73} \text { unit }\)

Inclination of \(\vec{v}\) with the x-axis, \(\theta=\tan ^{-1}\left(\frac{-8}{3}\right)=-69.4^{\circ}\)