WBCHSE Class 11 Physics Notes For Subtraction Of Two Vectors

Subtraction Of Two Vectors

The general rule for subtraction in algebra can be represented in two ways: 5-3 = 2 and 5 + (-3) = 2. This implies that the result of subtraction of a positive quantity from a number is the same as the result of adding a negative quantity of the same value to the number. This rule is followed in vector subtraction too.

If the difference between two vectors \(\vec{a}\) and \(\vec{b}\) is \(\vec{d}\), then \(\vec{a}\) – \(\vec{b}\) = \(\vec{d}\) and also \(\vec{a}\) + (-\(\vec{b}\)) = \(\vec{d}\), where –\(\vec{b}\) is the opposite vector of \(\vec{b}\)

Line segments OA and OB represent the magnitudes and directions of the vectors \(\vec{a}\) and \(\vec{b}\) respectively. BO is extended up to D, so that OD = BO.

This makes \(\overrightarrow{O D}\) and \(\overrightarrow{O B}\) two opposite vectors i.e., \(\overrightarrow{O D}\)  = –\(\overrightarrow{O B}\) = -b .

In the parallelogram OAED, the diagonal \(\overrightarrow{O E}\) gives the resultant of \(\vec{a}\) and –\(\vec{b}\) which is the difference of \(\vec{a}\) and \(\vec{b}\).

Hence, \(\overrightarrow{O A}+\overrightarrow{O D}=\overrightarrow{O E}\)

or, \(\vec{a}+(-\vec{b})=\vec{d}\)

or, \(\vec{a}-\vec{b}=\vec{d}\)

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Relation Between The Resultant And The Difference Of Two Vectors: The diagonal \(\overrightarrow{O F}\) of the parallelogram OAFB represents the resultant of two vectors \(\vec{a}\) and \(\vec{a}\).

Now parallelograms OAFB and OAED are congruent. Therefore, it is evident that \(\overrightarrow{O E}\) and \(\overrightarrow{O F}\) can be represented by the diagonals (\(\overrightarrow{B A}\) and \(\overrightarrow{O F}\)) of the same parallelogram.

Hence, we may conclude that, if a diagonal of a parallelogram represents the resultant of two vectors, the other diagonal would represent the difference between them, with proper assignment of directions.

Relation Between The Resultant And The Difference Of Two Vectors Special Cases:

1. When the angle between the two vectors \(\vec{a}\) and \(\vec{b}\) is 90° i.e., they are a pair of orthogonal vectors, the parallelograms change into rectangles, for which the diagonals are equal and make equal angles with the base vector \(\vec{a}\).

Hence, the resultant and the difference of these two vectors are equal in magnitude and they are inclined at the same angle as the base vector but in opposite directions.

Analytically, \(\left.\begin{array}{l}
c^2=a^2+b^2 \text { or, } c=\sqrt{a^2+b^2} \\
d^2=a^2+b^2 \text { or, } d=\sqrt{a^2+b^2}
\end{array}\right\}\)….(1)

⇒ \(\left.\begin{array}{r}
\tan \theta=\frac{b}{a} \\
\tan \theta^{\prime}=-\frac{b}{a}
\end{array}\right\}\)…..(2)

∴ θ = θ’

2. For two orthogonal vectors of equal value, i.e., when a = b, the parallelogram changes into a square for which the diagonals are of equal length and are mutu¬ally perpendicular. In this case, the resultant and the difference of the two vectors are of equal magnitude; they are mutually orthogonal as well.

Putting a = b in equation (2), tanθ =1 or, θ = 45° and tanθ’ = -1 or, θ = -45°

Hence, the angle between the resultant and the difference of the two vectors =45° + 45° = 90°.

Zero Of Null Vector: A vector having a magnitude zero (0) and no fixed direction is a null vector. If the initial and the terminal points of a line segment representing a vector coincide, then the vector is called a zero or null vector.

Indispensability Of The Use Of A Null Vector: if a particle starting from a point returns to the same point after some time, its displacement, which is a vector quantity, is a zero vector. Also, two equal and opposite forces acting on a body have a zero resultant, which is a zero vector \(\vec{0}\), i.e., \(\vec{F}-\vec{F}=\overrightarrow{0}\). Hence the difference of two equal vectors is a zero vector.

Again, the acceleration of a particle moving with uniform velocity is zero. This is also another example of a zero or null vector, as acceleration is a vector quantity. Hence, the use of zero or null vectors is indispensable in vector algebra.

The following two operations give rise to a zero vector.

1. When a vector is multiplied by zero, the result is a zero vector. Thus, \(0(\vec{A})=\overrightarrow{0}.\)
2. When the negative of a vector is added to the vector itself, the result is a zero vector, thus \(\vec{A}+(-\vec{A})=\overrightarrow{0}\).

Properties Of Zero Or Null Vector:

1. The addition or subtraction of a zero vector from a vector results in the same vector. Thus \(\vec{A} \pm \overrightarrow{0}=\vec{A}.\)
2. The multiplication of a non-zero real number with a zero vector is again a zero vector. If n is a non-zero real number, then \(n(\overrightarrow{0})=\overrightarrow{0}.\)
3. If n₁ and n₂ are two different non-zero real numbers, then the relation \(n_1 \vec{A}=n_2 \vec{B}\) can hold only if both \(\vec{A}\) and \(\vec{B}\) are zero vector.

Position And Displacement Vectors: In the chapter One-dimensional Motion, we defined the position and the displacement vectors as two important properties related to the motion of a particle. Here, we shall discuss some important points about them vectorically.

Position vector Definition: A vector used to specify the position of a point with reference to the origin of the coordinate system is called a position vector.

Consider a point P having coordinates (x, y, z). If O is the origin then \(\overrightarrow{O P}\) is called the position vector, \(\vec{r}\). The distance between the origin and the point gives the magnitude of the position vector. Here, \(\overrightarrow{O P}=\vec{r}\) (position vector).

Thus, \(\vec{r}=x \hat{i}+y \hat{j}+z \hat{k}\)]

Hence, magnitude of \(\vec{r}\) is given by \(|\vec{r}|=\sqrt{x^2+y^2+z^2}\)

Displacement Definition: The vector connecting the initial and final positions of a particle is called the displacement vector.

Suppose a particle is moved from point A(x₁, y₁, z₁) to another point B(x₂, y₂, z₂).

The shortest distance between the points A and B is the displacement. So \(\overrightarrow{A B}\) represents the displacement vector, \(\vec{r}\) as shown.

Applying triangle law of vectors to the ΔOAB we get, \(\overrightarrow{O A}+\overrightarrow{A B}=\overrightarrow{O B} \quad \text { or, } \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A}=\vec{r}_2-\vec{r}_1=\vec{s}\)

The length of the straight line AB gives the magnitude of the displacement vector. The direction of the vector is along the direction of motion of the particle.

With respect to the origin O, \(\vec{r}_1=x_1 \hat{i}+y_1 \hat{j}+z_1 \hat{k}\) = initial position vector

and \(\vec{r}_2=x_2 \hat{i}+y_2 \hat{j}+z_2 \hat{k}\) = final position vector

∴ \(\overrightarrow{A B}=\vec{s}=\left(\vec{r}_2-\vec{r}_1\right)=\left(x_2-x_1\right) \hat{i}+\left(y_2-y_1\right) \hat{j}+\left(z_2-z_1\right) \hat{k}\)

The magnitude of the displacement vector is independent of the choice of origin of the cartesian coordinate system.

∴ \(|\overrightarrow{A B}|=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}\)

The points to be noted here are:

1. The displacement of a particle is the vector difference between its final and initial position vectors.
2. The direction of \(\overrightarrow{A B}\) is different from that of \(\overrightarrow{O A}\) and \(\overrightarrow{O B}\). So, in general, the displacement vector is directed neither along the initial nor along the final position vector of a particle. In other words, generally ‘position’ and ‘change of position’ are vectors that are not in the same direction.
3. Velocity is defined as the rate of displacement with time, i. e., the rate of change of position with time. There is nothing like a ‘change of displacement’; hence, the velocity vector always lies along the displacement vector. The average velocity of the particle during the time spent while going from A to B, is directed along \(\overrightarrow{A B}\).

Position And Displacement Vectors Numerical Examples

Example 1. A car is traveling towards the east at 10 m · s^-1. It takes 10 seconds to change its direction of motion to the north and continues with the same magnitude of velocity. Find the magnitude and direction of the average acceleration of the car.
Solution:

Change in the velocity of the car

= final velocity – initial velocity

= (10 m • s^-1 towards north) – (10 m • s^-1 towards east)

= \(\overrightarrow{A B}-\overrightarrow{O A}=\overrightarrow{A B}+\overrightarrow{A O}=\overrightarrow{A C}\) [from parallelogram law of vector addition]

∴ AC² =AB² + BC² =AB² + AO²

= 10² + 10² = 100 + 100 = 200

∴ AC = 10√2 m · s^-1

Average acceleration = \(\frac{\text { change in velocity }}{\text { time }}\)

= \(\frac{10 \sqrt{2}}{10}=\sqrt{2} \mathrm{~m} \cdot \mathrm{s}^{-2}\)

It is directed along \(\overrightarrow{A C}\) that makes angle 6 with \(\overrightarrow{A B}\), and \(\tan \theta=\frac{B C}{A B}=\frac{A O}{A B}=\frac{10}{10}=1=\tan 45^{\circ}\)

∴ \(\theta=45^{\circ}\)

Hence, average acceleration is directed towards the north-west.

Example 2. A boy runs 210 m along the corridor of his school and turns; right at the end of the corridor runs 180 m to the end of the building and then turns right and runs 30 m.

1. Construct a vector diagram that represents this motion. Indicate your choice of unit vectors,
2. What is the direction and magnitude of the straight line between start and finish?

Solution:

1. We choose, \(\hat{i}\) = unit vector along x-axis, \(\hat{j}\) = unit vector along y-axis,

The consecutive displacements are, \(\overrightarrow{A B}=210 \hat{i} \mathrm{~m}, \overrightarrow{B C}=-180 \hat{j} \mathrm{~m}, \overrightarrow{C D}=-30 \hat{i} \mathrm{~m}\)

2. Resultant, \(\overrightarrow{A D}=\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C D}=(180 \hat{i}-180 \hat{j}) \mathrm{m}\)

∴ \(\tan \theta=\frac{E D}{A E}=\frac{B C}{A E}=\frac{-180}{180}=-1\)

or, θ = 45°

So \(\overrightarrow{A D}\) is inclined at 45° with the x-axis in the negative direction.

∴ \(|\overrightarrow{A D}|=\sqrt{(180)^2+(-180)^2}=180 \sqrt{2} \mathrm{~m}\)

Example 3. A particle is moving in a circular path with a uniform speed v. Show that, when the particle traverses through an angle of 120°, the change in its velocity is √3v.
Solution:

The initial velocity \(\overrightarrow{v_1}\) and the final velocity \(\overrightarrow{v_2}\) are shown.

They are drawn from the same initial point.

Here, \(\left|\overrightarrow{v_1}\right|=\left|\vec{v}_2\right|=v\)

Change in velocity = \(\vec{v}_2-\vec{v}_1\)

∴ \(\left|\vec{v}_2-\vec{v}_1\right|=\sqrt{v_1^2+v_2^2-2 v_1 v_2 \cos 120^{\circ}}\)

= \(\sqrt{v^2+v^2-2 v \cdot v \cdot\left(-\frac{1}{2}\right)}=\sqrt{3 v^2}=\sqrt{3} \nu\)