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,
- \(\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}\) and
- \(\cos A=\frac{b^2+c^2-a^2}{2 b c}\).
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, \(\frac{A}{O P}+\frac{B}{O Q}=\frac{R}{O S}\).