Differentiation (Part-38) | S N De

In this article, we will discuss the solutions of Very Short Answer (VSA) Type Questions in the chapter Product of Two Vectors (Ex-2A) as given in the Chhaya Publication Book by S.N. Dey. In the previous article, we discussed basic vector concepts.

Vector Product Solutions: S.N. Dey Class 12 (Ex-2A)

1. Define scalar product of two vectors. Show that the scalar product of vectors satisfies the commutative and distributive laws.

Definition:

The scalar product (or dot product) of two vectors $\vec{a}$ and $\vec{b}$ is denoted by $\vec{a} \cdot \vec{b}$. It is a scalar and is defined as:

$$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$$

where $\theta$ is the angle between vectors $\vec{a}$ and $\vec{b}$, and $|\vec{a}|, |\vec{b}|$ are the moduli of vectors $\vec{a}$ and $\vec{b}$ respectively.

Since $\cos\theta > 0$ when $\theta$ is acute and $\cos\theta < 0$ when $\theta$ is obtuse, the scalar product $\vec{a} \cdot \vec{b}$ is positive when $\theta$ is acute and negative when $\theta$ is obtuse.

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Proof 1: Scalar product is commutative

We know that:

$$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$$

$$\vec{b} \cdot \vec{a} = |\vec{b}| |\vec{a}| \cos\theta$$

Since $|\vec{a}| |\vec{b}| = |\vec{b}| |\vec{a}|$, it follows that:

$$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$$

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Proof 2: Scalar product is distributive with respect to vector addition

Let $\vec{OA} = \vec{a}$, $\vec{OB} = \vec{b}$, and $\vec{BC} = \vec{c}$, where $|\vec{a}| = \overline{OA}$, $|\vec{b}| = \overline{OB}$, and $|\vec{c}| = \overline{BC}$.

Fig 1. Distributive property of vector scalar product

From points $B$ and $C$, draw perpendiculars $BM$ and $CN$ onto $OA$. By the triangle law of addition:

$$\vec{b} + \vec{c} = \vec{OB} + \vec{BC} = \vec{OC}$$

Therefore:

$$\vec{a} \cdot (\vec{b} + \vec{c}) = |\vec{a}| \times (\text{scalar component of } (\vec{b} + \vec{c}) \text{ along } \vec{a})$$

$$= |\vec{a}| \overline{ON} = |\vec{a}| (\overline{OM} + \overline{MN})$$

$$= |\vec{a}| \overline{OM} + |\vec{a}| \overline{MN}$$

$$= |\vec{a}|(\text{scalar component of } \vec{b} \text{ along } \vec{a}) + |\vec{a}|(\text{scalar component of } \vec{c} \text{ along } \vec{a})$$

$$= \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$$

Hence, distributive law $\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$ is proved.

2. Find the scalar product of the following pair of vectors and the angle between them:

(i) $\hat{i} + \hat{j}$ and $\hat{j} + \hat{k}$

Solution:

$$(\hat{i} + \hat{j}) \cdot (\hat{j} + \hat{k}) = \hat{i}\cdot\hat{j} + \hat{i}\cdot\hat{k} + \hat{j}\cdot\hat{j} + \hat{j}\cdot\hat{k} = 0 + 0 + 1 + 0 = 1 \quad \text{--- (1)}$$

If $\theta$ is the angle between the vectors:

$$\cos\theta = \frac{(\hat{i} + \hat{j}) \cdot (\hat{j} + \hat{k})}{|\hat{i} + \hat{j}| |\hat{j} + \hat{k}|} = \frac{1}{\sqrt{1^2 + 1^2} \sqrt{1^2 + 1^2}} = \frac{1}{\sqrt{2}\sqrt{2}} = \frac{1}{2}$$

$$\implies \theta = \frac{\pi}{3}$$

Scalar Product = $1$, Angle $\theta = \frac{\pi}{3}$

(ii) $2\hat{i} - \hat{j} + 2\hat{k}$ and $\hat{i} + \hat{j} + \hat{k}$

Solution:

$$(2\hat{i} - \hat{j} + 2\hat{k}) \cdot (\hat{i} + \hat{j} + \hat{k}) = 2(1) + (-1)(1) + 2(1) = 3 \quad \text{--- (1)}$$

If $\theta$ is the angle between the vectors:

$$\cos\theta = \frac{3}{\sqrt{2^2 + (-1)^2 + 2^2} \sqrt{1^2 + 1^2 + 1^2}} = \frac{3}{\sqrt{9}\sqrt{3}} = \frac{3}{3\sqrt{3}} = \frac{1}{\sqrt{3}}$$

$$\implies \theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$$

Scalar Product = $3$, Angle $\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right)$

(iii) $\vec{a} = 2\hat{i} + 3\hat{j} - 4\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$

Solution:

$$\vec{a} \cdot \vec{b} = 2(1) + 3(2) + (-4)(1) = 2 + 6 - 4 = 4 \quad \text{--- (1)}$$

Magnitudes:

$$|\vec{a}| = \sqrt{2^2 + 3^2 + (-4)^2} = \sqrt{29}$$

$$|\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}$$

Angle calculation:

$$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = \frac{4}{\sqrt{29}\sqrt{6}} = \frac{4}{\sqrt{174}}$$

Scalar Product = $4$, Angle $\theta = \cos^{-1}\left(\frac{4}{\sqrt{174}}\right)$

(iv) $\vec{a} = 2\hat{i} - 5\hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} - 2\hat{j} - 4\hat{k}$

Solution:

$$\vec{a} \cdot \vec{b} = 2(1) + (-5)(-2) + 3(-4) = 2 + 10 - 12 = 0$$

$$\cos\theta = \frac{0}{|\vec{a}||\vec{b}|} = 0 \implies \theta = \frac{\pi}{2}$$

Scalar Product = $0$, Angle $\theta = \frac{\pi}{2}$

3. Show that the following pairs of vectors are perpendicular to each other:

(i) $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} + 3\hat{k}$

Solution:

Vectors are perpendicular if $\vec{a} \cdot \vec{b} = 0$.

$$\vec{a} \cdot \vec{b} = (2\hat{i} + \hat{j} - \hat{k}) \cdot (\hat{i} + \hat{j} + 3\hat{k}) = 2(1) + 1(1) + (-1)(3) = 2 + 1 - 3 = 0$$

Since $\vec{a} \cdot \vec{b} = 0$, the vectors are perpendicular.

(ii) $\vec{a} = (3, 2, -1)$ and $\vec{b} = (-2, 0, -6)$

Solution:

Writing in component form: $\vec{a} = 3\hat{i} + 2\hat{j} - \hat{k}$, $\vec{b} = -2\hat{i} - 6\hat{k}$

$$\vec{a} \cdot \vec{b} = 3(-2) + 2(0) + (-1)(-6) = -6 + 0 + 6 = 0$$

Since $\vec{a} \cdot \vec{b} = 0$, the vectors are perpendicular.

(iii) $\vec{\alpha} = -3\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{\beta} = -2\hat{i} + \hat{j} - 4\hat{k}$

Solution:

$$\vec{\alpha} \cdot \vec{\beta} = (-3)(-2) + (-2)(1) + 1(-4) = 6 - 2 - 4 = 0$$

Since $\vec{\alpha} \cdot \vec{\beta} = 0$, the vectors are perpendicular.

4. Show that the following trios of vectors are mutually perpendicular to each other:

(i) $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$, $\vec{b} = \hat{i} + \hat{j} - 3\hat{k}$, and $\vec{c} = 7\hat{i} - 4\hat{j} + \hat{k}$

Solution:

$$\vec{a} \cdot \vec{b} = 1(1) + 2(1) + 1(-3) = 1 + 2 - 3 = 0$$

$$\vec{b} \cdot \vec{c} = 1(7) + 1(-4) + (-3)(1) = 7 - 4 - 3 = 0$$

$$\vec{c} \cdot \vec{a} = 7(1) + (-4)(2) + 1(1) = 7 - 8 + 1 = 0$$

Since $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 0$, all 3 vectors are mutually perpendicular.

(ii) $\vec{a} = 2\hat{i} + 3\hat{j} + 6\hat{k}$, $\vec{b} = 3\hat{i} - 6\hat{j} + 2\hat{k}$, and $\vec{c} = 6\hat{i} + 2\hat{j} - 3\hat{k}$

Solution:

$$\vec{a} \cdot \vec{b} = 2(3) + 3(-6) + 6(2) = 6 - 18 + 12 = 0$$

$$\vec{b} \cdot \vec{c} = 3(6) + (-6)(2) + 2(-3) = 18 - 12 - 6 = 0$$

$$\vec{c} \cdot \vec{a} = 6(2) + 2(3) + (-3)(6) = 12 + 6 - 18 = 0$$

Since $\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{c} = \vec{c}\cdot\vec{a} = 0$, all 3 vectors are mutually perpendicular.

5. In each of the following cases, two vectors are perpendicular to each other. Find $m$:

(i) $m\hat{i} - 2\hat{j} + \hat{k}$ and $3\hat{i} - 2\hat{j} - 7\hat{k}$

Solution:

Set dot product to zero:

$$(m\hat{i} - 2\hat{j} + \hat{k}) \cdot (3\hat{i} - 2\hat{j} - 7\hat{k}) = 0$$

$$\implies 3m + (-2)(-2) + 1(-7) = 0 \implies 3m + 4 - 7 = 0$$

$$\implies 3m = 3 \implies m = 1$$

$m = 1$

(ii) $\vec{a} = 2\hat{j} - 5\hat{k}$ and $\vec{b} = -3\hat{i} - m\hat{j} + 4\hat{k}$

Solution:

$$\vec{a} \cdot \vec{b} = 0 \implies 0(-3) + 2(-m) + (-5)(4) = 0$$

$$\implies -2m - 20 = 0 \implies -2m = 20 \implies m = -10$$

$m = -10$

(iii) $\vec{a} = (1, 1, m)$ and $\vec{b} = (4, m, -3)$

Solution:

$$\vec{a} \cdot \vec{b} = 0 \implies 1(4) + 1(m) + m(-3) = 0$$

$$\implies 4 + m - 3m = 0 \implies 4 - 2m = 0 \implies m = 2$$

$m = 2$

6. If $\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{b} = 3\hat{i} - \hat{j} + 2\hat{k}$, show that $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ are perpendicular to each other.

Solution:

$$\vec{a} + \vec{b} = (\hat{i} + 2\hat{j} - 3\hat{k}) + (3\hat{i} - \hat{j} + 2\hat{k}) = 4\hat{i} + \hat{j} - \hat{k} \quad \text{--- (1)}$$

$$\vec{a} - \vec{b} = (\hat{i} + 2\hat{j} - 3\hat{k}) - (3\hat{i} - \hat{j} + 2\hat{k}) = -2\hat{i} + 3\hat{j} - 5\hat{k} \quad \text{--- (2)}$$

Taking scalar product:

$$(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 4(-2) + 1(3) + (-1)(-5) = -8 + 3 + 5 = 0$$

Since $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0$, they are perpendicular.

7. If $\vec{a} = 3\hat{i} + 2\hat{j} + 9\hat{k}$ and $\vec{b} = \hat{i} + \lambda\hat{j} + 3\hat{k}$, find $\lambda$ so that $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$ are perpendicular to each other.

Solution:

$$\vec{a} + \vec{b} = 4\hat{i} + (2 + \lambda)\hat{j} + 12\hat{k}$$

$$\vec{a} - \vec{b} = 2\hat{i} + (2 - \lambda)\hat{j} + 6\hat{k}$$

Since $(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 0$:

$$4(2) + (2 + \lambda)(2 - \lambda) + 12(6) = 0$$

$$8 + (4 - \lambda^2) + 72 = 0 \implies 84 - \lambda^2 = 0$$

$$\lambda^2 = 84 \implies \lambda = \pm\sqrt{84} = \pm 2\sqrt{21}$$

$\lambda = \pm 2\sqrt{21}$

8. If $\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} + \hat{j} - 2\hat{k}$, find the angle between $(\vec{a} + \vec{b})$ and $(\vec{a} - \vec{b})$.

Solution:

$$\vec{a} + \vec{b} = 5\hat{i} + 0\hat{j} + \hat{k} \implies |\vec{a} + \vec{b}| = \sqrt{5^2 + 1^2} = \sqrt{26}$$

$$\vec{a} - \vec{b} = -\hat{i} - 2\hat{j} + 5\hat{k} \implies |\vec{a} - \vec{b}| = \sqrt{(-1)^2 + (-2)^2 + 5^2} = \sqrt{30}$$

Dot product:

$$(\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = 5(-1) + 0(-2) + 1(5) = -5 + 0 + 5 = 0$$

$$\cos\theta = \frac{0}{\sqrt{26}\sqrt{30}} = 0 \implies \theta = \frac{\pi}{2}$$

Angle $\theta = \frac{\pi}{2}$

9. Find the scalar and vector components of $\vec{a}$ in the direction of $\vec{b}$:

(i) $\vec{a} = \hat{i} + \hat{j}$ and $\vec{b} = \hat{j} + \hat{k}$

Solution:

$$\vec{a} \cdot \vec{b} = 0(0) + 1(1) + 1(1) = 1, \quad |\vec{b}| = \sqrt{1^2 + 1^2} = \sqrt{2}$$

Scalar component of $\vec{a}$ along $\vec{b}$:

$$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{1}{\sqrt{2}}$$

Vector component of $\vec{a}$ along $\vec{b}$:

$$\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b} = \frac{1}{(\sqrt{2})^2}(\hat{j} + \hat{k}) = \frac{1}{2}(\hat{j} + \hat{k})$$

Scalar Component = $\frac{1}{\sqrt{2}}$, Vector Component = $\frac{1}{2}(\hat{j} + \hat{k})$

(ii) $\vec{a} = 3\hat{i} + \hat{j} + 3\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} - \hat{k}$

Solution:

$$\vec{a} \cdot \vec{b} = 3(1) + 1(-1) + 3(-1) = 3 - 1 - 3 = -1$$

$$|\vec{b}| = \sqrt{1^2 + (-1)^2 + (-1)^2} = \sqrt{3}$$

Scalar component of $\vec{a}$ along $\vec{b}$:

$$\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{-1}{\sqrt{3}}$$

Vector component of $\vec{a}$ along $\vec{b}$:

$$\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b} = \frac{-1}{3}(\hat{i} - \hat{j} - \hat{k}) = -\frac{1}{3}(\hat{i} - \hat{j} - \hat{k})$$

Scalar Component = $-\frac{1}{\sqrt{2}}$, Vector Component = $-\frac{1}{3}(\hat{i} - \hat{j} - \hat{k})$

10. Find the scalar and vector projection of $\vec{b}$ on $\vec{a}$ where $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{j} + 2\hat{k}$.

Solution:

$$\vec{a} \cdot \vec{b} = 1(0) + 2(1) + 2(2) = 0 + 2 + 4 = 6$$

$$|\vec{a}| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3$$

Scalar projection of $\vec{b}$ on $\vec{a}$:

$$\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|} = \frac{6}{3} = 2$$

Vector projection of $\vec{b}$ on $\vec{a}$:

$$\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right)\vec{a} = \frac{6}{3^2}(\hat{i} + 2\hat{j} + 2\hat{k}) = \frac{2}{3}(\hat{i} + 2\hat{j} + 2\hat{k})$$

Scalar Projection = $2$, Vector Projection = $\frac{2}{3}(\hat{i} + 2\hat{j} + 2\hat{k})$

Related Solutions:

• S.N. Dey Solutions: Vector Algebra (Part 4)