Exercise 10.3 Class 12 Maths Vector Algebra

NCERT Solutions Class 12 Maths Chapter 10 Ex 10.3 | LearnCBSEHub

💡 Key Formulas

Dot Product: $\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta$
Angle $\theta$: $\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$
Projection of $\vec{a}$ on $\vec{b}$: $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$
Perpendicular Vectors: $\vec{a} \cdot \vec{b} = 0$

Question 01

Find the angle between two vectors with magnitudes $\sqrt{3}$ and $2$, having $\vec{a} \cdot \vec{b} = \sqrt{6}$.

$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{\sqrt{6}}{\sqrt{3} \times 2} = \frac{\sqrt{2} \times \sqrt{3}}{\sqrt{3} \times 2} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}$.
Since $\cos \theta = \frac{1}{\sqrt{2}}$, $\theta = \frac{\pi}{4}$.

Angle: $\pi/4$

Question 02

Find angle between vectors $\hat{i} – 2\hat{j} + 3\hat{k}$ and $3\hat{i} – 2\hat{j} + \hat{k}$.

Let $\vec{a} = \hat{i} – 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} – 2\hat{j} + \hat{k}$.
$\vec{a} \cdot \vec{b} = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10$.
$|\vec{a}| = \sqrt{1+4+9} = \sqrt{14}$. $|\vec{b}| = \sqrt{9+4+1} = \sqrt{14}$.
$\cos \theta = \frac{10}{\sqrt{14} \cdot \sqrt{14}} = \frac{10}{14} = \frac{5}{7}$.

Angle: $\cos^{-1}(5/7)$

Question 03

Find projection of $\hat{i} – \hat{j}$ on $\hat{i} + \hat{j}$.

Let $\vec{a} = \hat{i} – \hat{j}$ and $\vec{b} = \hat{i} + \hat{j}$.
Projection $= \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \frac{(1)(1) + (-1)(1)}{\sqrt{1^2 + 1^2}} = \frac{1-1}{\sqrt{2}} = 0$.

Projection: 0

Question 04

Find projection of $\hat{i} + 3\hat{j} + 7\hat{k}$ on $7\hat{i} – \hat{j} + 8\hat{k}$.

Let $\vec{a} = \hat{i} + 3\hat{j} + 7\hat{k}$ and $\vec{b} = 7\hat{i} – \hat{j} + 8\hat{k}$.
$\vec{a} \cdot \vec{b} = 1(7) + 3(-1) + 7(8) = 7 – 3 + 56 = 60$.
$|\vec{b}| = \sqrt{49 + 1 + 64} = \sqrt{114}$.
Projection $= \frac{60}{\sqrt{114}}$.

Question 05

Show given vectors are unit vectors and mutually perpendicular.

Let $\vec{a} = \frac{1}{7}(2\hat{i} + 3\hat{j} + 6\hat{k})$.
$|\vec{a}| = \frac{1}{7}\sqrt{4+9+36} = \frac{1}{7}\sqrt{49} = 1$. (Unit Vector)
Similarly verify for $\vec{b}$ and $\vec{c}$.
Check orthogonality: $\vec{a} \cdot \vec{b} = \frac{1}{49}(2(3) + 3(-6) + 6(2)) = \frac{1}{49}(6-18+12) = 0$.
Similarly $\vec{b} \cdot \vec{c} = 0$ and $\vec{c} \cdot \vec{a} = 0$.

Verified

Question 06

Find $|\vec{a}|$ and $|\vec{b}|$ if $(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = 8$ and $|\vec{a}| = 8|\vec{b}|$.

$(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b}) = |\vec{a}|^2 – |\vec{b}|^2 = 8$.
Substitute $|\vec{a}| = 8|\vec{b}|$: $(8|\vec{b}|)^2 – |\vec{b}|^2 = 8 \implies 64|\vec{b}|^2 – |\vec{b}|^2 = 8$.
$63|\vec{b}|^2 = 8 \implies |\vec{b}| = \sqrt{\frac{8}{63}}$.
$|\vec{a}| = 8\sqrt{\frac{8}{63}}$.

Question 07

Evaluate $(3\vec{a} – 5\vec{b}) \cdot (2\vec{a} + 7\vec{b})$.

Use distributive law:
$= 3\vec{a} \cdot 2\vec{a} + 3\vec{a} \cdot 7\vec{b} – 5\vec{b} \cdot 2\vec{a} – 5\vec{b} \cdot 7\vec{b}$
$= 6|\vec{a}|^2 + 21(\vec{a} \cdot \vec{b}) – 10(\vec{a} \cdot \vec{b}) – 35|\vec{b}|^2$
$= 6|\vec{a}|^2 + 11(\vec{a} \cdot \vec{b}) – 35|\vec{b}|^2$.

Question 08

Find magnitude of two vectors having same magnitude, angle $60^\circ$, scalar product $1/2$.

Let $|\vec{a}| = |\vec{b}| = x$.
$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos 60^\circ \implies \frac{1}{2} = x \cdot x \cdot \frac{1}{2}$.
$x^2 = 1 \implies x = 1$.

Magnitude: 1

Question 09

Find $|\vec{x}|$ if for unit vector $\vec{a}$, $(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a}) = 12$.

$(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a}) = |\vec{x}|^2 – |\vec{a}|^2 = 12$.
Since $\vec{a}$ is unit vector, $|\vec{a}| = 1$.
$|\vec{x}|^2 – 1 = 12 \implies |\vec{x}|^2 = 13 \implies |\vec{x}| = \sqrt{13}$.

Question 10

Find $\lambda$ if $\vec{a} + \lambda\vec{b}$ is perpendicular to $\vec{c}$.

$\vec{a} = 2\hat{i}+2\hat{j}+3\hat{k}, \quad \vec{b} = -\hat{i}+2\hat{j}+\hat{k}, \quad \vec{c} = 3\hat{i}+\hat{j}$.
$\vec{a} + \lambda\vec{b} = (2-\lambda)\hat{i} + (2+2\lambda)\hat{j} + (3+\lambda)\hat{k}$.
Dot with $\vec{c}$ is 0:
$(2-\lambda)(3) + (2+2\lambda)(1) + (3+\lambda)(0) = 0$
$6 – 3\lambda + 2 + 2\lambda = 0 \implies 8 – \lambda = 0 \implies \lambda = 8$.

Question 11

Show $|\vec{a}|\vec{b} + |\vec{b}|\vec{a}$ is perpendicular to $|\vec{a}|\vec{b} – |\vec{b}|\vec{a}$.

Dot product:
$(|\vec{a}|\vec{b} + |\vec{b}|\vec{a}) \cdot (|\vec{a}|\vec{b} – |\vec{b}|\vec{a})$
$= |\vec{a}|^2 (\vec{b}\cdot\vec{b}) – |\vec{a}||\vec{b}|(\vec{b}\cdot\vec{a}) + |\vec{b}||\vec{a}|(\vec{a}\cdot\vec{b}) – |\vec{b}|^2(\vec{a}\cdot\vec{a})$
$= |\vec{a}|^2 |\vec{b}|^2 – |\vec{b}|^2 |\vec{a}|^2 = 0$. (Since vectors commute).

Verified

Question 12

If $\vec{a}\cdot\vec{a}=0$ and $\vec{a}\cdot\vec{b}=0$, conclude about $\vec{b}$.

$\vec{a}\cdot\vec{a}=0 \implies |\vec{a}|^2=0 \implies |\vec{a}|=0 \implies \vec{a}=\vec{0}$.

If $\vec{a}$ is the zero vector, then $\vec{a}\cdot\vec{b}=0$ is true for any vector $\vec{b}$.

Question 13

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors and $\vec{a}+\vec{b}+\vec{c}=\vec{0}$, find $\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}$.

$|\vec{a}+\vec{b}+\vec{c}|^2 = 0$. Expand:
$|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{c} + \vec{c}\cdot\vec{a}) = 0$
$1 + 1 + 1 + 2(\text{Value}) = 0 \implies 3 + 2(\text{Value}) = 0$.
Value $= -3/2$.

Question 14

If $\vec{a}=0$ or $\vec{b}=0 \implies \vec{a}\cdot\vec{b}=0$. Is converse true?

Converse is False.

Example: Let $\vec{a} = \hat{i}$ and $\vec{b} = \hat{j}$. Both are non-zero.
However, $\vec{a} \cdot \vec{b} = 1(0) + 0(1) = 0$.
Dot product is zero because they are perpendicular, not because one is zero.

Question 15

Vertices A(1,2,3), B(-1,0,0), C(0,1,2). Find $\angle ABC$.

$\angle ABC$ is angle between vectors $\vec{BA}$ and $\vec{BC}$.
$\vec{BA} = \vec{a} – \vec{b} = (1-(-1))\hat{i} + (2-0)\hat{j} + (3-0)\hat{k} = 2\hat{i} + 2\hat{j} + 3\hat{k}$.
$\vec{BC} = \vec{c} – \vec{b} = (0-(-1))\hat{i} + (1-0)\hat{j} + (2-0)\hat{k} = \hat{i} + \hat{j} + 2\hat{k}$.
$\vec{BA} \cdot \vec{BC} = 2(1) + 2(1) + 3(2) = 10$.
$|\vec{BA}| = \sqrt{4+4+9} = \sqrt{17}$. $|\vec{BC}| = \sqrt{1+1+4} = \sqrt{6}$.
$\cos \theta = \frac{10}{\sqrt{17}\sqrt{6}} = \frac{10}{\sqrt{102}} \implies \theta = \cos^{-1}(\frac{10}{\sqrt{102}})$.

Question 16

Show points A(1, 2, 7), B(2, 6, 3), C(3, 10, -1) are collinear.

$|\vec{AB}| = \sqrt{(2-1)^2 + (6-2)^2 + (3-7)^2} = \sqrt{1+16+16} = \sqrt{33}$.
$|\vec{BC}| = \sqrt{(3-2)^2 + (10-6)^2 + (-1-3)^2} = \sqrt{1+16+16} = \sqrt{33}$.
$|\vec{AC}| = \sqrt{(3-1)^2 + (10-2)^2 + (-1-7)^2} = \sqrt{4+64+64} = \sqrt{132} = 2\sqrt{33}$.
Since $|\vec{AC}| = |\vec{AB}| + |\vec{BC}|$, points are collinear.

Question 17

Show vectors form vertices of right angled triangle.

Let $\vec{a} = 2\hat{i}-\hat{j}+\hat{k}, \vec{b} = \hat{i}-3\hat{j}-5\hat{k}, \vec{c} = 3\hat{i}-4\hat{j}-4\hat{k}$.
Sides: $\vec{AB} = \vec{b}-\vec{a} = -\hat{i}-2\hat{j}-6\hat{k}$. $|\vec{AB}|^2 = 1+4+36 = 41$.
$\vec{BC} = \vec{c}-\vec{b} = 2\hat{i}-\hat{j}+\hat{k}$. $|\vec{BC}|^2 = 4+1+1 = 6$.
$\vec{CA} = \vec{a}-\vec{c} = -\hat{i}+3\hat{j}+5\hat{k}$. $|\vec{CA}|^2 = 1+9+25 = 35$.
$6 + 35 = 41 \implies |\vec{BC}|^2 + |\vec{CA}|^2 = |\vec{AB}|^2$. (Pythagoras holds).

Question 18

If $\vec{a}$ is non-zero magnitude ‘a’, $\lambda$ non-zero scalar, then $\lambda \vec{a}$ is unit vector if:

$|\lambda \vec{a}| = 1 \implies |\lambda| |\vec{a}| = 1$.
$|\lambda| a = 1 \implies a = \frac{1}{|\lambda|}$.

Correct Option: (D)