Exercise 10.4 Class 12 Maths Vector Algebra

NCERT Solutions Class 12 Maths Chapter 10 Ex 10.4 | LearnCBSEHub

💡 Cross Product Formulas

Determinant Form: $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
Magnitude: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta$
Area of Triangle: $\frac{1}{2} |\vec{a} \times \vec{b}|$ (adjacent sides)
Area of Parallelogram: $|\vec{a} \times \vec{b}|$ (adjacent sides)

Question 01

Find $|\vec{a} \times \vec{b}|$, if $\vec{a} = \hat{i} – 7\hat{j} + 7\hat{k}$ and $\vec{b} = 3\hat{i} – 2\hat{j} + 2\hat{k}$.

Step 1: Calculate Cross Product

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix}$
$= \hat{i}(-14 – (-14)) – \hat{j}(2 – 21) + \hat{k}(-2 – (-21))$
$= \hat{i}(0) – \hat{j}(-19) + \hat{k}(19) = 19\hat{j} + 19\hat{k}$.

Step 2: Calculate Magnitude

$|\vec{a} \times \vec{b}| = \sqrt{0^2 + 19^2 + 19^2} = \sqrt{2 \times 19^2} = 19\sqrt{2}$.

Result: $19\sqrt{2}$

Question 02

Find a unit vector perpendicular to both $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$, where $\vec{a} = 3\hat{i} + 2\hat{j} + 2\hat{k}$ and $\vec{b} = \hat{i} + 2\hat{j} – 2\hat{k}$.

Let $\vec{p} = \vec{a}+\vec{b} = 4\hat{i} + 4\hat{j} + 0\hat{k}$.
Let $\vec{q} = \vec{a}-\vec{b} = 2\hat{i} + 0\hat{j} + 4\hat{k}$.
Vector perpendicular to both is $\vec{n} = \vec{p} \times \vec{q}$.
$\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 0 \\ 2 & 0 & 4 \end{vmatrix} = \hat{i}(16) – \hat{j}(16) + \hat{k}(-8) = 16\hat{i} – 16\hat{j} – 8\hat{k}$.
$|\vec{n}| = \sqrt{16^2 + (-16)^2 + (-8)^2} = \sqrt{256+256+64} = \sqrt{576} = 24$.
Unit Vector $\hat{n} = \pm \frac{16\hat{i} – 16\hat{j} – 8\hat{k}}{24} = \pm (\frac{2}{3}\hat{i} – \frac{2}{3}\hat{j} – \frac{1}{3}\hat{k})$.

Question 03

If unit vector $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}$, $\frac{\pi}{4}$ with $\hat{j}$ and acute angle $\theta$ with $\hat{k}$, find $\theta$ and components of $\vec{a}$.

Direction cosines: $l = \cos\frac{\pi}{3} = \frac{1}{2}$, $m = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$, $n = \cos\theta$.
Since $|\vec{a}|=1$, $l^2 + m^2 + n^2 = 1$.
$(\frac{1}{2})^2 + (\frac{1}{\sqrt{2}})^2 + \cos^2\theta = 1 \implies \frac{1}{4} + \frac{1}{2} + \cos^2\theta = 1$.
$\frac{3}{4} + \cos^2\theta = 1 \implies \cos^2\theta = \frac{1}{4} \implies \cos\theta = \pm \frac{1}{2}$.
Since $\theta$ is acute, $\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}$.
Components: $(l, m, n) = (\frac{1}{2}, \frac{1}{\sqrt{2}}, \frac{1}{2})$.

Question 04

Show that $(\vec{a} – \vec{b}) \times (\vec{a} + \vec{b}) = 2(\vec{a} \times \vec{b})$.

LHS $= \vec{a} \times \vec{a} + \vec{a} \times \vec{b} – \vec{b} \times \vec{a} – \vec{b} \times \vec{b}$
We know $\vec{a} \times \vec{a} = 0$ and $\vec{b} \times \vec{b} = 0$.
Also, $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$.
$= 0 + \vec{a} \times \vec{b} – (-(\vec{a} \times \vec{b})) – 0$
$= \vec{a} \times \vec{b} + \vec{a} \times \vec{b} = 2(\vec{a} \times \vec{b}) =$ RHS.

Verified

Question 05

Find $\lambda$ and $\mu$ if $(2\hat{i} + 6\hat{j} + 27\hat{k}) \times (\hat{i} + \lambda\hat{j} + \mu\hat{k}) = \vec{0}$.

$\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 6 & 27 \\ 1 & \lambda & \mu \end{vmatrix} = \hat{i}(6\mu – 27\lambda) – \hat{j}(2\mu – 27) + \hat{k}(2\lambda – 6) = \vec{0}$.
Compare components:
1. $2\mu – 27 = 0 \implies \mu = 27/2$.
2. $2\lambda – 6 = 0 \implies \lambda = 3$.
Check: $6(\frac{27}{2}) – 27(3) = 81 – 81 = 0$. Verified.

$\lambda = 3, \mu = 27/2$

Question 06

Given $\vec{a} \cdot \vec{b} = 0$ and $\vec{a} \times \vec{b} = \vec{0}$. What can you conclude?

$\vec{a} \cdot \vec{b} = 0$ implies $\vec{a} \perp \vec{b}$ (or one is zero).

$\vec{a} \times \vec{b} = \vec{0}$ implies $\vec{a} \parallel \vec{b}$ (or one is zero).

Two non-zero vectors cannot be both perpendicular AND parallel simultaneously. Therefore, either $\vec{a} = \vec{0}$ or $\vec{b} = \vec{0}$.

Question 07

Show $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$ using component form.

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, etc.

$\vec{b} + \vec{c} = (b_1+c_1)\hat{i} + (b_2+c_2)\hat{j} + (b_3+c_3)\hat{k}$.
LHS (i-component): $a_2(b_3+c_3) – a_3(b_2+c_2) = (a_2b_3 – a_3b_2) + (a_2c_3 – a_3c_2)$.
RHS (i-component): $(a_2b_3 – a_3b_2) + (a_2c_3 – a_3c_2)$.
Since components match, the distributive law holds.

Question 08

If $\vec{a}=\vec{0}$ or $\vec{b}=\vec{0}$, then $\vec{a} \times \vec{b} = \vec{0}$. Is converse true?

No, the converse is not true.

If $\vec{a} \times \vec{b} = \vec{0}$, it implies $\vec{a}$ and $\vec{b}$ are parallel (collinear), not necessarily zero.

Example: $\vec{a} = 2\hat{i}$ and $\vec{b} = 4\hat{i}$. $\vec{a} \times \vec{b} = 0$, but neither vector is zero.

Question 09

Find area of triangle with vertices A(1, 1, 2), B(2, 3, 5), C(1, 5, 5).

$\vec{AB} = (2-1)\hat{i} + (3-1)\hat{j} + (5-2)\hat{k} = \hat{i} + 2\hat{j} + 3\hat{k}$.
$\vec{AC} = (1-1)\hat{i} + (5-1)\hat{j} + (5-2)\hat{k} = 0\hat{i} + 4\hat{j} + 3\hat{k}$.
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} = \hat{i}(6-12) – \hat{j}(3-0) + \hat{k}(4-0) = -6\hat{i} – 3\hat{j} + 4\hat{k}$.
Magnitude $= \sqrt{36 + 9 + 16} = \sqrt{61}$.
Area $= \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{\sqrt{61}}{2}$ sq. units.

Question 10

Find area of parallelogram with adjacent sides $\vec{a} = \hat{i} – \hat{j} + 3\hat{k}$ and $\vec{b} = 2\hat{i} – 7\hat{j} + \hat{k}$.

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 3 \\ 2 & -7 & 1 \end{vmatrix} = \hat{i}(-1+21) – \hat{j}(1-6) + \hat{k}(-7+2)$
$= 20\hat{i} + 5\hat{j} – 5\hat{k}$.
Area $= |\vec{a} \times \vec{b}| = \sqrt{400 + 25 + 25} = \sqrt{450} = 15\sqrt{2}$ sq. units.

Question 11

Let $|\vec{a}|=3, |\vec{b}|=\frac{\sqrt{2}}{3}$. If $\vec{a} \times \vec{b}$ is unit vector, find angle.

$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta = 1$.
$3 \cdot \frac{\sqrt{2}}{3} \sin\theta = 1 \implies \sqrt{2}\sin\theta = 1 \implies \sin\theta = \frac{1}{\sqrt{2}}$.
$\theta = \frac{\pi}{4}$.

Correct Option: (B) $\pi/4$

Question 12

Area of rectangle with vertices A, B, C, D (position vectors given).

Given: A ($-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$), B ($\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$), C ($\hat{i} – \frac{1}{2}\hat{j} + 4\hat{k}$), D ($-\hat{i} – \frac{1}{2}\hat{j} + 4\hat{k}$).

$\vec{AB} = \vec{b} – \vec{a} = (1 – (-1))\hat{i} + 0\hat{j} + 0\hat{k} = 2\hat{i}$. Length $|\vec{AB}| = 2$.
$\vec{AD} = \vec{d} – \vec{a} = 0\hat{i} + (-1/2 – 1/2)\hat{j} + 0\hat{k} = -1\hat{j}$. Length $|\vec{AD}| = 1$.
Area $= \text{Length} \times \text{Breadth} = 2 \times 1 = 2$.

Correct Option: (C) 2