Exercise 11.2 Class 12 Maths Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 Ex 11.2 | LearnCBSEHub

💡 Key Formulas

Vector Eq: $\vec{r} = \vec{a} + \lambda\vec{b}$ (Passes through $\vec{a}$, parallel to $\vec{b}$)
Cartesian Eq: $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$
Shortest Distance (Skew Lines): $d = \left| \frac{(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} – \vec{a_1})}{|\vec{b_1} \times \vec{b_2}|} \right|$
Perpendicular Condition: $a_1a_2 + b_1b_2 + c_1c_2 = 0$ (Using DRs) or $l_1l_2 + m_1m_2 + n_1n_2 = 0$ (Using DCs)

Question 01

Show that lines with direction cosines $\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13}$; $\frac{4}{13}, \frac{12}{13}, \frac{3}{13}$; $\frac{3}{13}, \frac{-4}{13}, \frac{12}{13}$ are mutually perpendicular.

For perpendicular lines, $l_1l_2 + m_1m_2 + n_1n_2 = 0$.
Line 1 & 2: $\frac{12}{13}(\frac{4}{13}) + \frac{-3}{13}(\frac{12}{13}) + \frac{-4}{13}(\frac{3}{13}) = \frac{48 – 36 – 12}{169} = 0$.
Line 2 & 3: $\frac{4}{13}(\frac{3}{13}) + \frac{12}{13}(\frac{-4}{13}) + \frac{3}{13}(\frac{12}{13}) = \frac{12 – 48 + 36}{169} = 0$.
Line 3 & 1: $\frac{3}{13}(\frac{12}{13}) + \frac{-4}{13}(\frac{-3}{13}) + \frac{12}{13}(\frac{-4}{13}) = \frac{36 + 12 – 48}{169} = 0$.
Since all pairs dot to 0, they are mutually perpendicular.

Question 02

Show line through (1, -1, 2), (3, 4, -2) is perpendicular to line through (0, 3, 2), (3, 5, 6).

DRs of Line 1 ($AB$): $(3-1, 4-(-1), -2-2) = (2, 5, -4)$.
DRs of Line 2 ($CD$): $(3-0, 5-3, 6-2) = (3, 2, 4)$.
Product Sum: $a_1a_2 + b_1b_2 + c_1c_2 = 2(3) + 5(2) + (-4)(4) = 6 + 10 – 16 = 0$.
Hence, lines are perpendicular.

Question 03

Show line through (4, 7, 8), (2, 3, 4) is parallel to line through (-1, -2, 1), (1, 2, 5).

DRs Line 1: $(2-4, 3-7, 4-8) = (-2, -4, -4)$.
DRs Line 2: $(1-(-1), 2-(-2), 5-1) = (2, 4, 4)$.
Ratio: $\frac{-2}{2} = \frac{-4}{4} = \frac{-4}{4} = -1$.
Since direction ratios are proportional, the lines are parallel.

Question 04

Find equation of line passing through (1, 2, 3) parallel to $3\hat{i} + 2\hat{j} – 2\hat{k}$.

Point $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$. Direction $\vec{b} = 3\hat{i} + 2\hat{j} – 2\hat{k}$.
Vector Eq: $\vec{r} = \vec{a} + \lambda\vec{b} \implies \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(3\hat{i} + 2\hat{j} – 2\hat{k})$.
Cartesian Eq: $\frac{x-1}{3} = \frac{y-2}{2} = \frac{z-3}{-2}$.

Question 05

Eq of line through $2\hat{i} – \hat{j} + 4\hat{k}$, parallel to $\hat{i} + 2\hat{j} – \hat{k}$.

Vector Eq: $\vec{r} = (2\hat{i} – \hat{j} + 4\hat{k}) + \lambda(\hat{i} + 2\hat{j} – \hat{k})$.
Cartesian Eq: $\frac{x-2}{1} = \frac{y+1}{2} = \frac{z-4}{-1}$.

Question 06

Cartesian eq of line through (-2, 4, -5) parallel to $\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}$.

The given line has direction ratios (3, 5, 6).
Required line passes through (-2, 4, -5) with same DRs.
Equation: $\frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6}$.

Question 07

Convert $\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}$ to vector form.

Point ($x_1, y_1, z_1$) = (5, -4, 6) $\implies \vec{a} = 5\hat{i} – 4\hat{j} + 6\hat{k}$.
Direction ratios (3, 7, 2) $\implies \vec{b} = 3\hat{i} + 7\hat{j} + 2\hat{k}$.
Vector Form: $\vec{r} = (5\hat{i} – 4\hat{j} + 6\hat{k}) + \lambda(3\hat{i} + 7\hat{j} + 2\hat{k})$.

Question 08

Find angle between pairs of lines.

(i)

$\vec{b_1} = 3\hat{i} + 2\hat{j} + 6\hat{k}$, $\vec{b_2} = \hat{i} + 2\hat{j} + 2\hat{k}$.
$\cos\theta = |\frac{\vec{b_1}\cdot\vec{b_2}}{|\vec{b_1}||\vec{b_2}|}| = |\frac{3(1)+2(2)+6(2)}{\sqrt{9+4+36}\sqrt{1+4+4}}| = \frac{19}{7 \times 3} = \frac{19}{21}$.
$\theta = \cos^{-1}(\frac{19}{21})$.

(ii)

$\vec{b_1} = \hat{i} – \hat{j} – 2\hat{k}$, $\vec{b_2} = 3\hat{i} – 5\hat{j} – 4\hat{k}$.
$\vec{b_1}\cdot\vec{b_2} = 3 + 5 + 8 = 16$.
$|\vec{b_1}| = \sqrt{6}, |\vec{b_2}| = \sqrt{50} = 5\sqrt{2}$.
$\cos\theta = \frac{16}{\sqrt{6} \cdot 5\sqrt{2}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}$.

Question 09

Find angle between lines (Cartesian form).

(i) $\frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3}$ and $\frac{x+2}{-1} = \frac{y-4}{8} = \frac{z-5}{4}$

DRs: (2, 5, -3) and (-1, 8, 4).
$\cos\theta = |\frac{-2 + 40 – 12}{\sqrt{4+25+9}\sqrt{1+64+16}}| = \frac{26}{\sqrt{38}\sqrt{81}} = \frac{26}{9\sqrt{38}}$.

Question 10

Find $p$ so lines are perpendicular.

Line 1: $\frac{1-x}{3} \dots \implies \frac{x-1}{-3} = \frac{y-2}{2p/7} = \frac{z-3}{2}$. DRs: $(-3, 2p/7, 2)$.
Line 2: $\frac{7-7x}{3p} \dots \implies \frac{x-1}{-3p/7} = \frac{y-5}{1} = \frac{z-6}{-5}$. DRs: $(-3p/7, 1, -5)$.
Perpendicular Condition: $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
$(-3)(\frac{-3p}{7}) + (\frac{2p}{7})(1) + 2(-5) = 0$.
$\frac{9p}{7} + \frac{2p}{7} – 10 = 0 \implies \frac{11p}{7} = 10 \implies p = \frac{70}{11}$.

Question 11

Show lines $\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1}$ and $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ are perpendicular.

DRs: (7, -5, 1) and (1, 2, 3).
Product: $7(1) + (-5)(2) + 1(3) = 7 – 10 + 3 = 0$.
Since dot product is 0, lines are perpendicular.

Question 12

Find shortest distance between lines $\vec{r} = (\hat{i}+2\hat{j}+\hat{k}) + \lambda(\hat{i}-\hat{j}+\hat{k})$ and $\vec{r} = (2\hat{i}-\hat{j}-\hat{k}) + \mu(2\hat{i}+\hat{j}+2\hat{k})$.

$\vec{a_1} = (1, 2, 1), \vec{a_2} = (2, -1, -1) \implies \vec{a_2}-\vec{a_1} = (1, -3, -2)$.
$\vec{b_1} = (1, -1, 1), \vec{b_2} = (2, 1, 2)$.
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{vmatrix} = -3\hat{i} + 0\hat{j} + 3\hat{k}$. Magnitude $= \sqrt{9+9} = 3\sqrt{2}$.
Distance $= |\frac{(\vec{b_1}\times\vec{b_2})\cdot(\vec{a_2}-\vec{a_1})}{|\vec{b_1}\times\vec{b_2}|}| = |\frac{(-3)(1) + 0(-3) + 3(-2)}{3\sqrt{2}}| = |\frac{-9}{3\sqrt{2}}| = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$.

Question 13

Shortest distance between $\frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1}$ and $\frac{x-3}{1} = \frac{y-5}{-2} = \frac{z-7}{1}$.

$\vec{a_1} = (-1, -1, -1), \vec{a_2} = (3, 5, 7) \implies \vec{a_2}-\vec{a_1} = (4, 6, 8)$.
$\vec{b_1} = (7, -6, 1), \vec{b_2} = (1, -2, 1)$.
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} = -4\hat{i} – 6\hat{j} – 8\hat{k}$.
Magnitude $= \sqrt{16+36+64} = \sqrt{116} = 2\sqrt{29}$.
Dot Product: $(-4)(4) + (-6)(6) + (-8)(8) = -16 – 36 – 64 = -116$.
Distance $= |\frac{-116}{2\sqrt{29}}| = \frac{58}{\sqrt{29}} = 2\sqrt{29}$.

Question 14

Shortest distance between parallel lines $\vec{r} = (\hat{i}+2\hat{j}+3\hat{k}) + \lambda(\hat{i}-\hat{j}+\hat{k})$ and $\vec{r} = (4\hat{i}+5\hat{j}+6\hat{k}) + \mu(2\hat{i}-2\hat{j}+2\hat{k})$.

$\vec{b} = \hat{i}-\hat{j}+\hat{k}$ (Direction is same, factor 2 doesn’t matter).
$\vec{a_2}-\vec{a_1} = (3, 3, 3)$.
$\vec{b} \times (\vec{a_2}-\vec{a_1}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 3 & 3 & 3 \end{vmatrix} = -6\hat{i} + 0\hat{j} + 6\hat{k}$.
Magnitude $= \sqrt{36+36} = 6\sqrt{2}$.
$|\vec{b}| = \sqrt{3}$.
Distance $= \frac{|\vec{b} \times (\vec{a_2}-\vec{a_1})|}{|\vec{b}|} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt{6}$.

Question 15

Shortest distance between lines involving parameters $t$ and $s$.

Convert to standard form.
Line 1: $\vec{r} = (1-t)\hat{i} + (t-2)\hat{j} + (3-2t)\hat{k} = (\hat{i}-2\hat{j}+3\hat{k}) + t(-\hat{i}+\hat{j}-2\hat{k})$.
Line 2: $\vec{r} = (s+1)\hat{i} + (2s-1)\hat{j} + (-2s-1)\hat{k} = (\hat{i}-\hat{j}-\hat{k}) + s(\hat{i}+2\hat{j}-2\hat{k})$.
$\vec{a_1} = (1, -2, 3), \vec{a_2} = (1, -1, -1) \implies \vec{a_2}-\vec{a_1} = (0, 1, -4)$.
$\vec{b_1} = (-1, 1, -2), \vec{b_2} = (1, 2, -2)$.
$\vec{b_1} \times \vec{b_2} = 2\hat{i} – 4\hat{j} – 3\hat{k}$. Magnitude $= \sqrt{29}$.
Distance $= |\frac{0 – 4 + 12}{\sqrt{29}}| = \frac{8}{\sqrt{29}}$.