Miscellaneous Exercise Class 12 Maths Three Dimensional Geometry

NCERT Solutions Class 12 Maths Chapter 11 Miscellaneous | LearnCBSEHub

💡 Key Concepts Recap

Angle between lines: $\cos\theta = |\frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}|$.
Perpendicular Lines: $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
Shortest Distance: $d = |\frac{(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2} – \vec{a_1})}{|\vec{b_1} \times \vec{b_2}|}|$.

Question 01

Find angle between lines with direction ratios $a, b, c$ and $b-c, c-a, a-b$.

Let DRs of line 1 be $a, b, c$.
Let DRs of line 2 be $b-c, c-a, a-b$.
Dot Product of direction vectors:
$a(b-c) + b(c-a) + c(a-b)$
$= ab – ac + bc – ba + ca – cb = 0$.
Since the dot product is 0, the lines are perpendicular. Angle $\theta = 90^\circ$.

Angle: $90^\circ$

Question 02

Equation of line parallel to x-axis passing through origin.

Line passes through origin $(0, 0, 0)$. Position vector $\vec{a} = \vec{0}$.
Parallel to x-axis means direction vector $\vec{b} = \hat{i}$ (or direction ratios $1, 0, 0$).
Vector Equation: $\vec{r} = \vec{0} + \lambda\hat{i} \implies \vec{r} = \lambda\hat{i}$.
Cartesian Equation: $\frac{x-0}{1} = \frac{y-0}{0} = \frac{z-0}{0}$.

Question 03

If lines $\frac{x-1}{-3} = \frac{y-2}{2k} = \frac{z-3}{2}$ and $\frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5}$ are perpendicular, find $k$.

Line 1 DRs: $a_1=-3, b_1=2k, c_1=2$.
Line 2 DRs: $a_2=3k, b_2=1, c_2=-5$.
Perpendicular Condition: $a_1a_2 + b_1b_2 + c_1c_2 = 0$.
$(-3)(3k) + (2k)(1) + (2)(-5) = 0$
$-9k + 2k – 10 = 0 \implies -7k = 10 \implies k = -\frac{10}{7}$.

Value: $k = -10/7$

Question 04

Find shortest distance between lines:
$\vec{r} = 6\hat{i}+2\hat{j}+2\hat{k} + \lambda(\hat{i}-2\hat{j}+2\hat{k})$
$\vec{r} = -4\hat{i}-\hat{k} + \mu(3\hat{i}-2\hat{j}-2\hat{k})$

$\vec{a_1} = (6, 2, 2), \vec{a_2} = (-4, 0, -1) \implies \vec{a_2}-\vec{a_1} = (-10, -2, -3)$.
$\vec{b_1} = (1, -2, 2), \vec{b_2} = (3, -2, -2)$.
$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{vmatrix} = \hat{i}(4+4) – \hat{j}(-2-6) + \hat{k}(-2+6) = 8\hat{i} + 8\hat{j} + 4\hat{k}$.
$|\vec{b_1} \times \vec{b_2}| = \sqrt{64+64+16} = \sqrt{144} = 12$.
$(\vec{b_1} \times \vec{b_2}) \cdot (\vec{a_2}-\vec{a_1}) = 8(-10) + 8(-2) + 4(-3) = -80 – 16 – 12 = -108$.
Distance $d = |\frac{-108}{12}| = |-9| = 9$ units.

Shortest Distance: 9 units

Question 05

Find vector eq. of line through $(1, 2, -4)$ perpendicular to:
$\frac{x-8}{3} = \frac{y+19}{-16} = \frac{z-10}{7}$ and $\frac{x-15}{3} = \frac{y-29}{8} = \frac{z-5}{-5}$.

Let required line direction be $\vec{b}$. Since it’s perp to both lines, $\vec{b} = \vec{b_1} \times \vec{b_2}$.
$\vec{b_1} = 3\hat{i} – 16\hat{j} + 7\hat{k}$.
$\vec{b_2} = 3\hat{i} + 8\hat{j} – 5\hat{k}$.
$\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -16 & 7 \\ 3 & 8 & -5 \end{vmatrix} = \hat{i}(80-56) – \hat{j}(-15-21) + \hat{k}(24+48)$
$= 24\hat{i} + 36\hat{j} + 72\hat{k} = 12(2\hat{i} + 3\hat{j} + 6\hat{k})$.
Use simpler direction vector $2\hat{i} + 3\hat{j} + 6\hat{k}$.
Point $\vec{a} = \hat{i} + 2\hat{j} – 4\hat{k}$.
Equation: $\vec{r} = (\hat{i} + 2\hat{j} – 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$.