Class 12 Maths NCERT Solutions Chapter 11 Ex 11.2 – Equation of a Line, Angle Between Lines, Shortest Distance | Boundless Maths
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Ex 11.2 Class 12 Maths NCERT Solutions

Class 12 Maths NCERT Solutions Chapter 11 Ex 11.2 – Equation of a Line, Angle Between Lines, Shortest Distance

This Class 12 Maths NCERT Solutions Chapter 11 Ex 11.2 page covers all 15 questions, solved step-by-step. This is the exercise where a line in 3D space finally gets an equation — in vector form \vec{r}=\vec{a}+\lambda\vec{b} and in Cartesian form \frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c} — and where you learn to compare two lines: are they parallel, perpendicular, or skew, and how far apart do they sit.

Questions 1–3 apply the direction-ratio tests from Exercise 11.1 to prove lines are perpendicular or parallel — the dot product going to zero, or the ratios matching. Questions 4–7 are pure construction: writing down a line's equation from a point and a direction, and converting cleanly between vector and Cartesian form, which is a skill every later question in this chapter assumes you already have. Questions 8–11 find the angle between two lines using both the vector dot-product formula and the Cartesian direction-ratio formula, including solving for an unknown when two lines are given to be perpendicular — a classic CBSE 2-mark question. Questions 12–15 are the exercise's centrepiece: the shortest-distance formula for skew lines, d=\left|\dfrac{(\vec{b_1}\times\vec{b_2})\cdot(\vec{a_2}-\vec{a_1})}{|\vec{b_1}\times\vec{b_2}|}\right|, worked in both vector and Cartesian form.

15Questions
Easy–HardDifficulty Mix
2026-27CBSE Syllabus

Class 12 Maths NCERT Solutions Chapter 11 Ex 11.2 — All 15 Questions

1

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

Medium +
Solution

Two lines with direction cosines are perpendicular when l_1l_2+m_1m_2+n_1n_2=0.

Lines 1 & 2: \dfrac{1}{169}[12(4)+(-3)(12)+(-4)(3)] = \dfrac{1}{169}[48-36-12] = 0.

Lines 2 & 3: \dfrac{1}{169}[4(3)+12(-4)+3(12)] = \dfrac{1}{169}[12-48+36] = 0.

Lines 1 & 3: \dfrac{1}{169}[12(3)+(-3)(-4)+(-4)(12)] = \dfrac{1}{169}[36+12-48] = 0.

Answer: every pair gives 0, so all three lines are mutually perpendicular.
2

Show that the line through the points (1, −1, 2), (3, 4, −2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Easy +
Solution

Direction ratios of the first line: (3-1,\,4-(-1),\,-2-2) = (2,5,-4).

Direction ratios of the second line: (3-0,\,5-3,\,6-2) = (3,2,4).

(2)(3)+(5)(2)+(-4)(4) = 6+10-16 = 0.

Answer: the dot product is 0, so the lines are perpendicular.
3

Show that the line through the points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (−1, −2, 1), (1, 2, 5).

Easy +
Solution

Direction ratios of the first line: (2-4,\,3-7,\,4-8) = (-2,-4,-4).

Direction ratios of the second line: (1-(-1),\,2-(-2),\,5-1) = (2,4,4).

Since (2,4,4) = -1\times(-2,-4,-4), the direction ratios are proportional.

Answer: proportional direction ratios mean the two lines are parallel.
4

Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3\hat{i}+2\hat{j}-2\hat{k}.

Easy +
Solution

Vector equation: \vec{r} = (\hat{i}+2\hat{j}+3\hat{k}) + \lambda(3\hat{i}+2\hat{j}-2\hat{k}).

Cartesian equation, using point (1, 2, 3) and direction ratios (3, 2, −2): \dfrac{x-1}{3} = \dfrac{y-2}{2} = \dfrac{z-3}{-2}.

Answer: r = (î + 2ĵ + 3k̂) + λ(3î + 2ĵ − 2k̂); Cartesian: (x−1)/3 = (y−2)/2 = (z−3)/−2.
5

Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2\hat{i}-\hat{j}+4\hat{k} and is in the direction \hat{i}+2\hat{j}-\hat{k}.

Easy +
Solution

Vector equation: \vec{r} = (2\hat{i}-\hat{j}+4\hat{k}) + \lambda(\hat{i}+2\hat{j}-\hat{k}).

Cartesian equation, using point (2, −1, 4) and direction ratios (1, 2, −1): \dfrac{x-2}{1} = \dfrac{y+1}{2} = \dfrac{z-4}{-1}.

Answer: r = (2î − ĵ + 4k̂) + λ(î + 2ĵ − k̂); Cartesian: (x−2)/1 = (y+1)/2 = (z−4)/−1.
6

Find the cartesian equation of the line which passes through the point (−2, 4, −5) and parallel to the line given by \dfrac{x+3}{3} = \dfrac{y-4}{5} = \dfrac{z+8}{6}.

Easy +
Solution

The given line has direction ratios (3, 5, 6). A line parallel to it has the same direction ratios.

Using point (−2, 4, −5): \dfrac{x-(-2)}{3} = \dfrac{y-4}{5} = \dfrac{z-(-5)}{6}.

Answer: (x+2)/3 = (y−4)/5 = (z+5)/6.
7

The cartesian equation of a line is \dfrac{x-5}{3} = \dfrac{y+4}{7} = \dfrac{z-6}{2}. Write its vector form.

Easy +
Solution

The line passes through (5, −4, 6) with direction ratios (3, 7, 2).

Answer: r = (5î − 4ĵ + 6k̂) + λ(3î + 7ĵ + 2k̂).
8

Find the angle between the following pairs of lines: (i) \vec{r}=2\hat{i}-5\hat{j}+\hat{k}+\lambda(3\hat{i}+2\hat{j}+6\hat{k}) and \vec{r}=7\hat{i}-6\hat{k}+\mu(\hat{i}+2\hat{j}+2\hat{k}) (ii) \vec{r}=3\hat{i}+\hat{j}-2\hat{k}+\lambda(\hat{i}-\hat{j}-2\hat{k}) and \vec{r}=2\hat{i}-\hat{j}-56\hat{k}+\mu(3\hat{i}-5\hat{j}-4\hat{k})

Medium +
Solution

(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 = \left|\dfrac{\vec{b_1}\cdot\vec{b_2}}{|\vec{b_1}||\vec{b_2}|}\right| = \left|\dfrac{3+4+12}{\sqrt{9+4+36}\sqrt{1+4+4}}\right| = \dfrac{19}{7\times3} = \dfrac{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}.

\cos\theta = \left|\dfrac{3+5+8}{\sqrt{1+1+4}\sqrt{9+25+16}}\right| = \dfrac{16}{\sqrt6\cdot5\sqrt2} = \dfrac{16}{5\sqrt{12}} = \dfrac{8\sqrt3}{15}.

Answer: (i) θ = cos⁻¹(19/21). (ii) θ = cos⁻¹(8√3/15).
9

Find the angle between the following pair of lines: (i) \dfrac{x-2}{2}=\dfrac{y-1}{5}=\dfrac{z+3}{-3} and \dfrac{x+2}{-1}=\dfrac{y-4}{8}=\dfrac{z-5}{4} (ii) \dfrac{x}{2}=\dfrac{y}{2}=\dfrac{z}{1} and \dfrac{x-5}{4}=\dfrac{y-2}{1}=\dfrac{z-3}{8}

Medium +
Solution

(i) Direction ratios: (2,5,-3) and (-1,8,4).

\cos\theta = \left|\dfrac{2(-1)+5(8)+(-3)(4)}{\sqrt{4+25+9}\sqrt{1+64+16}}\right| = \left|\dfrac{-2+40-12}{\sqrt{38}\cdot9}\right| = \dfrac{26}{9\sqrt{38}}.

(ii) Direction ratios: (2,2,1) and (4,1,8).

\cos\theta = \left|\dfrac{2(4)+2(1)+1(8)}{\sqrt{4+4+1}\sqrt{16+1+64}}\right| = \dfrac{18}{3\times9} = \dfrac{2}{3}.

Answer: (i) θ = cos⁻¹(26/9√38). (ii) θ = cos⁻¹(2/3).
10

Find the values of p so that the lines \dfrac{1-x}{3}=\dfrac{7y-14}{2p}=\dfrac{z-3}{2} and \dfrac{7-7x}{3p}=\dfrac{y-5}{1}=\dfrac{6-z}{5} are at right angles.

Hard +
Solution

Rewrite the first line with positive denominators: \dfrac{1-x}{3}=\dfrac{x-1}{-3}, and \dfrac{7y-14}{2p}=\dfrac{y-2}{2p/7}. So its direction ratios are \left(-3,\,\dfrac{2p}{7},\,2\right).

Similarly, the second line: \dfrac{7-7x}{3p}=\dfrac{x-1}{-3p/7}, and \dfrac{6-z}{5}=\dfrac{z-6}{-5}. So its direction ratios are \left(-\dfrac{3p}{7},\,1,\,-5\right).

For perpendicularity: (-3)\left(-\dfrac{3p}{7}\right) + \dfrac{2p}{7}(1) + 2(-5) = 0.

\dfrac{9p}{7}+\dfrac{2p}{7}-10=0 \Rightarrow \dfrac{11p}{7}=10.

Answer: p = 70/11.
11

Show that the lines \dfrac{x-5}{7}=\dfrac{y+2}{-5}=\dfrac{z}{1} and \dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3} are perpendicular to each other.

Easy +
Solution

Direction ratios: (7,-5,1) and (1,2,3).

(7)(1)+(-5)(2)+(1)(3) = 7-10+3 = 0.

Answer: dot product is 0, so the lines are perpendicular.
12

Find the shortest distance between the 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}).

Medium +
Solution

\vec{a_1}=\hat{i}+2\hat{j}+\hat{k}, \vec{b_1}=\hat{i}-\hat{j}+\hat{k}; \vec{a_2}=2\hat{i}-\hat{j}-\hat{k}, \vec{b_2}=2\hat{i}+\hat{j}+2\hat{k}.

\vec{a_2}-\vec{a_1} = \hat{i}-3\hat{j}-2\hat{k}.

\vec{b_1}\times\vec{b_2} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&1\\2&1&2\end{vmatrix} = \hat{i}(-2-1)-\hat{j}(2-2)+\hat{k}(1+2) = -3\hat{i}+0\hat{j}+3\hat{k}.

|\vec{b_1}\times\vec{b_2}| = \sqrt{9+0+9} = 3\sqrt2.

(\vec{b_1}\times\vec{b_2})\cdot(\vec{a_2}-\vec{a_1}) = (-3)(1)+(0)(-3)+(3)(-2) = -3-6 = -9.

d = \dfrac{|-9|}{3\sqrt2} = \dfrac{3}{\sqrt2} = \dfrac{3\sqrt2}{2}.

Answer: shortest distance = 3/√2 = (3√2)/2 units.
13

Find the shortest distance between the lines \dfrac{x+1}{7}=\dfrac{y+1}{-6}=\dfrac{z+1}{1} and \dfrac{x-3}{1}=\dfrac{y-5}{-2}=\dfrac{z-7}{1}.

Hard +
Solution

Point on line 1: (−1, −1, −1), direction ratios (7, −6, 1). Point on line 2: (3, 5, 7), direction ratios (1, −2, 1).

x_2-x_1=4, y_2-y_1=6, z_2-z_1=8.

Numerator determinant: \begin{vmatrix}4&6&8\\7&-6&1\\1&-2&1\end{vmatrix} = 4(-6+2)-6(7-1)+8(-14+6) = -16-36-64 = -116.

Denominator terms: b_1c_2-b_2c_1 = -6-(-2) = -4; c_1a_2-c_2a_1 = 1-7 = -6; a_1b_2-a_2b_1 = -14+6 = -8.

\sqrt{(-4)^2+(-6)^2+(-8)^2} = \sqrt{16+36+64} = \sqrt{116} = 2\sqrt{29}.

d = \dfrac{|-116|}{2\sqrt{29}} = \dfrac{58}{\sqrt{29}} = 2\sqrt{29}.

Answer: shortest distance = 2√29 units.
14

Find the shortest distance between the lines whose vector equations are \vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda(\hat{i}-3\hat{j}+2\hat{k}) and \vec{r}=4\hat{i}+5\hat{j}+6\hat{k}+\mu(2\hat{i}+3\hat{j}+\hat{k}).

Medium +
Solution

\vec{a_2}-\vec{a_1} = 3\hat{i}+3\hat{j}+3\hat{k}.

\vec{b_1}\times\vec{b_2} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-3&2\\2&3&1\end{vmatrix} = \hat{i}(-3-6)-\hat{j}(1-4)+\hat{k}(3+6) = -9\hat{i}+3\hat{j}+9\hat{k}.

|\vec{b_1}\times\vec{b_2}| = \sqrt{81+9+81} = \sqrt{171} = 3\sqrt{19}.

(\vec{b_1}\times\vec{b_2})\cdot(\vec{a_2}-\vec{a_1}) = (-9)(3)+(3)(3)+(9)(3) = -27+9+27 = 9.

d = \dfrac{|9|}{3\sqrt{19}} = \dfrac{3}{\sqrt{19}} = \dfrac{3\sqrt{19}}{19}.

Answer: shortest distance = 3/√19 = (3√19)/19 units.
15

Find the shortest distance between the lines whose vector equations are \vec{r}=(1-t)\hat{i}+(t-2)\hat{j}+(3-2t)\hat{k} and \vec{r}=(s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}.

Hard +
Solution

Rewriting the first line as \vec{r}=(\hat{i}-2\hat{j}+3\hat{k})+t(-\hat{i}+\hat{j}-2\hat{k}): \vec{a_1}=\hat{i}-2\hat{j}+3\hat{k}, \vec{b_1}=-\hat{i}+\hat{j}-2\hat{k}.

Rewriting the second as \vec{r}=(\hat{i}-\hat{j}-\hat{k})+s(\hat{i}+2\hat{j}-2\hat{k}): \vec{a_2}=\hat{i}-\hat{j}-\hat{k}, \vec{b_2}=\hat{i}+2\hat{j}-2\hat{k}.

\vec{a_2}-\vec{a_1} = 0\hat{i}+\hat{j}-4\hat{k}.

\vec{b_1}\times\vec{b_2} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&1&-2\\1&2&-2\end{vmatrix} = \hat{i}(-2+4)-\hat{j}(2+2)+\hat{k}(-2-1) = 2\hat{i}-4\hat{j}-3\hat{k}.

|\vec{b_1}\times\vec{b_2}| = \sqrt{4+16+9} = \sqrt{29}.

(\vec{b_1}\times\vec{b_2})\cdot(\vec{a_2}-\vec{a_1}) = (2)(0)+(-4)(1)+(-3)(-4) = -4+12 = 8.

Answer: shortest distance = 8/√29 = (8√29)/29 units.

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Common Questions

FAQs — Class 12 Maths NCERT Solutions Chapter 11 Ex 11.2

How many questions are there in Exercise 11.2?
Exercise 11.2 has 15 questions on the equation of a line in vector and Cartesian form, the angle between two lines, checking perpendicularity and parallelism, and finding the shortest distance between skew or parallel lines.

What concept does Exercise 11.2 test?
It tests writing the equation of a line as r = a + λb (vector form) or (x−x₁)/a = (y−y₁)/b = (z−z₁)/c (Cartesian form), finding the angle between two lines from their direction ratios or direction vectors, and computing the shortest distance between two lines using the cross product formula for skew lines.

Where can I find the official NCERT textbook for this exercise?
Exercise 11.2 is from Chapter 11, Three Dimensional Geometry, in the NCERT Class 12 Mathematics textbook (Part I), published by the National Council of Educational Research and Training (NCERT) and prescribed by CBSE. You can download the official textbook PDF directly from ncert.nic.in, NCERT's official website — the solutions on this page follow the questions exactly as they appear there.

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