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NCERT Solutions · Class 12 · Chapter 11: Three Dimensional Geometry

NCERT Solutions for Class 12 Maths Chapter 11 Exercise 11.2

Exercise 11.2: Lines: equations, angles and shortest distance. Line through \(\vec a\) parallel to \(\vec b\): \(\vec r = \vec a + \lambda\vec b\), i.e. \(\dfrac{x - x_1}{a} = \dfrac{y - y_1}{b} = \dfrac{z - z_1}{c}\). Angle between lines: \(\cos\theta = \dfrac{|\vec b_1 \cdot \vec b_2|}{|\vec b_1||\vec b_2|}\). Shortest distance between skew lines: \(\left|\dfrac{(\vec b_1 \times \vec b_2) \cdot (\vec a_2 - \vec a_1)}{|\vec b_1 \times \vec b_2|}\right|\); for parallel lines use \(\dfrac{|\vec b \times (\vec a_2 - \vec a_1)|}{|\vec b|}\).

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Exercise 11.2 questions and solutions

Exercise 11.2, Question 1

Show the lines with direction cosines \(\tfrac{12}{13}, -\tfrac{3}{13}, -\tfrac{4}{13}\); \(\tfrac{4}{13}, \tfrac{12}{13}, \tfrac{3}{13}\); \(\tfrac{3}{13}, -\tfrac{4}{13}, \tfrac{12}{13}\) are mutually perpendicular.
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  1. \(l_1l_2 + m_1m_2 + n_1n_2\) for each pair: \(\tfrac{48 - 36 - 12}{169}\), \(\tfrac{12 - 48 + 36}{169}\), \(\tfrac{36 + 12 - 48}{169}\), all 0.
Answer: Shown

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Exercise 11.2, Question 2

Show the line through \((1, -1, 2)\), \((3, 4, -2)\) is perpendicular to the line through \((0, 3, 2)\), \((3, 5, 6)\).
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  1. Direction ratios \(2, 5, -4\) and \(3, 2, 4\): \(6 + 10 - 16 = 0\).
Answer: Shown

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Exercise 11.2, Question 3

Show the line through \((4, 7, 8)\), \((2, 3, 4)\) is parallel to the line through \((-1, -2, 1)\), \((1, 2, 5)\).
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  1. Direction ratios \(-2, -4, -4\) and \(2, 4, 4\): proportional.
Answer: Shown

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Exercise 11.2, Question 4

Find the equation of the line through \((1, 2, 3)\) parallel to \(3\hat i + 2\hat j - 2\hat k\).
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  1. \(\vec a = \hat i + 2\hat j + 3\hat k\), \(\vec b = 3\hat i + 2\hat j - 2\hat k\).
Answer: \[\vec r = (\hat i + 2\hat j + 3\hat k) + \lambda(3\hat i + 2\hat j - 2\hat k)\]

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Exercise 11.2, Question 5

Find the vector and cartesian equations of the line through \(2\hat i - \hat j + 4\hat k\) in the direction \(\hat i + 2\hat j - \hat k\).
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  1. \(\vec r = \vec a + \lambda\vec b\); cartesian: point \((2, -1, 4)\), direction ratios \(1, 2, -1\).
Answer: \[\vec r = (2\hat i - \hat j + 4\hat k) + \lambda(\hat i + 2\hat j - \hat k)\]; \[\begin{aligned}\dfrac{x - 2}{1} &= \dfrac{y + 1}{2} \\ &= \dfrac{z - 4}{-1}\end{aligned}\]

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Exercise 11.2, Question 6

Find the cartesian equation of the line through \((-2, 4, -5)\) parallel to \(\dfrac{x + 3}{3} = \dfrac{y - 4}{5} = \dfrac{z + 8}{6}\).
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  1. Parallel lines share direction ratios \(3, 5, 6\).
Answer: \[\begin{aligned}\dfrac{x + 2}{3} &= \dfrac{y - 4}{5} \\ &= \dfrac{z + 5}{6}\end{aligned}\]

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Exercise 11.2, Question 7

The line \(\dfrac{x - 5}{3} = \dfrac{y + 4}{7} = \dfrac{z - 6}{2}\): write its vector form.
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  1. Point \((5, -4, 6)\), direction \(3, 7, 2\).
Answer: \[\vec r = (5\hat i - 4\hat j + 6\hat k) + \lambda(3\hat i + 7\hat j + 2\hat k)\]

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Exercise 11.2, Question 8

Find the angle between the pair of lines.
(i) \(\vec r = 2\hat i - 5\hat j + \hat k + \lambda(3\hat i + 2\hat j + 6\hat k)\) and \(\vec r = 7\hat i - 6\hat k + \mu(\hat i + 2\hat j + 2\hat k)\)
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  1. \[\begin{aligned}\vec b_1 \cdot \vec b_2 &= 3 + 4 + 12 \\ &= 19\end{aligned}\], \(|\vec b_1| = 7\), \(|\vec b_2| = 3\).
Answer: \(\theta = \cos^{-1}\dfrac{19}{21}\)
(ii) \(\vec r = 3\hat i + \hat j - 2\hat k + \lambda(\hat i - \hat j - 2\hat k)\) and \(\vec r = 2\hat i - \hat j - 56\hat k + \mu(3\hat i - 5\hat j - 4\hat k)\)
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  1. \(\vec b_1 \cdot \vec b_2 = 3 + 5 + 8 = 16\), \(|\vec b_1| = \sqrt6\), \(|\vec b_2| = 5\sqrt2\): \[\begin{aligned}\cos\theta &= \dfrac{16}{5\sqrt{12}} \\ &= \dfrac{8}{5\sqrt3}\end{aligned}\]
Answer: \(\theta = \cos^{-1}\dfrac{8}{5\sqrt3}\)

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Exercise 11.2, Question 9

Find the angle between the pair of lines.
(i) \(\dfrac{x - 2}{2} = \dfrac{y - 1}{5} = \dfrac{z + 3}{-3}\) and \(\dfrac{x + 2}{-1} = \dfrac{y - 4}{8} = \dfrac{z - 5}{4}\)
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  1. Direction ratios \(2, 5, -3\) and \(-1, 8, 4\): \(-2 + 40 - 12 = 26\); magnitudes \(\sqrt{38}\) and 9.
Answer: \(\theta = \cos^{-1}\dfrac{26}{9\sqrt{38}}\)
(ii) \(\dfrac{x}{2} = \dfrac{y}{2} = \dfrac{z}{1}\) and \(\dfrac{x - 5}{4} = \dfrac{y - 2}{1} = \dfrac{z - 3}{8}\)
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  1. Direction ratios \(2, 2, 1\) and \(4, 1, 8\): \(8 + 2 + 8 = 18\); magnitudes 3 and 9.
Answer: \(\theta = \cos^{-1}\dfrac23\)

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Exercise 11.2, Question 10

Find p so that \(\dfrac{1 - x}{3} = \dfrac{7y - 14}{2p} = \dfrac{z - 3}{2}\) and \(\dfrac{7 - 7x}{3p} = \dfrac{y - 5}{1} = \dfrac{6 - z}{5}\) are at right angles.
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  1. Standard form: \[\begin{aligned}\dfrac{x - 1}{-3} &= \dfrac{y - 2}{2p/7} \\ &= \dfrac{z - 3}{2}\end{aligned}\] and \[\begin{aligned}\dfrac{x - 1}{-3p/7} &= \dfrac{y - 5}{1} \\ &= \dfrac{z - 6}{-5}\end{aligned}\]
  2. \[\begin{aligned}(-3)\left(-\tfrac{3p}{7}\right) + \tfrac{2p}{7}(1) + 2(-5) &= \tfrac{11p}{7} - 10 \\ &= 0\end{aligned}\]
Answer: \(p = \tfrac{70}{11}\)

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Exercise 11.2, Question 11

Show \(\dfrac{x - 5}{7} = \dfrac{y + 2}{-5} = \dfrac{z}{1}\) and \(\dfrac{x}{1} = \dfrac{y}{2} = \dfrac{z}{3}\) are perpendicular.
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  1. \(7 - 10 + 3 = 0\).
Answer: Shown

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Exercise 11.2, Question 12

Find the shortest distance between \(\vec r = (\hat i + 2\hat j + \hat k) + \lambda(\hat i - \hat j + \hat k)\) and \(\vec r = 2\hat i - \hat j - \hat k + \mu(2\hat i + \hat j + 2\hat k)\).
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  1. \[\vec b_1 \times \vec b_2 = -3\hat i + 3\hat k\]; \[\vec a_2 - \vec a_1 = \hat i - 3\hat j - 2\hat k\]
  2. \(d = \dfrac{|-3 - 6|}{3\sqrt2}\).
Answer: \(\dfrac{3}{\sqrt2}\) units

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Exercise 11.2, Question 13

Find the shortest distance between \(\dfrac{x + 1}{7} = \dfrac{y + 1}{-6} = \dfrac{z + 1}{1}\) and \(\dfrac{x - 3}{1} = \dfrac{y - 5}{-2} = \dfrac{z - 7}{1}\).
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  1. \[\vec b_1 \times \vec b_2 = -4\hat i - 6\hat j - 8\hat k\], magnitude \(\sqrt{116}\); \[\vec a_2 - \vec a_1 = 4\hat i + 6\hat j + 8\hat k\]
  2. \[\begin{aligned}d &= \dfrac{|-16 - 36 - 64|}{\sqrt{116}} \\ &= \dfrac{116}{\sqrt{116}}\end{aligned}\]
Answer: \(2\sqrt{29}\) units

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Exercise 11.2, Question 14

Find the shortest distance between \(\vec r = (\hat i + 2\hat j + 3\hat k) + \lambda(\hat i - 3\hat j + 2\hat k)\) and \(\vec r = 4\hat i + 5\hat j + 6\hat k + \mu(2\hat i + 3\hat j + \hat k)\).
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  1. \[\vec b_1 \times \vec b_2 = -9\hat i + 3\hat j + 9\hat k\], magnitude \(3\sqrt{19}\); \[\vec a_2 - \vec a_1 = 3\hat i + 3\hat j + 3\hat k\]
  2. \(d = \dfrac{|-27 + 9 + 27|}{3\sqrt{19}}\).
Answer: \(\dfrac{3}{\sqrt{19}}\) units

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Exercise 11.2, Question 15

Find the shortest distance between \(\vec r = (1 - t)\hat i + (t - 2)\hat j + (3 - 2t)\hat k\) and \(\vec r = (s + 1)\hat i + (2s - 1)\hat j - (2s + 1)\hat k\).
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  1. Rewrite: \[\vec r = (\hat i - 2\hat j + 3\hat k) + t(-\hat i + \hat j - 2\hat k)\] and \[\vec r = (\hat i - \hat j - \hat k) + s(\hat i + 2\hat j - 2\hat k)\]
  2. \[\vec b_1 \times \vec b_2 = 2\hat i - 4\hat j - 3\hat k\], magnitude \(\sqrt{29}\); \(\vec a_2 - \vec a_1 = \hat j - 4\hat k\): \(d = \dfrac{|0 - 4 + 12|}{\sqrt{29}}\).
Answer: \(\dfrac{8}{\sqrt{29}}\) units

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Done the NCERT exercises? The board paper asks more

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Three Dimensional Geometry in our sample papers: Sample paper 1 (questions 14, 19, 30, 35) · Sample paper 2 (questions 17, 18, 25, 30) · Sample paper 3 (questions 15, 19, 30, 35) · Sample paper 4 (questions 17, 18, 30, 34) · Sample paper 5 (questions 14, 15, 30, 35).

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Textbook: NCERT Mathematics Class 12, Parts I and II (rationalised edition, 2023-24 reprint onward), free from ncert.nic.in. Question statements are shortened to the minimum needed; the solutions and tips are our own. CBSE Math Revision is independent and not affiliated with NCERT or CBSE. Spotted a slip? Tell us and it goes in the corrections log.