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Solutions

Overview

Step-by-step NCERT solutions for Three Dimensional Geometry (Chapter 11, CBSE Class 12 Mathematics) — every question and answer worked out in full, not just the final result. You can also read the Three Dimensional Geometry textbook chapter.

Solved

What these solutions cover

All 25 questions in Three Dimensional Geometry are solved in the PDF. Here's what's inside, exercise by exercise:

Exercise 11.1

  1. If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.
  2. Find the direction cosines of a line which makes equal angles with the coordinate axes.
  3. If a line has the direction ratios -18, 12, -4, then what are its direction cosines?
  4. Show that the points (2, 3, 4), (– 1, – 2, 1), (5, 8, 7) are collinear.
  5. Find the direction cosines of the sides of the triangle whose vertices are (3, 5, – 4), (– 1, 1, 2) and (– 5, – 5, – 2).

Exercise 11.2

  1. Show that the three lines with direction cosines 12/13, -3/13, -4/13; 4/13, 12/13, 3/13; 3/13, -4/13, 12/13 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).
  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).
  4. Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector 3 î + 2 ĵ – 2 k̂.
  5. Find the equation of the line in vector and in cartesian form that passes through the point with position vector 2 î – ĵ + 4 k̂ and is in the direction î + 2ĵ – k̂.
  6. Find the cartesian equation of the line which passes through the point (– 2, 4, – 5) and parallel to the line given by (x + 3)/3 = (y – 4)/5 = (z + 8)/6.
  7. The cartesian equation of a line is (x – 5)/3 = (y + 4)/7 = (z – 6)/2. Write its vector form.
  8. Find the angle between the following pairs of lines:
    • (i) r = 2î – 5ĵ + k̂ + λ(3î + 2ĵ + 6k̂) and r = 7î – 6k̂ + μ(î + 2ĵ + 2k̂)
    • (ii) r = 3î + ĵ – 2k̂ + λ(î – ĵ – 2k̂) and r = 2î – ĵ – 56k̂ + μ(3î – 5ĵ – 4k̂)
  9. Find the angle between the following pair of lines:
    • (i) (x – 2)/2 = (y – 1)/5 = (z + 3)/(–3) and (x + 2)/(–1) = (y – 4)/8 = (z – 5)/4
    • (ii) x/2 = y/2 = z/1 and (x – 5)/4 = (y – 2)/1 = (z – 3)/8
  10. Find the values of p so that the lines (1 – x)/3 = (7y – 14)/(2p) = (z – 3)/2 and (7 – 7x)/(3p) = (y – 5)/1 = (6 – z)/5 are at right angles.
  11. Show that the lines (x – 5)/7 = (y + 2)/(–5) = z/1 and x/1 = y/2 = z/3 are perpendicular to each other.
  12. Find the shortest distance between the lines r = (î + 2ĵ + k̂) + λ(î – ĵ + k̂) and r = 2î – ĵ – k̂ + μ(2î + ĵ + 2k̂)
  13. Find the shortest distance between the lines (x + 1)/7 = (y + 1)/(–6) = (z + 1)/1 and (x – 3)/1 = (y – 5)/(–2) = (z – 7)/1.
  14. Find the shortest distance between the lines whose vector equations are r = (î + 2ĵ + 3k̂) + λ(î – 3ĵ + 2k̂) and r = 4î + 5ĵ + 6k̂ + μ(2î + 3ĵ + k̂).
  15. Find the shortest distance between the lines whose vector equations are r = (1 – t) î + (t – 2) ĵ + (3 – 2t) k̂ and r = (s + 1) î + (2s – 1) ĵ – (2s + 1) k̂.

Miscellaneous Exercise

  1. Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.
  2. Find the equation of a line parallel to x-axis and passing through the origin.
  3. If the lines (x – 1)/(–3) = (y – 2)/(2k) = (z – 3)/2 and (x – 1)/(3k) = (y – 1)/1 = (z – 6)/(–5) are perpendicular, find the value of k.
  4. Find the shortest distance between lines r = 6î + 2ĵ + 2k̂ + λ(î – 2ĵ + 2k̂) and r = –4î – k̂ + μ(3î – 2ĵ – 2k̂).
  5. Find the vector equation of the line passing through the point (1, 2, – 4) and perpendicular to the two lines: (x – 8)/3 = (y + 19)/(–16) = (z – 10)/7 and (x – 15)/3 = (y – 29)/8 = (z – 5)/(–5).
Sample

Worked examples

3 questions from Three Dimensional Geometry, solved step by step.

Exercise 11.1

If a line makes angles 90°, 135°, 45° with the x, y and z-axes respectively, find its direction cosines.

Let the direction cosines be l, m, n corresponding to angles α = 90°, β = 135°, γ = 45°.

l = cos 90° = 0

m = cos 135° = -1/√2

n = cos 45° = 1/√2

Verification: l² + m² + n² = 0 + 1/2 + 1/2 = 1. ✓

Answer: l = 0, m = -1/√2, n = 1/√2

Exercise 11.2

Show that the three lines with direction cosines 12/13, -3/13, -4/13; 4/13, 12/13, 3/13; 3/13, -4/13, 12/13 are mutually perpendicular.

Let L1, L2, L3 have the given direction cosines. Two lines are perpendicular if the sum of products of their direction cosines is zero.

L1 · L2 = (12/13)(4/13) + (-3/13)(12/13) + (-4/13)(3/13)

= (48 - 36 - 12)/169 = 0/169 = 0.

L2 · L3 = (4/13)(3/13) + (12/13)(-4/13) + (3/13)(12/13)

= (12 - 48 + 36)/169 = 0/169 = 0.

L1 · L3 = (12/13)(3/13) + (-3/13)(-4/13) + (-4/13)(12/13)

= (36 + 12 - 48)/169 = 0/169 = 0.

Since each pair of dot products equals zero, the three lines are mutually perpendicular.

Answer: All three dot products equal zero, so the lines are mutually perpendicular.

Miscellaneous Exercise

Find the angle between the lines whose direction ratios are a, b, c and b – c, c – a, a – b.

Let d1 = (a, b, c) and d2 = (b-c, c-a, a-b).

Compute the dot product:

d1 · d2 = a(b-c) + b(c-a) + c(a-b)

= ab – ac + bc – ab + ca – cb

= 0.

Since d1 · d2 = 0, cos θ = 0, so θ = 90°.

The lines are perpendicular to each other.

Answer: The angle between the lines is 90°.

All 25 questions in Three Dimensional Geometry are solved step by step in the Solutions PDF above.

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More NCERT Solutions for Class 12

Read the Three Dimensional Geometry textbook chapter / PDF, or browse all CBSE Class 12 Mathematics solutions.

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