Class 12 Maths Chapter 11: Three Dimensional Geometry worksheet with answer key

Answer: (b) 45°

OP = √(1 + 1 + 2) = 2, so cos γ = √22 = 1√2 and γ = 45°. Option (a) takes cos γ = 12 from the x or y coordinate.

Answer: (d) is parallel to the y-axis and passes through the point (3, 0, −2)

In →r = →a + λ→b, →a = 3î − 2k̂ is the position vector of a point on the line, (3, 0, −2), and →b = ĵ gives direction ratios 0, 1, 0, the direction of the y-axis. Option (c) swaps the roles of →a and →b; (a) is wrong because the direction is that of the y-axis itself.

Answer: (b) skew lines

Lines in space which are neither parallel nor intersecting are skew lines; they lie in different planes, so (a) is wrong. Perpendicular lines may intersect, and concurrent lines meet.

Answer: (a) 13

(2k)² + (−k)² + (2k)² = 1 gives 9k² = 1, so k = 13. Option (b) stops at k²; (c) uses 3k² = 1; (d) inverts.

Answer: (c) −2, −3, −6

Any set of direction ratios of the line is k(2, −3, 6) with k ≠ 0: (a) is k = −2, (b) is k = 12, (d) is k = 3. In (c) the first number needs k = −1 but the second needs k = 1, so (c) is not proportional to 2, −3, 6.

Direction ratios of AB: 2, a − 2, 4; of AC: 6, 9, b − 1. For collinear points these are proportional: 62 = 3, so 9 = 3(a − 2) and b − 1 = 3 × 4. Hence a = 5 and b = 13.

• direction ratios of AB and AC, set proportional (1)
• a = 5, b = 13 (1)

(a) Direction ratios 2, −1, −2; √(4 + 1 + 4) = 3, so the direction cosines are 23, −13, −23 (or their negatives). (b) A general point is (1 + 2λ, −3 − λ, 4 − 2λ). 4 − 2λ = 0 gives λ = 2: the point is (5, −5, 0).

• direction cosines 23, −13, −23 (1)
• λ = 2, point (5, −5, 0) (1)

The line is x − 42 = y − 1−2 = z1, direction ratios 2, −2, 1. The y-axis has direction ratios 0, 1, 0. cos θ = |0 − 2 + 0|3 × 1 = 23, so θ = cos⁻¹(23).

• direction ratios 2, −2, 1 (sign of y fixed) (1)
• θ = cos⁻¹(23) (1)

l₁: →r = î − k̂ + λ(î + ĵ + 2k̂); l₂: →r = 2ĵ + k̂ + μ(î + (k − 1)k̂). The direction ratios are not proportional, so the lines meet exactly when (→b₁ × →b₂) · (→a₂ − →a₁) = 0. →b₁ × →b₂ = |î ĵ k̂; 1 1 2; 1 0 k − 1| = (k − 1)î + (3 − k)ĵ − k̂, and →a₂ − →a₁ = −î + 2ĵ + 2k̂, so the product is −(k − 1) + 2(3 − k) − 2 = 5 − 3k = 0, k = 5/3. Then (1 + λ, λ, −1 + 2λ) = (μ, 2, 1 + 23μ) gives λ = 2, μ = 3, and z agrees (3 = 3). The lines meet at (3, 2, 3).

• condition (→b₁ × →b₂) · (→a₂ − →a₁) = 0 giving 5 − 3k = 0 (1)
• k = 53 (1)
• λ = 2, μ = 3, point (3, 2, 3) (1)

General points: (1 + 2λ, −1 + 3λ, 2 − λ) and (2 + μ, 4 − 2μ, −2 + 3μ). Equating x and y: 2λ − μ = 1 and 3λ + 2μ = 5, so λ = 1, μ = 1. Then z: 2 − 1 = 1 and −2 + 3 = 1, equal, so the lines intersect. The point is (3, 2, 1).

• general points of both lines (1)
• λ = 1, μ = 1 from two coordinates, third coordinate verified (1)
• point of intersection (3, 2, 1) (1)

Direction ratios of BA: 2, 1, −2; of BC: 2, 0, 2. 2 × 2 + 1 × 0 + (−2) × 2 = 0, so BA ⟂ BC and ∠B = 90°. At A: AB has direction ratios −2, −1, 2 (magnitude 3) and AC has 0, −1, 4 (magnitude √17). cos A = 0 + 1 + 83√17 = 3√17.

• direction ratios of BA and BC (1)
• a₁a₂ + b₁b₂ + c₁c₂ = 0, so ∠B = 90° (1)
• cos A = 3√17 (1)