Maths · Vectors and 3-D geometry · chapter 11 of 13

Class 12 Maths Chapter 11: Three Dimensional Geometry case study question with answers

One case-based question of the kind the board asks: a passage, then 3 parts worth 4 marks in all, with the answer and the marking scheme for each. Written from the NCERT chapter.

Case study · 4 marks

For a wedding shamiana on flat ground in Jaipur, the organisers fix a coordinate system with the origin at one corner of the ground, the x-axis and y-axis along two edges of the ground and the z-axis vertically upward; all lengths are in metres. The top of a vertical pole is at A(2, 3, 6). A straight guy rope runs from A to a peg at B(8, 5, 3), and a second straight rope runs from A to a hook at C(−1, 9, 4) on a neighbouring frame. Each rope is treated as a segment of a straight line.

Part 1 · 1 mark

The length of the rope AB is

(a) 7 m(b) 11 m(c) √41 m(d) 49 m

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Answer: (a) 7 m

AB = √(6² + 2² + (−3)²) = √49 = 7 m. (b) adds the absolute differences; (d) forgets the square root.

Part 2 · 1 mark

Using the length of AB, the direction cosines of the rope AB, directed from A to B, are

(a) 611, 211, −311(b) 6, 2, −3(c) −67, −27, 37(d) 67, 27, −37

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Answer: (d) 67, 27, −37

Direction ratios 6, 2, −3; dividing by AB = 7 m from part (i) gives the direction cosines from A to B, 67, 27, −37. (b) are direction ratios; (c) is directed from B to A; (a) divides by the sum of the absolute values.

Part 3 · 2 marks

Find the acute angle between the rope AB and the pole.

Answer and marking scheme

The pole is vertical, along k̂ (direction ratios 0, 0, 1). AB has direction ratios 6, 2, −3 and magnitude 7. cos θ = |−3|7 × 1 = 37, so θ = cos⁻¹(37).

pole along 0, 0, 1 and AB along 6, 2, −3 (1)
θ = cos^{−1}(\frac{3}{7}) (1)

Show that the ropes AB and AC are perpendicular, and write the vector equation of the line AC.

Answer and marking scheme

AC has direction ratios −3, 6, −2. 6(−3) + 2(6) + (−3)(−2) = −18 + 12 + 6 = 0, so AB ⟂ AC. AC: →r = 2î + 3ĵ + 6k̂ + μ(−3î + 6ĵ − 2k̂).

a₁a₂ + b₁b₂ + c₁c₂ = 0 for AB and AC (1)
→r = 2î + 3ĵ + 6k̂ + μ(−3î + 6ĵ − 2k̂) (1)

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