Three Dimensional Geometry: 3 marks Questions (CBSE Class 12)

2 different 3 marks questions on Three Dimensional Geometry from CBSE Class 12 Maths board exams 2026, newest first.

Find a point on the line $\frac{x - 2}{3} = \frac{1 - y}{2} = \frac{z - 3}{2}$ at a distance of $2$ units from the point (1, 2, 3).

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Answer: (2, 1, 3) or

(\frac{4}{17}, \frac{37}{17}, \frac{31}{17})

Write the line as $\frac{x - 2}{3} = \frac{y - 1}{- 2} = \frac{z - 3}{2} = λ$ ; a general point is $(3 λ + 2, 1 - 2 λ, 2 λ + 3)$ .
Distance from (1, 2, 3): $(3 λ + 1)^{2} + (- 1 - 2 λ)^{2} + (2 λ)^{2} = 2$ .
$17 λ^{2} + 10 λ = 0 \Rightarrow λ = 0$ or $λ = - \frac{10}{17}$ .
Points: $(2, 1, 3)$ and $(\frac{4}{17}, \frac{37}{17}, \frac{31}{17})$ .

Find the shortest distance between the lines
$r = (4 + λ) i^+ (2 λ - 1) j^- 3 λ k^$
$r = (1 + 2 μ) i^+ (2 - 5 μ) k^+ (4 μ - 1) j^$ .

Show answer & solution

Answer:

\frac{6}{5}

units

Line 1: $a_{1} = 4 i^- j^$ , $b_{1} = i^+ 2 j^- 3 k^$ .
Line 2: $a_{2} = i^- j^+ 2 k^$ , $b_{2} = 2 i^+ 4 j^- 5 k^$ .
$b_{1} \times b_{2} = 2 i^- j^$ , $∣ b_{1} \times b_{2} ∣ = 5$ .
$a_{2} - a_{1} = - 3 i^+ 2 k^$ ; $(a_{2} - a_{1}) \cdot (b_{1} \times b_{2}) = - 6$ .
SD $= \frac{∣ - 6∣}{5} = \frac{6}{5}$ units.

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