JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Three Dimensional Geometry

The shortest distance between the line

\frac{x - 3}{2} = \frac{y + 15}{- 7} = \frac{z - 9}{5}

and

\frac{x + 1}{2} = \frac{y - 1}{1} = \frac{z - 9}{- 3}

(A) $63$
(B) $83$
(C) $53$
(D) $43$ ✓ Correct

Answer: (D)

Solution

d_{1} \times d_{2} = \hat{i} 22 \hat{j} - 7 1 \hat{k} 5 - 3 = 16 \hat{i} + 16 \hat{j} + 16 \hat{k}

, magnitude

163

a_{2} - a_{1} = (- 4, 16, 0)

, and

(a_{2} - a_{1}) \cdot (d_{1} \times d_{2}) = - 64 + 256 = 192

shortest distance = \frac{192}{16 3} = \frac{12}{3} = 43 .

JEE Main · 2024 · 06 Apr 2024 · Shift 1 · Single correct option

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