JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Three Dimensional Geometry

The perpendicular distance, of the line

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

from the point

P (2, - 10, 1)

, is:

(A)6
(B) $52$
(C) $35$ ✓ Correct
(D) $43$

Answer: (C)

Solution

The line

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

has direction vector

d = (2, - 1, 2)

and passes through

A_{0} = (1, - 2, - 3)

A_{0} P = (2 - 1, - 10 - (- 2), 1 - (- 3)) = (1, - 8, 4) .

Perpendicular distance from P to the line:

d = \frac{A _{0} P \times d}{∣ d ∣} .

A_{0} P \times d = \hat{i} 12 \hat{j} - 8 - 1 \hat{k} 42 = \hat{i} (- 16 + 4) - \hat{j} (2 - 8) + \hat{k} (- 1 + 16) = (- 12, 6, 15) .

∣ (- 12, 6, 15) ∣ = 144 + 36 + 225 = 405 = 95 .

Since

∣ d ∣ = 4 + 1 + 4 = 3

d = \frac{9 5}{3} = 35 .

JEE Main · 2025 · 22 Jan 2025 · Shift 2 · Single correct option

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