JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Three Dimensional Geometry

The distance, of the point

(7, - 2, 11)

from the line

\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}

along the line

\frac{x - 5}{2} = \frac{y - 1}{- 3} = \frac{z - 5}{6}

, is :

(A)12
(B)14✓ Correct
(C)18
(D)21

Answer: (B)

Solution

Let

A = (7, - 2, 11)

. A point on the line through

A

in the direction of the second line is

B = (7 + 2 λ, - 2 - 3 λ, 11 + 6 λ)

. For

B

to lie on the first line,

\frac{x - 6}{1} = \frac{y - 4}{0} = \frac{z - 8}{3}

, the middle ratio having denominator

0

forces

y - 4 = 0

- 2 - 3 λ - 4 = 0 \Rightarrow λ = - 2 \Rightarrow B = (3, 4, - 1)

A B = (7 - 3)^{2} + (- 2 - 4)^{2} + (11 + 1)^{2} = 16 + 36 + 144 = 196 = 14

JEE Main · 2024 · 27 Jan 2024 · Shift 1 · Single correct option

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