Let the plane P : $8 \mathrm{x}+\alpha_{1} \mathrm{y}+\alpha_{2} \mathrm{z}+12=0$ be parallel to the line $\mathrm{L}: \frac{\mathrm{x}+2}{2}=\frac{\mathrm{y}-3}{3}=\frac{\mathrm{z}+4}{5}$. If the intercept of P on the y-axis is 1, then the distance between P and L is :

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Three Dimensional Geometry

Let the plane P :

8 x + α_{1} y + α_{2} z + 12 = 0

be parallel to the line

L : \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}

. If the intercept of P on the y-axis is 1, then the distance between P and L is :

(A) $14$ ✓ Correct
(B) $\frac{6}{14}$
(C) $\frac{2}{7}$
(D) $\frac{7}{2}$

Answer: (A)

Solution

Parallelism requires the normal

(8, α_{1}, α_{2})

to be perpendicular to the direction

(2, 3, 5)

16 + 3 α_{1} + 5 α_{2} = 0.

y

-intercept of

1

means

(0, 1, 0)

lies on

P

, so

α_{1} + 12 = 0

, i.e.

α_{1} = - 12

, whence

α_{2} = 4.

The plane

8 x - 12 y + 4 z + 12 = 0

reduces to

2 x - 3 y + z + 3 = 0.

The point

(- 2, 3, - 4)

L

is at distance

\frac{∣ - 4 - 9 - 4 + 3∣}{4 + 9 + 1} = \frac{14}{14} = 14

from it.

JEE Main · 2023 · 31 Jan 2023 · Shift 2 · Single correct option

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