JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Three Dimensional Geometry

Let N be the foot of perpendicular from the point P

(1, - 2, 3)

on the line passing through the points

(4, 5, 8)

and

(1, - 7, 5)

. Then the distance of N from the plane

2 x - 2 y + z + 5 = 0

Answer: (C)

Solution

The line through

(4, 5, 8)

and

(1, - 7, 5)

has direction vector

(3, 12, 3)

and can be written as:

\frac{x - 4}{3} = \frac{y - 5}{12} = \frac{z - 8}{3} = λ

So a general point on the line is

N (3 λ + 4, 12 λ + 5, 3 λ + 8)

.

The vector

P N = (3 λ + 3, 12 λ + 7, 3 λ + 5)

must be perpendicular to the direction vector

(3, 12, 3)

3 (3 λ + 3) + 12 (12 λ + 7) + 3 (3 λ + 5) = 0

9 λ + 9 + 144 λ + 84 + 9 λ + 15 = 0

162 λ + 108 = 0 \Rightarrow λ = - \frac{2}{3}

N (2, - 3, 6)

.

The distance from N to the plane

2 x - 2 y + z + 5 = 0

is:

d = \frac{∣2 ( 2 ) - 2 ( - 3 ) + 6 + 5∣}{4 + 4 + 1} = \frac{∣4 + 6 + 6 + 5∣}{3} = \frac{21}{3} = 7

JEE Main · 2023 · 13 Apr 2023 · Shift 2 · Single correct option