Let the foot of the perpendicular from the point $(1,2,4)$ on the line $\frac{x+2}{4}=\frac{y-1}{2}=\frac{z+1}{3}$ be $P$. Then the distance of P from the plane $3x+4y+12z+23=0$

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Three Dimensional Geometry

Let the foot of the perpendicular from the point

(1, 2, 4)

on the line

\frac{x + 2}{4} = \frac{y - 1}{2} = \frac{z + 1}{3}

P

. Then the distance of P from the plane

3 x + 4 y + 12 z + 23 = 0

Answer: (A)

Let the line be parameterized as

\frac{x + 2}{4} = \frac{y - 1}{2} = \frac{z + 1}{3} = λ

, so a general point on the line is

P (4 λ - 2, 2 λ + 1, 3 λ - 1)

.

The direction vector of the line is

b = 4 \hat{i} + 2 \hat{j} + 3 \hat{k}

.

The vector from

(1, 2, 4)

P

A P = (4 λ - 3) \hat{i} + (2 λ - 1) \hat{j} + (3 λ - 5) \hat{k}

.

For

P

to be the foot of perpendicular,

A P \cdot b = 0

4 (4 λ - 3) + 2 (2 λ - 1) + 3 (3 λ - 5) = 0

16 λ - 12 + 4 λ - 2 + 9 λ - 15 = 0

29 λ = 29 \Rightarrow λ = 1

Thus

P = (2, 3, 2)

.

Distance from

P

to plane

3 x + 4 y + 12 z + 23 = 0

is:

d = \frac{∣3 ( 2 ) + 4 ( 3 ) + 12 ( 2 ) + 23∣}{9 + 16 + 144} = \frac{∣6 + 12 + 24 + 23∣}{169} = \frac{65}{13} = 5

JEE Main · 2022 · 27 Jun 2022 · Shift 2 · Single correct option