If the foot of the perpendicular from the point $\mathrm{A}(-1,4,3)$ on the plane $\mathrm{P}: 2\mathrm{x}+\mathrm{my}+\mathrm{nz}=4$, is $\left(-2, \frac{7}{2}, \frac{3}{2}\right)$, then the distance of the point A from the plane P, measured parallel to a line with direction ratios $3,-1,-4$, is equal to :

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Three Dimensional Geometry

If the foot of the perpendicular from the point

A (- 1, 4, 3)

on the plane

P : 2 x + my + nz = 4

, is

(- 2, \frac{7}{2}, \frac{3}{2})

, then the distance of the point A from the plane P, measured parallel to a line with direction ratios

3, - 1, - 4

, is equal to :

Answer: (B)

The direction ratios of AB (perpendicular to plane) are

⟨ - 1, - \frac{1}{2}, - \frac{3}{2} ⟩

⟨ 2, 1, 3 ⟩

.

Since AB is perpendicular to the plane

2 x + m y + n z = 4

, we have

m = 1, n = 3

.

The line through A with direction ratios

⟨ 3, - 1, - 4 ⟩

is:

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

Point C on this line:

(3 λ - 1, - λ + 4, - 4 λ + 3)

C lies on plane

2 x + y + 3 z = 4

6 λ - 2 - λ + 4 - 12 λ + 9 = 4

λ = 1

C = (2, 3, - 1)

and

A C = 9 + 1 + 16 = 26

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Single correct option