JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Three Dimensional Geometry

Let the line passing through the points,

P (2, - 1, 2)

and

Q (5, 3, 4)

meet the plane

x - y + z = 4

at the point R. Then the distance of the point R from the plane

x + 2 y + 3 z + 2 = 0

measured parallel to the line

\frac{x - 7}{2} = \frac{y + 3}{2} = \frac{z - 2}{1}

is equal to

Answer: (D)

Solution

The line through P(2,-1,2) and Q(5,3,4) has direction ratios (3,4,2). Its equation is:

\frac{x - 5}{3} = \frac{y - 3}{4} = \frac{z - 4}{2} = λ

Point on line:

R (3 λ + 5, 4 λ + 3, 2 λ + 4)

. Substituting into plane

x - y + z = 4

(3 λ + 5) - (4 λ + 3) + (2 λ + 4) = 4

λ + 6 = 4 \Rightarrow λ = - 2

R \equiv (- 1, - 5, 0)

.

The line parallel to

\frac{x - 7}{2} = \frac{y + 3}{2} = \frac{z - 2}{1}

through R is:

\frac{x + 1}{2} = \frac{y + 5}{2} = \frac{z}{1} = μ

Point T on this line:

(2 μ - 1, 2 μ - 5, μ)

. Substituting into plane

x + 2 y + 3 z + 2 = 0

(2 μ - 1) + 2 (2 μ - 5) + 3 μ + 2 = 0

9 μ - 9 = 0 \Rightarrow μ = 1

T = (1, - 3, 1)

.

Distance

R T = (1 + 1)^{2} + (- 3 + 5)^{2} + (1 - 0)^{2} = 4 + 4 + 1 = 3

JEE Main · 2023 · 11 Apr 2023 · Shift 2 · Single correct option