JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Three Dimensional Geometry

If the line

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

intersects the plane

2 x + 3 y - z + 13 = 0

at a point

P

and the plane

3 x + y + 4 z = 16

at a point

Q

, then

P Q

is equal to :

(A) $214$ ✓ Correct
(B)14
(C) $27$
(D) $14$

Answer: (A)

Solution

Write the line in parametric form using the

z

-direction ratio

- 1

x = 3 λ + 2

y = 2 λ - 1

z = 1 - λ

. To locate

P

, put these into

2 x + 3 y - z + 13 = 0

2 (3 λ + 2) + 3 (2 λ - 1) - (1 - λ) + 13 = 13 λ + 13 = 0

, so

λ = - 1

and

P = (- 1, - 3, 2)

. To locate

Q

, put them into

3 x + y + 4 z = 16

3 (3 λ + 2) + (2 λ - 1) + 4 (1 - λ) = 7 λ + 9 = 16

, so

λ = 1

and

Q = (5, 1, 0)

. Hence

P Q = (5 + 1)^{2} + (1 + 3)^{2} + (0 - 2)^{2} = 36 + 16 + 4 = 56 = 214

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option

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