JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Three Dimensional Geometry

Let P be a plane passing through the points

(2, 1, 0), (4, 1, 1)

and

(5, 0, 1)

and R be any point

(2, 1, 6)

. Then the image of R in the plane P is:

(A) $(6, 5, 2)$
(B) $(4, 3, 2)$
(C) $(6, 5, - 2)$ ✓ Correct
(D) $(3, 4, - 2)$

Answer: (C)

Solution

A plane through

(2, 1, 0), (4, 1, 1), (5, 0, 1)

has normal

n = (2, 0, 1) \times (3, - 1, 1) = (1, 1, - 2)

, giving

x + y - 2 z = 3

. For

R (2, 1, 6)

\frac{n \cdot R - 3}{∣ n ∣ ^{2}} = \frac{2 + 1 - 12 - 3}{6} = - 2

, so the image is

R - 2 (- 2) n = R + 4 n = (2 + 4, 1 + 4, 6 - 8) = (6, 5, - 2)

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Single correct option

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