JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Three Dimensional Geometry

A plane which bisects the angle between the two given planes

2 x - y + 2 z - 4 = 0

and

x + 2 y + 2 z - 2 = 0

, passes through the point:

(A) $(1, 4, - 1)$
(B) $(2, - 4, 1)$ ✓ Correct
(C) $(2, 4, 1)$
(D) $(1, - 4, 1)$

Answer: (B)

Solution

Both given planes have normals of length

3

, so their bisecting planes are

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

. These give

x - 3 y - 2 = 0

and

3 x + y + 4 z - 6 = 0

. Testing

(2, - 4, 1)

in the second,

3 (2) + (- 4) + 4 (1) - 6 = 0

, so a bisecting plane passes through

(2, - 4, 1)

JEE Main · 2019 · 12 Apr 2019 · Shift 2 · Single correct option

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