JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Three Dimensional Geometry

The equation of the plane passing through the point

(1, 2, - 3)

and perpendicular to the planes

3 x + y - 2 z = 5

and

2 x - 5 y - z = 7

, is:

(A) $6 x - 5 y + 2 z + 10 = 0$
(B) $3 x - 10 y - 2 z + 11 = 0$
(C) $11 x + y + 17 z + 38 = 0$ ✓ Correct
(D) $6 x - 5 y - 2 z - 2 = 0$

Answer: (C)

Solution

The required plane is perpendicular to both given planes, so its normal vector is parallel to the cross product of the normals of the two given planes.

Normal to first plane:

n_{1} = (3, 1, - 2)

Normal to second plane:

n_{2} = (2, - 5, - 1)

Cross product:

n = n_{1} \times n_{2} = i 32 j 1 - 5 k - 2 - 1 = i (- 1 - 10) - j (- 3 + 4) + k (- 15 - 2)

= - 11 i - j - 17 k = (- 11, - 1, - 17)

Or equivalently

(11, 1, 17)

.

Equation of plane through

(1, 2, - 3)

with normal

(11, 1, 17)

11 (x - 1) + 1 (y - 2) + 17 (z + 3) = 0

11 x - 11 + y - 2 + 17 z + 51 = 0

11 x + y + 17 z + 38 = 0

JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Single correct option

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JEE Main · 2021 · 24 Feb 2021 · Shift 1 · Three Dimensional Geometry

Solution

More JEE Main questions on Three Dimensional Geometry