JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Three Dimensional Geometry

The direction ratios of normal to the plane through the points

(0, - 1, 0)

and

(0, 0, 1)

and making an angle

\frac{π}{4}

with the plane

y - z + 5 = 0

(A) $2, - 1, 1$
(B) $2, 2, - 2$
(C) $2, 1, - 1$ ✓ Correct
(D) $23, 1, - 1$

Answer: (C)

Solution

The plane passes through

(0, - 1, 0)

and

(0, 0, 1)

, so the vector

(0, 1, 1)

lies in it and is perpendicular to the normal

(a, b, c)

, giving

b + c = 0

. The angle with

y - z + 5 = 0

, whose normal is

(0, 1, - 1)

, is

\frac{π}{4}

, so

cos \frac{π}{4} = \frac{∣ b - c ∣}{a ^{2} + b ^{2} + c ^{2} 2} = \frac{1}{2}

, which gives

a^{2} + b^{2} + c^{2} = 2 (b - c)^{2}

. With

c = - b

this is

a^{2} = 2 b^{2}

, so

a = \pm 2 b

, and taking

b = 1, c = - 1

gives direction ratios

2, 1, - 1

JEE Main · 2019 · 11 Jan 2019 · Shift 1 · Single correct option

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