JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Vector Algebra

A vector v in the first octant is inclined to the x-axis at

6 0^{\circ}

, to the y-axis at

4 5^{\circ}

and to the z-axis at an acute angle. If a plane passing through the points

(2, - 1, 1)

and

(a, b, c)

, is normal to

v

, then

Answer: (C)

Solution

The direction cosines satisfy

cos^{2} 6 0^{\circ} + cos^{2} 4 5^{\circ} + cos^{2} γ = 1

, so

\frac{1}{4} + \frac{1}{2} + cos^{2} γ = 1

and

cos^{2} γ = \frac{1}{4}

. As

γ

is acute,

cos γ = \frac{1}{2}

, giving

v ∥ (\frac{1}{2}, \frac{1}{2}, \frac{1}{2}) ∥ (1, 2, 1)

. The plane normal to

v

through

(2, - 1, 1)

1 (x - 2) + 2 (y + 1) + 1 (z - 1) = 0

, i.e.

x + 2 y + z = 1

. Since

(a, b, c)

lies on this plane,

a + 2 b + c = 1

JEE Main · 2023 · 30 Jan 2023 · Shift 2 · Single correct option