JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Three Dimensional Geometry

Let a line with direction ratios a,

- 4 a, - 7

be perpendicular to the lines with direction ratios 3,

- 1, 2 b

and

b, a, - 2

. If the point of intersection of the line

\frac{x + 1}{a ^{2} + b ^{2}} = \frac{y - 2}{a ^{2} - b ^{2}} = \frac{z}{1}

and the plane

x - y + z = 0

(α, β, γ)

, then

α + β + γ

is equal to
____ .

Answer: 10

Solution

Perpendicularity conditions:

(a, - 4 a, - 7) \cdot (3, - 1, 2 b) = 0 \Rightarrow 3 a + 4 a - 14 b = 0 \Rightarrow 7 a = 14 b \Rightarrow a = 2 b

(a, - 4 a, - 7) \cdot (b, a, - 2) = 0 \Rightarrow ab - 4 a^{2} + 14 = 0

Substitute

a = 2 b

2 b^{2} - 16 b^{2} + 14 = 0 \Rightarrow - 14 b^{2} = - 14 \Rightarrow b^{2} = 1

a^{2} = 4

.

The line is

\frac{x + 1}{5} = \frac{y - 2}{3} = \frac{z}{1} = k

, giving

(x, y, z) = (5 k - 1, 3 k + 2, k)

.

Substitute into the plane equation:

(5 k - 1) - (3 k + 2) + k = 0 \Rightarrow 3 k - 3 = 0 \Rightarrow k = 1

(α, β, γ) = (4, 5, 1)

, and

α + β + γ = 10

JEE Main · 2022 · 29 Jul 2022 · Shift 1 · Integer answer