JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let the plane $P: 8x + a_1y + a_2z + 12 = 0$ be parallel to the line $L: \frac{x + 2}{2} = \frac{y - 3}{3} = \frac{z + 4}{5}$ . If the intercept of $P$ on the $y$ -axis is $1$ , then the distance between $P$ and $L$ is:

Choose the correct answer:

Explanation

Solution:

Y-intercept: At $y$ -axis, $x=0, z=0$ . So $a_1(1) + 12 = 0 \implies a_1 = -12$ .
Parallel Condition: Normal vector of plane $\vec{n} = (8, a_1, a_2)$ is perpendicular to line direction $\vec{v} = (2, 3, 5)$ .

$8(2) + a_1(3) + a_2(5) = 0 \implies 16 - 36 + 5a_2 = 0 \implies 5a_2 = 20 \implies a_2 = 4$ .
Plane P: $8x - 12y + 4z + 12 = 0 \implies 2x - 3y + z + 3 = 0$ .
Distance: Point on line $L$ is $(-2, 3, -4)$ .

$d = \frac{|2(-2) - 3(3) + 1(-4) + 3|}{\sqrt{2^2 + (-3)^2 + 1^2}} = \frac{|-4 - 9 - 4 + 3|}{\sqrt{14}} = \frac{14}{\sqrt{14}} = \sqrt{14}$ .

Correct Option: (4)