JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let the image of the point $P(2, -1, 3)$ in the plane $x + 2y - z = 0$ be $Q$ . Then the distance of the plane $3x + 2y + z + 29 = 0$ from the point $Q$ is

Choose the correct answer:

Explanation

Solution

Step 1: Find the Coordinates of Point $Q$ .

Using the image formula $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \frac{-2(ax_1 + by_1 + cz_1 + d)}{a^2 + b^2 + c^2}$ :

$\frac{x-2}{1} = \frac{y+1}{2} = \frac{z-3}{-1} = \frac{-2(2 + 2(-1) - 3)}{1^2 + 2^2 + (-1)^2} = \frac{-2(-3)}{6} = 1$ .

$x-2 = 1 \implies x = 3$ ; $y+1 = 2 \implies y = 1$ ; $z-3 = -1 \implies z = 2$ .

So, $Q = (3, 1, 2)$ .

Step 2: Find Distance from $Q$ to the Second Plane.

Plane: $3x + 2y + z + 29 = 0$ .

$d = \frac{|3(3) + 2(1) + 2 + 29|}{\sqrt{3^2 + 2^2 + 1^2}} = \frac{|9 + 2 + 2 + 29|}{\sqrt{14}} = \frac{42}{\sqrt{14}}$ .

$\frac{42}{\sqrt{14}} = \frac{3 \times 14}{\sqrt{14}} = 3\sqrt{14}$ .

Correct Option: (3)

Choose the correct answer:

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