JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

The distance of the point $(7, -3, -4)$ from the plane passing through the points $(2, -3, 1)$ , $(-1, 1, -2)$ and $(3, -4, 2)$ is :

Choose the correct answer:

Explanation

Step 1: Plane ki equation nikalna

Teen points $A(2, -3, 1)$ , $B(-1, 1, -2)$ aur $C(3, -4, 2)$ se guzarne wale plane ki equation determinant method se nikali jati hai:

\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0

Value rakhne par:

\begin{vmatrix} x-2 & y+3 & z-1 \\ -1-2 & 1+3 & -2-1 \\ 3-2 & -4+3 & 2-1 \end{vmatrix} = 0 \implies \begin{vmatrix} x-2 & y+3 & z-1 \\ -3 & 4 & -3 \\ 1 & -1 & 1 \end{vmatrix} = 0

Solve karne par:

$(x-2)(4-3) - (y+3)(-3+3) + (z-1)(3-4) = 0$

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

$x - 2 - z + 1 = 0$

Plane ki equation: $x - z - 1 = 0$

Step 2: Point $(7, -3, -4)$ se distance nikalna

Distance formula: $d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$

$d = \frac{|1(7) + 0(-3) - 1(-4) - 1|}{\sqrt{1^2 + 0^2 + (-1)^2}}$

$d = \frac{|7 + 4 - 1|}{\sqrt{1 + 1}}$

$d = \frac{10}{\sqrt{2}} = 5\sqrt{2}$

Sahi Option: (C) $5\sqrt{2}$

Choose the correct answer:

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