JEE 2023 — Mathematics PYQ

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Question

JEE 2023 — Mathematics PYQ

JEE | Mathematics | 2023

Let $N$ be the foot of the perpendicular from the point $P(1, -2, 3)$ on the line passing through the points $(4, 5, 8)$ and $(1, -7, 5)$ . Then, the distance of $N$ from the plane $2x - 2y + z + 5 = 0$ is:

Choose the correct answer:

Explanation

Solution

1. Line ki equation likhna:

Points $A(4, 5, 8)$ aur $B(1, -7, 5)$ se guzarne wali line:

Direction ratios: $(4-1, 5-(-7), 8-5) = (3, 12, 3)$ ya $(1, 4, 1)$ .

Line: $\frac{x-1}{1} = \frac{y+7}{4} = \frac{z-5}{1} = \lambda$

Kisi bhi point $N$ ke coordinates: $( \lambda+1, 4\lambda-7, \lambda+5 )$ .

2. $\lambda$ ki value nikalna:

Vector $\vec{PN}$ line ke perpendicular hai. $P = (1, -2, 3)$ .

$\vec{PN} = ((\lambda+1)-1, (4\lambda-7)-(-2), (\lambda+5)-3) = ( \lambda, 4\lambda-5, \lambda+2 )$ .

Perpendicularity condition ( $\vec{PN} \cdot \vec{v} = 0$ ):

1(\lambda) + 4(4\lambda-5) + 1(\lambda+2) = 0

\lambda + 16\lambda - 20 + \lambda + 2 = 0 \implies 18\lambda = 18 \implies \mathbf{\lambda = 1}

3. Point $N$ aur Distance:

$N = (1+1, 4(1)-7, 1+5) = \mathbf{(2, -3, 6)}$ .

Plane $2x - 2y + z + 5 = 0$ se $N$ ka distance:

d = \frac{|2(2) - 2(-3) + (6) + 5|}{\sqrt{2^2 + (-2)^2 + 1^2}}

d = \frac{|4 + 6 + 6 + 5|}{\sqrt{9}} = \frac{21}{3} = \mathbf{7}

Sahi Option: (2) 7