The distance of the point (7,10,11) from the line (x-4)/(1)=(y-4)/(0)=(z-2)/(3) along the line (x-9)/(2)=(y-13)/(3)=(z-17)/(6) is

Correct answer: B. 14 line P Q is parallel to line (x-9)/(2)=(y-3)/(3)=(z-17)/(6) aligned & (λ-3)/(2)=(-6)/(3)=(3 λ-9)/(6) λ=-1 & Q=(3,4,-1) & P Q= 16+36+144 =14 aligned

JEE Main 20253 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 3 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2025 · 3 April · Evening Shift · Mathematics · Three Dimensional Geometry

The distance of the point (7,10,11) from the line...

The distance of the point

(7,10,11)

from the line

\frac{x-4}{1}=\frac{y-4}{0}=\frac{z-2}{3}

along the line

\frac{x-9}{2}=\frac{y-13}{3}=\frac{z-17}{6}

Options

A $18$
B $14$
C $12$
D $16$

Correct answer

B. $14$

Step-by-step solution

\because

line

P Q

is parallel to line

\frac{x-9}{2}=\frac{y-3}{3}=\frac{z-17}{6}

\begin{aligned}<br/>& \therefore \frac{\lambda-3}{2}=\frac{-6}{3}=\frac{3 \lambda-9}{6} \Rightarrow \lambda=-1 \\ & Q=(3,4,-1) \\ & \therefore P Q=\sqrt{16+36+144}=14<br/>\end{aligned}

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