Let P be the point (10,-2,-1) and Q be the foot of the perpendicular drawn from the point R (1,7,6) on the line passing through the points (2,-5,11) and (-6,7,-5) . Then the length of the line segment PQ is equal to

Mathematics · 3D Geometry

JEE Main 2024 — 6 April, Shift 1 — Question 29

Let $P$ be the point $(10,-2,-1)$ and $Q$ be the foot of the perpendicular drawn from the point $\mathrm{R}(1,7,6)$ on the line passing through the points $(2,-5,11)$ and $(-6,7,-5)$ . Then the length of the line segment PQ is equal to $\qquad$

Answer: 13

Numerical answer — enter this value.

Step-by-step solution

Line : $\frac{x+6}{-8}=\frac{y-7}{12}=\frac{z+5}{-16}$

$\frac{x+6}{2}=\frac{y-7}{-3}=\frac{z+5}{4}=\lambda$

$\mathrm{Q}(2 \lambda-6,7-3 \lambda, 4 \lambda-5)$

$\overline{\mathrm{QR}}(2 \lambda-7,-3 \lambda, 4 \lambda-11)$

$\overline{\mathrm{QR}} \cdot$ dr's of line $=0$

$4 \lambda-14+9 \lambda+16 \lambda-44=0$

$29 \lambda=58 \Rightarrow \lambda=2$

$\mathrm{Q}(-2,1,3)$

$\mathrm{PQ}=\sqrt{144+9+16}=\sqrt{169}=13$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 6 April, Shift 1
Subject: Mathematics
Chapter: 3D Geometry
Topic: Intersection of lines, line & plane.

The shortest distance between the lines x-3/2=y+15/-7=z-9/5 and x+1/2=y-1/1=z-9/-3 is

Answer: 13

Step-by-step solution

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