Mathematics · 3D Geometry

JEE Main 2024 — 29 January, Shift 1 — Question 29

A line with direction ratios 2,1,22,1,2 meets the lines x=y+2=z\mathrm{x}=\mathrm{y}+2=\mathrm{z} and x+2=2y=2z\mathrm{x}+2=2 \mathrm{y}=2 \mathrm{z} respectively at the point P and Q . if the length of the perpendicular from the point (1,2,12)(1,2,12) to the line PQ is ll, then l2l^{2} is

Answer: 65

Numerical answer — enter this value.

Step-by-step solution

Let P(t,t−2,t)P(t, t-2, t) and Q(2s−2,s,s)Q(2 s-2, s, s) D.R's of PQ are 2,1,22, 1, 2

2 s−2−t2=s−t+21=s−t2\frac{2 \mathrm{~s}-2-\mathrm{t}}{2}=\frac{\mathrm{s}-\mathrm{t}+2}{1}=\frac{\mathrm{s}-\mathrm{t}}{2}

⇒t=6\Rightarrow \mathrm{t}=6 and s=2\mathrm{s}=2

⇒P(6,4,6)\Rightarrow \mathrm{P}(6,4,6) and Q(2,2,2)\mathrm{Q}(2,2,2)

PQ:x−22=y−21=z−22=λ\mathrm{PQ}: \frac{\mathrm{x}-2}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-2}{2}=\lambda

Let F(2λ+2,λ+2,2λ+2)\mathrm{F}(2 \lambda+2, \lambda+2,2 \lambda+2)

A(1,2,12)\mathrm{A}(1,2,12)

AF→⋅PQ→=0\overrightarrow{\mathrm{AF}} \cdot \overrightarrow{\mathrm{PQ}}=0

∴λ=2\therefore \lambda=2

So F(6,4,6)F(6,4,6) and AF=65A F=\sqrt{65}

figure

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
3D Geometry
Topic
Intersection of lines, line & plane.
A line with direction ratios 2,1,2 meets the lines x = y +2= z and x… | JEE Main 2024 PYQ with Solution · DhiX AI