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QJEEM24Ma5f8b: JEE Main 2024 Mathematics - Class 12 JEEM24Ma5f8b Solution: A line with direction ratios meets the lines and respectively…

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

Step-by-Step Verified Solution

Let P(t,t−2,t)\mathrm{P}(\mathrm{t}, \mathrm{t}-2, \mathrm{t}) and Q(2 s−2, s, s)\mathrm{Q}(2 \mathrm{~s}-2, \mathrm{~s}, \mathrm{~s})

D.R's of PQ are 2, 1, 2

amp;2 s−2−t2=s−t+21=s−t2amp;⇒t=6 and s=2amp;⇒P(6,4,6) and Q(2,2,2)amp;PQ:x−22=y−21=z−22=λ\begin{aligned}& \frac{2 \mathrm{~s}-2-\mathrm{t}}{2}=\frac{\mathrm{s}-\mathrm{t}+2}{1}=\frac{\mathrm{s}-\mathrm{t}}{2} \\& \Rightarrow \mathrm{t}=6 \text { and } \mathrm{s}=2 \\& \Rightarrow \mathrm{P}(6,4,6) \text { and } \mathrm{Q}(2,2,2) \\& \mathrm{PQ}: \frac{\mathrm{x}-2}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-2}{2}=\lambda\end{aligned}

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}

JEE Main 2024 (Online) 29th January Morning Shift Mathematics - 3D Geometry Question 45 English Explanation

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