Verified Solution • 2025–26 Session moderate

QJEEM25Ma242c: JEE Main 2025 Mathematics - Class 12 JEEM25Ma242c Solution: Let the line passing through the points and parallel to the…

Let the line passing through the points (−1,2,1)(-1,2,1) and parallel to the line x−12=y+13=z4\frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4} intersect the line x+23=y−32=z−41\frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1} at the point PP . Then the distance of PP from the point Q(4,−5,1)Q(4,-5,1) is

Multiple Choice Options

  • A 565 \sqrt{6}
  • B 55
  • C 555 \sqrt{5}
  • D 1010

Step-by-Step Verified Solution

Equation of line through point (−1,2,1)(-1,2,1) is →\rightarrow

JEE Main 2025 (Online) 24th January Morning Shift Mathematics - 3D Geometry Question 4 English Explanation

amp;⇒x+12=y−23=z−14−(2)=λamp; So, [x=2λ−1y=3λ+2z=4λ+1\begin{aligned}&\Rightarrow \frac{x+1}{2}=\frac{y-2}{3}=\frac{z-1}{4}-(2)=\lambda\\&\text { So, }\left[\begin{array}{l}x=2 \lambda-1 \\y=3 \lambda+2 \\z=4 \lambda+1\end{array}\right.\end{aligned}

By (1)→x+23=y−32=z−41=μ((1) \rightarrow \frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}=\mu( Let ))

So, [x=3μ−2y=2μ+3z=μ+4\left[\begin{array}{l}x=3 \mu-2 \\ y=2 \mu+3 \\ z=\mu+4\end{array}\right.

For intersection point 'P'

amp;x=2λ−1=3μ−2amp;y=3λ+2=2μ+3[λ=1μ=1]amp;z=4λ+1=μ+4amp; So, point P(x,y,z)=(1,5,5)amp;&Q(4,−5,1)amp;∴PQ=9+100+16amp;=125=55\begin{aligned}& \mathrm{x}=2 \lambda-1=3 \mu-2 \\& \mathrm{y}=3 \lambda+2=2 \mu+3 \quad\left[\begin{array}{l}\lambda=1 \\\mu=1\end{array}\right] \\& \mathrm{z}=4 \lambda+1=\mu+4 \\& \text { So, point } \mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(1,5,5) \\& \& \mathrm{Q}(4,-5,1) \\& \therefore \mathrm{PQ}=\sqrt{9+100+16} \\& =\sqrt{125}=5 \sqrt{5}\end{aligned}

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