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JEE Main 202524 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Three Dimensional Geometry

Let the line passing through the points (-1,2,1) and parallel to...

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

Options

  1. A55
  2. B555 \sqrt{5}
  3. C565 \sqrt{6}
  4. D1010

Correct answer

B. 555 \sqrt{5}

Step-by-step solution

Equation of line through point (−1,2,1)(-1,2,1) is →\rightarrow⇒x+12=y−23=z−14=λ−(2)So, [x=2λ−1y=3λ+2z=4λ+1\begin{aligned} & \Rightarrow \frac{x+1}{2}=\frac{y-2}{3}=\frac{z-1}{4}=\lambda-(2) \\ & \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(\operatorname{By}(1) \rightarrow \frac{x+2}{3}=\frac{y-3}{2}=\frac{z-4}{1}=\mu( Let )) So, \left[x=3μ−2y=2μ+3z=μ+4\begin{array}{l}x=3 \mu-2 \\ y=2 \mu+3 \\ z=\mu+4\end{array}\right. For intersection point ' P ' x=2λ−1=3μ−2y=3λ+2=2μ+3z=4λ+1=μ+4\begin{aligned} & \mathrm{x}=2 \lambda-1=3 \mu-2 \\ & \mathrm{y}=3 \lambda+2=2 \mu+3 \\ & \mathrm{z}=4 \lambda+1=\mu+4 \end{aligned} So, point P(x,y,z)=(1,5,5)\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})=(1,5,5) & Q(4,−5,1)\mathrm{Q}(4,-5,1)∴PQ=9+100+16=125=55\begin{aligned} & \therefore P Q=\sqrt{9+100+16} \\ & =\sqrt{125}=5 \sqrt{5} \end{aligned}$$
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