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JEE Main 202529 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
29 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2025 · 29 January · Evening Shift · Mathematics · Three Dimensional Geometry

Let a straight line L pass through the point P(2,-1,3) and be...

Let a straight line LL pass through the point P(2,−1,3)P(2,-1,3) and be perpendicular to the lines x−12=y+11=z−3−2\frac{x-1}{2}=\frac{y+1}{1}=\frac{z-3}{-2} and x−31=y−23=z+24\frac{x-3}{1}=\frac{y-2}{3}=\frac{z+2}{4}. If the line LL intersects the yzy z-plane at the point QQ, then the distance between the points PP and QQ is :

Options

  1. A10\sqrt{10}
  2. B232 \sqrt{3}
  3. C2
  4. D3

Correct answer

D. 3

Step-by-step solution

Vector parallel to LL=∣i^j^k^21−2134∣=10i^−10j^+5k^=5(2i^−2j^+k^)\begin{aligned} & =\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{array}\right|=10 \hat{i}-10 \hat{j}+5 \hat{k} \\ & =5(2 \hat{i}-2 \hat{j}+\hat{k}) \end{aligned}Equationof′Equation of 'L′'\frac{x-2}{2}=\frac{y+1}{-2}=\frac{z-3}{1}=\lambda(\text{ say })LetLetQ(2 \lambda+2,-2 \lambda-1, \lambda+3) ⇒2λ+2=0⇒λ=−1⇒Q(0,1,2)d(P,Q)=3\begin{aligned} & \Rightarrow \quad 2 \lambda+2=0 \Rightarrow \lambda=-1 \\ & \Rightarrow Q(0,1,2) \\ & d(P, Q)=3 \end{aligned}$
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