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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Three Dimensional Geometry

Let P(α, β, γ) be the point on the line (x-1)/(2)=(y+1)/(-3)=z at a...

Let P(α,β,γ)\mathrm{P}(\alpha, \beta, \gamma) be the point on the line x−12=y+1−3=z\frac{x-1}{2}=\frac{y+1}{-3}=z at a distance 4144 \sqrt{14} from the point (1,−1,0)(1,-1,0) and nearer to the origin. Then the shortest distance, between the lines x−α1=y−β2=z−γ3\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3} and x+52=y−101=z−31\frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}, is equal to

Options

  1. A4754 \sqrt{\frac{7}{5}}
  2. B2742 \sqrt{\frac{7}{4}}
  3. C7547 \sqrt{\frac{5}{4}}
  4. D4574 \sqrt{\frac{5}{7}}

Correct answer

A. 4754 \sqrt{\frac{7}{5}}

Step-by-step solution

Let P(2λ+1,−3λ−1,λ)P(2\lambda + 1, -3\lambda - 1, \lambda)
Then 4λ2+9λ2+λ2=16⋅14⇒λ=±4⇒−44\lambda^2 + 9\lambda^2 + \lambda^2 = 16 \cdot 14 \Rightarrow \lambda = \pm 4 \Rightarrow -4
(nearer to origin)
∴P(−7,11,−4)\therefore P(-7, 11, -4)
∴\therefore Shortest distance =∣2−17123211∣∣i^j^k∣∣123211∣= \dfrac{\begin{vmatrix} 2 & -1 & 7 \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix}}{\left| \hat{i} \quad \hat{j} \quad k \right| \begin{vmatrix} 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix}}
=281+25+9=475= \dfrac{28}{\sqrt{1 + 25 + 9}} = \dfrac{4\sqrt{7}}{\sqrt{5}}
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