Mathematics · Straight lines

JEE Main 2026 — 22 January, Morning Shift — Question 6

Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the point on the line $\frac{x-1}{2}=\frac{y+1}{-3}=z$ at a distance $4 \sqrt{14}$ from the point $(1,-1,0)$ and nearer to the origin. Then the shortest distance, between the lines $\frac{x-\alpha}{1}=\frac{y-\beta}{2}=\frac{z-\gamma}{3}$ and $\frac{x+5}{2}=\frac{y-10}{1}=\frac{z-3}{1}$ , is equal to

Option A:
$7 \sqrt{\frac{5}{4}}$
Option B:
$4 \sqrt{\frac{7}{5}}$
Correct
Option C:
$4 \sqrt{\frac{5}{7}}$
Option D:
$2 \sqrt{\frac{7}{4}}$

Answer: B

Step-by-step solution

Let $\; P(2\lambda + 1,\,-3\lambda - 1,\,\lambda)$

$\text{Then } 4\lambda^2 + 9\lambda^2 + \lambda^2 = 16$ $\Rightarrow 14\lambda^2 = 16$ $\Rightarrow \lambda = \pm \frac{4}{\sqrt{14}}$ $\Rightarrow \lambda = -4 \;$ nearer to origin

\therefore P(-7,\,11,\,-4)

\text{Shortest distance} = \frac{ \left| \begin{vmatrix} 2 & -1 & 7 \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix} \right| }{ \sqrt{ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 3 \\ 2 & 1 & 1 \end{vmatrix}^2 } }

= \frac{28}{\sqrt{1+25+9}} = \frac{28}{\sqrt{35}} = \frac{4\sqrt{7}}{\sqrt{5}}

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 22 January, Morning Shift
Subject: Mathematics
Chapter: Straight lines
Topic: Angle between lines, perpendicular distance & distance between parallel lines, foot, image

If the image of the point P(1,2, a) in the line frac x-63=frac y-72=frac7- z2 is Q(5, b, c) , then a^2+ b^2+ c^2 is equal to

Answer: B

Step-by-step solution

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