If the lines vec r =(hat i -hat j +hat k )+λ(3 hat j -hat k ) and vec r =(α hat i -hat j )+μ(2 hat j -3 hat k ) are co-planar, then the distance of the plane containing these two lines from the point (α, 0,0) is :

Mathematics · 3D Geometry

JEE Main 2025 — 4 April, Evening Shift — Question 31

If the lines $\vec{r}=(\hat{i}-\hat{j}+\hat{k})+\lambda(3 \hat{j}-\hat{k})$ and $\vec{r}=(\alpha \hat{i}-\hat{j})+\mu(2 \hat{j}-3 \hat{k})$ are co-planar, then the

distance of the plane containing these two lines from the point $(\alpha, 0,0)$ is :

Option A:
$\frac{2}{9}$
Option B:
$\frac{2}{11}$
Correct
Option C:
$\frac{4}{11}$
Option D:
$2$

Answer: B

Step-by-step solution

$\because \quad$ Both lines are coplanar, so

\begin{aligned} & \left|\begin{array}{ccc} \alpha-1 & 0 & -1\\ 0 & 3 & -1 \\2 & 0 & -3 \end{array}\right|=0 \\& \Rightarrow \quad \alpha=\frac{5}{3} \end{aligned}

Equation of plane containing both lines

$\left|\begin{array}{ccc} x-1 & y+1 & z-1\\ 0 & 3 & -1\\ 2 & 0 & -3 \end{array}\right|=0$

$\Rightarrow 9 x+2 y+6 z=13$

So, distance of $\left(\frac{5}{3}, 0,0\right)$ from this plane $=\frac{2}{\sqrt{81+4+36}}=\frac{2}{11}$

Answer key and solution verified before publishing.

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Exam: JEE Main 2025
Paper: 4 April, Evening Shift
Subject: Mathematics
Chapter: 3D Geometry
Topic: Skew lines & shortest distance between them

Let A be the point of intersection of the lines L 1: x-7/1=y-5/0=z-3/-1 and L 2: x-1/3=y+3/4=z+7/5 . Let B and C be the points on the lines…

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Step-by-step solution

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