Let the values of λ for which the shortest distance between the lines (x-1)/(2)=(y-2)/(3)=(z-3)/(4) and (x-λ)/(3)=(y-4)/(4)=(z-5)/(5) is 1 6 be λ_1 and λ_2 . Then the radius of the circle passing through the points

JEE Main 20258 AprilEvening Shift

Exam: JEE Main
Year: 2025
Date: 8 April
Shift: Evening Shift
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Main 2025 · 8 April · Evening Shift · Mathematics · Three Dimensional Geometry

Let the values of λ for which the shortest distance between the...

Let the values of

\lambda

for which the shortest distance between the lines

\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} and \frac{x-\lambda}{3}=\frac{y-4}{4}=\frac{z-5}{5}

\frac{1}{\sqrt{6}}

\lambda_1

and

\lambda_2

. Then the radius of the circle passing through the points

(0,0),\left(\lambda_1, \lambda_2\right)

and \left(\lambda_2, \lambda_1\right) is

Options

A $\frac{5 \sqrt{2}}{3}$
B4
C $\frac{\sqrt{2}}{3}$
D3

Correct answer

A. $\frac{5 \sqrt{2}}{3}$

Step-by-step solution

\begin{aligned} & \overrightarrow{\mathrm{p}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}, \overrightarrow{\mathrm{q}}=3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+5 \hat{\mathrm{k}} \\ & \Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{lll}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}} \\ & \mathrm{A} \equiv(1,2,3) B \equiv(\lambda, 4,5) \\ & \text{ Shortest Distance }=\left|\frac{\overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{q}}|}\right|\end{aligned}

\begin{aligned}& \frac{1}{\sqrt{6}}=\left|\frac{((\lambda-1) \hat{i}+2 \hat{j}+2 \hat{k}) \cdot(-\hat{i}+2 \hat{j}-\hat{k})}{\sqrt{6}}\right| \\ & \Rightarrow|-\lambda+1+4-2|=1 \Rightarrow|\lambda-3|=1 \\ & \Rightarrow \lambda=3 \pm 1=4,2\end{aligned}

Radius of circle passing through points

\begin{aligned}& (0,0),(4,2) &(2,4) \\ & =\frac{\text{ abc }}{4 \Delta}=\frac{\sqrt{20} \times \sqrt{20} \times \sqrt{8}}{4 \times \frac{1}{2}\left|\begin{array}{lll}1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4\end{array}\right|}=\frac{20 \times 2 \sqrt{2}}{2 \times 12} \\ & =\frac{5 \sqrt{2}}{3}\end{aligned}

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