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JEE Main 8-April-2025 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 8-April-2025 Paper - Shift 24 Marks
MCQ
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 $\frac{x-\lambda}{3}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{6}}$ be $\lambda_{1}$ 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 __________
  • ✓
    $\frac{5 \sqrt{2}}{3}$
  • B
    4
  • C
    $\frac{\sqrt{2}}{3}$
  • D
    3
✓

Answer

Correct option: A.
$\frac{5 \sqrt{2}}{3}$
(A) $\frac{5 \sqrt{2}}{3}$
$\quad \vec{p}=2 \hat{i}+3 \hat{j}+4 \hat{k}, \vec{q}=3 \hat{i}+4 \hat{j}+5 \hat{k}$
$\Rightarrow \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{q}}=\left|\begin{array}{ccc}\hat{\mathrm{i}} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}$
$\mathrm{A} \equiv(1,2,3) \mathrm{B} \equiv(\lambda, 4,5)$
Shortest Distance $=\left|\frac{\overrightarrow{\mathrm{AB}} \cdot(\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{q}})}{|\overrightarrow{\mathrm{P}} \times \overrightarrow{\mathrm{q}}|}\right|$
$\frac{1}{\sqrt{6}}=\left|\frac{((\lambda-1) \hat{i}+2 \hat{j}+2 \hat{k}) \cdot(-\hat{i}+2 \hat{j}-\hat{k})}{\sqrt{6}}\right|$
$\Rightarrow|-\lambda+1+4-2|=1 \Rightarrow|\lambda-3|=1$
$\Rightarrow \lambda=3 \pm 1=4,2$
Radius of circle passing through points $(0,0),(4,2) \&(2,4)$
$=\frac{\mathrm{abc}}{4 \Delta}=\frac{\sqrt{20} \times \sqrt{20} \times \sqrt{8}}{4 \times \frac{1}{2}\left|\begin{array}{lll}1 & 1 & 1 \\ 0 & 4 & 2 \\ 0 & 2 & 4\end{array}\right|}=\frac{20 \times 2 \sqrt{2}}{2 \times 12}$
$=\frac{5 \sqrt{2}}{3}$

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