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STD 12 - 11. three dimension geometry — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 11. three dimension geometry4 Marks
MCQ
If the shortest distance between the lines $\frac{x-\lambda}{-2}=\frac{y-2}{1}=\frac{z-1}{1}$ and $\frac{x-\sqrt{3}}{1}=\frac{y-1}{-2}=\frac{z-2}{1}$ is $1 ,$ then the sum of all possible values of $\lambda$ is :
  • A
    $0$
  • ✓
     $2 \sqrt{3}$
  • C
    $3 \sqrt{3}$
  • D
    $-2 \sqrt{3}$
✓

Answer

Correct option: B.
 $2 \sqrt{3}$
b
Passing points of lines $\mathrm{L}_1 \& \mathrm{~L}_2$ are $(\lambda, 2,1) \&(\sqrt{3}, 1,2)$

$(\lambda, 2,1) \&(\sqrt{3}, 1,2)$

$S.D$ $=\frac{\left|\begin{array}{ccc}\sqrt{3}-\lambda & -1 & 1 \\ -2 & 1 & 1 \\ 1 & -2 & 1\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -2 & 1 & 1 \\ 1 & -2 & 1\end{array}\right|}$

$\begin{aligned} & 1=\left|\frac{\sqrt{3}-\lambda}{\sqrt{3}}\right| \\ & \lambda=0, \lambda=2 \sqrt{3}\end{aligned}$

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