Three Dimensional Geometry Question 1
Question 1 - 2024 (01 Feb Shift 1)
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 :
(1) 0
(2) $2 \sqrt{3}$
(3) $3 \sqrt{3}$
(4) $-2 \sqrt{3}$
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Answer (2)
Solution
Passing points of lines $\mathrm{L}{1} & \mathrm{~L}{2}$ are $(\lambda, 2,1) &(\sqrt{3}, 1,2)$
$\mathrm{S} . \mathrm{D}=\frac{\left|\begin{array}{ccc}\sqrt{3}-\lambda & -1 & 1 \ -2 & 1 & 1 \ 1 & -2 & 1\end{array}\right|}{\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \ -2 & 1 & 1 \ 1 & -2 & 1\end{array}\right| \text { th }}$
$1=\left|\frac{\sqrt{3}-\lambda}{\sqrt{3}}\right|$
$\lambda=0, \lambda=2 \sqrt{3}$
