Let the shortest distance between the lines L : x -5 -2 = y - 0 =… - Vidyadip

MCQ

Let the shortest distance between the lines $L : \frac{ x -5}{-2}=\frac{ y -\lambda}{0}=\frac{ z +\lambda}{1}, \lambda \geq 0$ and $L _1: x +1= y -$ $1=4-z$ be $2 \sqrt{6}$. If $(\alpha, \beta, \gamma)$ lies on $L$, then which of the following is $\ce{NOT}$ possible?

A
$\alpha+2 \gamma=24$
B
$2 \alpha+\gamma=7$
C
$2 \alpha-\gamma=9$
D
$\alpha-2 \gamma=19$

✓

Answer

$\overline{ b _1} \times \overline{ b _2}=\left|\begin{array}{ccc}\hat{ i } & \hat{ j } & \hat{ k } \\ -2 & 0 & 1 \\ 1 & 1 & -1\end{array}\right|=-\hat{ i }-\hat{ j }-2 \hat{ k }$
$\overline{ a _2}-\overline{ a _1}=6 \hat{ i }+(\lambda-1) \hat{ j }+(-\lambda-4) \hat{ k }$
$2 \sqrt{6}=\left|\frac{-6-\lambda+1+2 \lambda+8}{\sqrt{1+1+4}}\right|$
$|\lambda+3|=12 $
$\Rightarrow \lambda=9,-15$
$\alpha=-2 k +5, \gamma= k -\lambda \text { where } k \in R$
$\Rightarrow \alpha+2 \gamma$
$=5-2 \lambda$
$=-13,35$

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