Question

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 NOT possible?

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

Answer 1

Answer: (A)

$\vec{ b _1} \times \vec{ b _2}=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ -2 & 0 & 1 \\ 1 & 1 & -1\end{vmatrix}=-\hat{ i }-\hat{ j }-2 \hat{ k }$

$ \vec{ a _2}-\vec{ 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$