Three Dimensional Geometry Question 25
Question 25 - 31 January - Shift 1
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?
(1) $\alpha+2 \gamma=24$
(2) $2 \alpha+\gamma=7$
(3) $2 \alpha-\gamma=9$
(4) $\alpha-2 \gamma=19$
Show Answer
Answer: (1)
Solution:
Formula: Skew Lines
$\overrightarrow{{}b_1} \times \overrightarrow{{}b_2}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -2 & 0 & 1 \\ 1 & 1 & -1\end{vmatrix} =-\hat{i}-\hat{j}-2 \hat{k}$
$\overrightarrow{{}a_2}-\overrightarrow{{}a_1}=6 \hat{i}+(\lambda-1) \hat{j}+(-\lambda-4) \hat{k}$
$2 \sqrt{6}=|\frac{-6-\lambda+1+2 \lambda+8}{\sqrt{1+1+4}}|$
$|\lambda+3|=12 \Rightarrow \lambda=9,-15$
$\alpha=-2 k+5, \gamma=k-\lambda$ where $k \in R$
$\Rightarrow \alpha+2 \gamma=5-2 \lambda=-13,35$
