Question

If the shortest distance between the lines $\overrightarrow{r_1}=\alpha \hat{i}+2\hat{j}+2\hat{k}+\lambda (\hat{i}-2\hat{j}+2\hat{k}),\lambda \in R,\alpha >0$ and $\overrightarrow{r_2}=-4\hat{i}-\hat{k}+\mu (3\hat{i}-2\hat{j}-2\hat{k}),\mu \in R$ is $9,$ then $\alpha$ is equal to

Answer 1

Let $\overrightarrow{r_1}=\overrightarrow{a_1}+\lambda \overrightarrow{b_1},\overrightarrow{r_2}=\overrightarrow{a_2}+\mu \overrightarrow{b_2}$
$\overrightarrow{a_1}-\overrightarrow{a_2}=(\alpha +4)\hat{i}+2\hat{j}+3\hat{k}$
$\overrightarrow{b_1}\times \overrightarrow{b_2}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -2& 2\\ 3& -2& -2\end{array}\right|=8\hat{i}+8\hat{j}+4\hat{k}\overrightarrow{b_1}\times \overrightarrow{b_2}=4(2\hat{i}+2\hat{j}+\hat{k})$

Shortest distance $\left(d\right)$ $=\left|\frac{(\overrightarrow{a_1}-\overrightarrow{a_2})\cdot (\overrightarrow{b_1}\times \overrightarrow{b_2})}{|\overrightarrow{b_1}\times \overrightarrow{b_2}|}\right|$
$=\left|\frac{2(\alpha +4)+4+3}{3}\right|=9$
$\Rightarrow (2\alpha +15)=27$
$\Rightarrow \alpha =6$