Question

If the shortest distance between the line joining the points $(1,2,3)$ and $(2,3,4)$, and the line $\frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-2}{0}$ is $\alpha$, then $28{\alpha }^2$ is equal to

Answer 1

Answer: 18

$\vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+\lambda (\hat{i}+\hat{j}+\hat{k})\vec{r}=\vec{a}+\lambda \vec{p}$
$\vec{r}=(+\hat{i}-\hat{j}+2\hat{k})+\mu (2\hat{i}-\hat{j})\vec{r}=\vec{b}+\mu \vec{q}$
$\vec{p}\times \vec{q}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2& -1& 0\end{array}\right|=\hat{i}+2\hat{j}-3\hat{k}$
$d=\left|\frac{(\vec{b}-\vec{a})\cdot (\vec{p}\times \vec{q})}{|\vec{p}\times \vec{q}|}\right|$
$d=\left|\frac{(-3\hat{j}-\hat{k})\cdot (\hat{i}+2\hat{j}-3\hat{k})}{\sqrt{14}}\right|$
$=\left|\frac{-6+3}{\sqrt{14}}\right|=\frac{3}{\sqrt{14}}$
$\alpha =\frac{3}{\sqrt{14}}$
Now, $28{\alpha }^2=/{28}^2\times \frac{9}{/14}=18$