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

$\vec{ r }=(\hat{ i }+2 \hat{ j }+3 \hat{ k })+\lambda(\hat{ i }+\hat{ j }+\hat{ k }) \vec{ r }=\vec{ a }+\lambda \vec{ p }$
$\vec{ r }=(+\hat{ i }-\hat{ j }+2 \hat{ k })+\mu(2 \hat{ i }-\hat{ j }) \vec{ r }=\vec{ b }+\mu \vec{ q }$
$\vec{ p } \times \vec{ q }=\begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \\ 1 & 1 & 1 \\ 2 & -1 & 0\end{vmatrix}=\hat{ i }+2 \hat{ j }-3 \hat{ k }$
$d =\left|\frac{(\vec{ b }-\vec{ a }) \cdot(\vec{ p } \times \vec{ q })}{|\vec{ p } \times \vec{ q }|}\right|$
$d =\left|\frac{(-3 \hat{ j }-\hat{ k }) \cdot(\hat{ i }+2 \hat{ j }-3 \hat{ k })}{\sqrt{14}}\right|$
$=\left|\frac{-6+3}{\sqrt{14}}\right|=\frac{3}{\sqrt{14}}$
$\alpha=\frac{3}{\sqrt{14}}$
Now, $28 \alpha^2=\not{28} ^2 \times \frac{9}{\not{14}}=18$