Question

A straight line $L$ intersects perpendicularly both the lines $\frac{x+2}{2}=\frac{y+6}{3}=\frac{z-34}{-10}$ and $\frac{x+6}{4}=\frac{y-7}{-3}=\frac{z-7}{-2}$ then the square of perpendicular distance of origin from $L$ is equal to

Answer 1

Answer: 5

so $\Vert$ vector is $\overrightarrow{b}=2\hat{i}+2\hat{j}+\hat{k}$
Let line perpendicular to $L_1$ \& $L_2$ is $L$ with $A$ and $B$ points respectively on $L_1\&L_2$
$A(2\lambda -2,3\lambda -6,-10\lambda +34)$
$B(4\mu -6,-3\mu +7,-2\mu +7)$
$AB$ is parallel to $2\hat{i}+2\hat{j}+\hat{k}$
$\therefore \frac{2\lambda -4\mu +4}{2}=\frac{3\lambda +3\mu -13}{2}=\frac{2\mu -10\lambda +27}{1}$
solving these,
$2\lambda -4\mu +4=3\lambda +3\mu -13$
$\Rightarrow \lambda +7\mu =17\dots (1)$
$\text{ and }3\lambda +3\mu -13=4\mu -20\lambda +54$
$\Rightarrow 23\lambda -\mu =67\dots (2)$
and $3\lambda +3\mu -13=4\mu -20\lambda +54$
$\to 23\lambda -\mu =67\text{. }$
solving (1) and (2), $\lambda =3$ and $\mu =2$
$\therefore A\equiv (4,3,4)\text{ and }B(2,1,3)$

$\cos \theta =\frac{\overrightarrow{AB}\cdot \overrightarrow{OB}}{|\overrightarrow{AB}||\overrightarrow{OB}|}$
$=\frac{4+2+3}{3\cdot \sqrt{14}}=\frac{3}{\sqrt{14}}$
$d=OB\sin \theta =\sqrt{14}\cdot \frac{\sqrt{5}}{\sqrt{14}}\Rightarrow d=\sqrt{5}\Rightarrow d^2=5$