Question

A line $\frac{x+2}{1}=\frac{y-3}{2}=\frac{z-k}{3}$ cuts the $y-z$ plane and the $x-y$ plane at $A$ and $B$ respectively. If $\angle AOB=\frac{\pi }{2}$, then $2k$, where $O$ is the origin, is

Answer 1

Answer: 9

$\frac{x+2}{1}=\frac{y-3}{2}=\frac{z-k}{3}=\lambda$
$\Rightarrow (\lambda -2,2\lambda +3,3\lambda +k)$
for $A,\lambda =2$
$A(0,7,6+k)$
$\Rightarrow$ for $B\lambda =-\frac{k}{3}$
$\Rightarrow B\left(-2-\frac{k}{3},3-\frac{2k}{3},0\right)$
$\angle AOB=90^{\circ }$
$\Rightarrow \overrightarrow{AO}\cdot \overrightarrow{OB}=0$
$\Rightarrow 7\left(-3+\frac{2k}{3}\right)=0$
or $k=\frac{9}{2}$
$\Rightarrow 2k=9$