Question

If the line,$\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane, $x+my-z=9$, then $l^2+m^2$ is equal to

Answer 1

Since the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane $x+my=9$.

Therefore we have $2I-m-3=0$

{ $\therefore$ normal will be perpendicular to the line}

$\Rightarrow 2I-m=3$

and $i-2m+4=9$

{ $\therefore$ point $(-3,-2-4$) lies on the plane

$\Rightarrow 3I-2m=5$

On solving equation $\left(1\right)$ and $\left(2\right)$ , we get

$I=1$ and $m=-1$

therefore , $l^2+m^2=2$