Discussion Forum

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

Answer:

Explanation:

Since, the line $\frac{x-3}{2}=\frac{y+2}{-1}=\frac{z+4}{3}$ lies in the plane, lx+my-z=9, therefore we have 2l-m-3=0

 [ $\because$ normal  will be perpendicular to the line]

   $\Rightarrow  2l-m=3$          .......(i)

   and  $ 3l-2m+4=9$

   $[\because point (3,-2,-4)$ lies  on the plane]

$\Rightarrow 3l-2m=5$     ........(ii)

On solvubg Eqs(i) and (ii) , we get

  l=1 and m=-1

$\therefore   l^{2}+m^{2}=2$