Question

Find the image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-1}$ in the plane $2x-y+2+3=0$.

Answer 1

Equation of normal on plane through $(1,3,4)$

$\Rightarrow \frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\lambda$

$\Rightarrow x=2\lambda +1,y=-\lambda +3,z=\lambda +4$

$\Rightarrow (2\lambda +1,3-\lambda ,\lambda +4)$ is any point on the normal.

$\Rightarrow 2(2\lambda +1)-(3-\lambda )+(\lambda +4)+3=0$

$\Rightarrow 4\lambda +2-3+\lambda +\lambda +4+3=0$

$\Rightarrow 6\lambda +6=0$

$\Rightarrow \lambda =-1$

$\therefore x=-1,y=4,z=3$

$\therefore$Equation of required line $\frac{x+1}{3}=\frac{y-4}{1}=\frac{z-3}{-1}$