The line $l_1$ has the equation $r=(6i+2j-2k)+\lambda(4i+5j-k)$ and the plane $\pi_1$ has the equation $2x-y+4z=4$. The line $l_2$ is the reflection of $l_1$ in the plane $\pi_1$. Find the exact vector equation of line $l_2$.
So the line intersects the plane when $\lambda=-2$, giving the point $(-2,-8,0)$ which will be common on $l_1$ and $l_2$. But I am unsure on how to find the direction vector for $l_2$. Any help would be appreciated.
Best Answer
Hint...Construct the line through $(6,2,-2)$ with direction the same as the normal to the plane, and find where this line meets the plane. This point will be the midpoint of the line joining $(6,2,-2)$ and its reflection in the plane. Once you have this reflection point you can form the line of reflection because you now have two points.
the answer I get is $r=-2i-8j+t(88i+103j-13k)$