Intersection of point normal to plane

Question

Hi if I have a point p (red dot) and I have a plane P (yellow) how do I find the intersection of the point normal to the plane? thanks!

Answer 1

Suppose the plane is given as $$ Ax + By + Cz = 0 $$ and the point $P$ is $(a, b, c)$. THen the ray from $P$ normal to the plane consists of points of the form $$ R(t) = (a, b, c) + t(A, B, C) $$ For what $t$ is $R(t)$ on the plane? We must have $$ A (a + tA) + B(b + tB) + C(c + tC) = 0 \\ aA + bB + cC + t(A^2 + B^2 + C^2) = 0 \\ t = -\frac{aA +bB + cC}{A^2+B^2+C^2} $$ Plugging this in as $t$ in $R(t)$ gets you the intersection point $Q$: $$ Q = (a, b, c) -\frac{aA +bB + cC}{A^2+B^2+C^2} (A, B, C). $$