Show that for all points P of the plane, the distance between Q and the point of intersection is less than or equal to the distance between Q and P Let $N$ be a non-zero vector, $c$ a number, and $Q$ a point. Let $P_0$ be the point of intersection of the line passing through Q, in the direction of N and the plane $X\cdot N=c$. Show that for all points P of the plane, we have $$||Q-P_0||\leq ||Q-P||$$
I'll leave my reasoning below just in case but I can't find a way to arrive at the conclusion.
$$X=Q+tN$$
$$(Q+tN)\cdot N=c$$
$$t=\frac{c-Q\cdot N}{N\cdot N}$$
$$P_0=Q+(\frac{c-Q\cdot N}{N\cdot N})\cdot N=cN$$
$$||Q-cN||\leq ||Q-P||$$
$$(X-P)\cdot N=0\implies P\cdot N=c$$

Since $$ \Vert\overrightarrow {QP} \Vert^2=\Vert\overrightarrow{QP_0}\Vert^2+\Vert \overrightarrow{P_0P}\Vert^2$$
 it follows that $\Vert\overrightarrow{QP}\Vert\ge\Vert\overrightarrow{QP_0}\Vert$
