Originally Posted by

**Plato** I think that you have not really understood the proof.

The usual formula for that distance is, using your diagram, $\displaystyle \frac{{\left| {\overrightarrow {PT} \cdot N} \right|}}{{\left\| N \right\|}}$.

But in that the $\displaystyle N$ is the **normal** to the plane.

That formula is derived by using the cosine of the angle between $\displaystyle \overrightarrow {PT}~\&~N$.

The distance is the length of one side of a right triangle with hypotenuse $\displaystyle \overrightarrow {PT}$.