Showing the distance between a point $P$ the line determined by a segment $AB$ is $d=\frac{||AP\times AB||}{||AB||}$

Question

Show that in $3$-space the distance $d$ from a point $P$ to the line $L$ through points A and B can be expressed as $$d=\frac{||AP\times AB||}{||AB||} .$$

My diagram of the situation:

My next thought was to find $\operatorname{proj}_{AB}AP$ and then use,

$d^2 = AP^2 - (\operatorname{proj}_{AB}AP)^2$

However, this seems to turn into a mess quite quickly and I'm not sure if I'm approaching the question in the best way possible.

Answer 1

Let $\theta$ be the angle between the vectors $\vec{AP}$ and $\vec{AB}$. Then, from the definition of cross product, we have:$$\vec{AP}\times\vec{AB}=|AP|.|AB|.\sin(\theta)\vec{n}$$where $\vec{n}$ is the unit vector at right angles to both $\vec{AP}$ and $\vec{AB}$.

We therefore get:$$|\vec{AP}\times\vec{AB}|=||AP|.|AB|.\sin(\theta)\vec{n}|=|AP|.|AB|.\sin(\theta)$$$$\therefore\frac{|\vec{AP}\times\vec{AB}|}{|AB|}=|AP|.\sin(\theta)$$

Hopefully you can see that this equals $d$