vector proof using cross product

**duaneg37** · February 8th 2011, 03:44 PM

Let A and B be 2 points on a line and P be a point off the line. Prove that the shortest distance from P to AB is given by the following equation:

$d=\frac{\parallel\vec{AP}\times\vec{AB}\parallel}{ \parallel\vec{AB}\parallel}$

I've tried to solve for d using the Pythagorean theorem and the projection of $\vec{AP}$ onto $\vec{AB}$ for one side and $\vec{AP}$ and d for the other sides, but I'm stumped.

I can't relate it to the cross product. Can anyone help? Thanks!

**Plato** · February 8th 2011, 04:48 PM

Originally Posted by duaneg37

Let A and B be 2 points on a line and P be a point off the line. Prove that the shortest distance from P to AB is given by the following equation:

$d=\frac{\parallel\vec{AP}\times\vec{AB}\parallel}{ \parallel\vec{AB}\parallel}$

You need to realize that $\left\|\vec{AP}\times\vec{AB}\right\|=\left\|\vec{ AB} \right\|\left\|\vec {AP} \right\|\sin(\theta)$ where $\theta$ is the angle between the vectors.

**duaneg37** · February 8th 2011, 05:14 PM

I feel stupid for not remembering that! It's plain as day now. Thank you so much!

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