How exactly is the distance between 2 parallel lines in 3D the cross product of the unit vector of the lines with a vector joining them both

3dvectors

According to my textbook, the formula for the distance between 2 parallel lines has been given as below:

enter image description here

Where PT is a vector from the first line that makes a perpendicular on the second line, vector B is a vector to which both the lines are parallel to and vector (a2 – a1) is a vector that joins one arbitrary point on the second line, to yet another arbitrary point on the other

This is what I am confused by. The book, along with the numerous threads I've scoured through already provide similar diagrams for the proof:

enter image description here

From what I understand, the crossing of ST with B should yield us a vector pointing OUT of the plane to which the lines (and in conjunction, ST) belong

How would that yield us TP/PT? TP/PT belongs to the same plane to which the lines and ST belong as well, so how'd crossing ST and B yield us PT?

I understand the end goal is to calculate the MAGNITUDE of the shortest vector joining both the lines, but I can't seem to understand how d is the magnitude of PT as opposed to being the magnitude of the vector jutting OUT of the plane

Best Answer

The formula uses only the magnitude of the the cross product. And the magnitude of $u \times v$ is $\|u\|$ times $\|v\|$ times the sine of the angle between $u$ and $v$. So, using the cross product is just a fancy way of getting the sine of an angle. And, if you do a bit of trig, you'll see that the sine of an angle is exactly what you need to calculate the desired distance. The direction of the cross product is irrelevant in all of this, so you don't need to worry about the fact that it's perpendicular to the plane in which all the action occurs.

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