# Why is a vector product perpendicular?

May 22, 2014

I'll assume you mean a vector cross product.

A dot product can't be perpendicular to anything because it's a scalar quantity, and thus has no direction. A cross product, however, can only be found for two 3-D vectors and is always another vector which is perpendicular to the plane formed by the two vectors. (right-hand rule)

The cross product can be thought of as a measurement of how perpendicular two vectors are. Its magnitude is defined as

|vec x × vec y| = |vec x|⋅|vec y|⋅sintheta

where $\theta$ is equal to the angle between the two vectors.

If $\theta$ is zero, then the vectors, no matter their magnitude, are parallel. And $\sin \theta$ is $0$, meaning the cross product is also zero. As $\theta$ increases, $\sin \theta$ approaches $1$. When $\theta$ reaches $90$ degrees, then the cross product will be at its maximum.

In short, the cross product of two vectors $\vec{x}$ and $\vec{y}$ ranges from $0$ to |vec x|⋅|vec y|, depending on how perpendicular $\vec{x}$ and $\vec{y}$ are.

Here's an applet which might help you visualize this:
http://mathinsight.org/applet/cross_product

To answer your question, the cross product is perpendicular to its multiplicands because if it weren't defined that way, it wouldn't be too useful.

Torque is a physics concept which is greatly simplified by the cross product. If you have a lever, and you apply a force to that lever, all you need to do is measure the distance vector from the pivot point to the point where force is applied, and cross that with the force vector. You suddenly have a measurement of torque.

Now you might realize why levers tend to rotate a lot faster when you apply a force that's perpendicular to them. If you pull on a lever at a smaller angle, then it will have less angular acceleration because the cross product won't be at its maximum. Likewise, if you just pull on one straight on, (your arm is parallel with the lever) then it won't move at all - you're applying zero torque, because $\theta$ is zero.