The other kind of product we often want to make is an
area based on the two vectors. Remember, the
area of a rectangle is the
product of its length and width. If we think of the length and the width as orthogonal vectors, we have a good example of the kind of "area product" I have in mind.
To see what happens for arbitrary angles, consider the following picture.
If we put a copy of a at the head of b, and vice versa, we make a parallelogram (shown in green). The area of this parallelogram is a useful kind of product to have between a and b.
The angular part of our product should be zero when the vectors make an angle of 0 or \pi, and maximum for an angle of \pi/2. That sounds a lot like the sine function! If we make this choice, f(\theta) = \sin(\theta), we have the "outer product" or "wedge product",
a \wedge b = |a| |b| \sin(\theta) \hat{D}_{(\wedge)}
Now all that's left is the direction. Once again, suppose we grab a and b, and spin them around any way we please -- as long as we don't change their lengths or the angle between them. The area will always be in the plane formed by the two vectors, and that plane
does change as we rotate them. So, \hat{D}_{(\wedge)} does contain some kind of information about directionality:
- It specifies a plane -- the plane that contains both a and b.
- It specifies an orientation -- the direction that turns a into b.
(Note that this last property implies that (a \wedge b) = -(b \wedge a) --
order matters!)
Many people are unfamiliar with the idea of an area being a kind of "direction", and don't have a good intuition for how to, e.g., add two areas that are in different planes. In 2D, there's no problem, because all the areas are coplanar (there's only one plane in 2D, after all!). In 3D there's a little trick. You can actually replace the plane with a
vector that's
perpendicular to it. This vector's magnitude is equal to the parallelogram's area, and its direction is chosen (by convention) as the direction that a right-handed screw would advance when turned from a to b.
This is, of course, nothing but the good old cross product a \times b!
Like I said: my two posts so far have been kind of handwavy. But I hope you've found them helpful. Let me know if I've made anything unclear, or you have more questions.
Cheers,
Chip