dot and cross products


A couple of notes that might be useful about dot and cross products. Recall that for the dot and the cross product we noted the relationships

a . b = |a| |b| cos(t), and
|a x b| = |a| |b| sin(t),
where t is the angle between the vectors a and b. Note that the second of these involves the magnitude of the cross product, not the cross product itself (which is a vector).
figure of triangle determined by two vectors a and b
figure 1: a, b, and the triangle determined by them.

We saw that the connection of the products to the lengths |b| cos(t) and |b| sin(t) is particularly useful, because they are the lengths of the sides of the triangle of vectors shown in figure 1 to the left. In particular, they allowed us to form the projection of b onto a the (see projections page), and give us the height of the parallelogram determined by the vectors a and b as shown in figure 2, below.

And if we have that height, then the area of the parallelogram is just (width)(height), or (|a|)(|b| sin(t)) = |a x b|. Pretty cool.

In our book this is put together in one other way, in section 13.4 (figure 3, equation 11), to find the volume of a parallelepiped (essentially a 3D parallelogram): they say "the volume of the parallelepiped is the area of the bottom parallelogram times the height of the parallelepiped." If the parallelepiped is determined by the vectors a, b and c, then the area of the bottom is just |b x c| = |b||c| sin(t), and the height looks mighty like |a| cos(s), where s is the angle between a and the cross product. So the volume is just (height)(area of base) = (|a| cos(s)) (|b||c| sin(t)) = a . (b x c). Super!

(There'a actually an absolute value thrown in there to make sure that the cosine is positive, but the idea is the same.)

figure of parallelogram determined by two vectors a and b
figure 2: a, b, and the parallelogram determined by them.

"So what?," you ask. Fair enough. What have we seen? We've seen what the dot and cross products are, and how to find them. We've seen that the dot product gives us the combination |b| cos(t), from which we can form the projection of b on another vector a. We've seen that the cross product gives us a vector perpendicular to the two vectors we start with. And we've seen that |a x b| gives us the combination |b| sin(t), which is useful for calculating areas of parallelograms as noted above.


Last modified: Thu Jan 15 17:04:35 EST 2004
©2004 Gavin LaRose, UM Math Dept.