11. The vector product
Unless one (or both) of the vectors is zero, it is not obvious how to get the formula to compute the vector product from the geometric definition of the vector product. If you are interested in how to obtain it, here is a possible way.
The procedure depends on looking at projections of the parallelograms spanned by the vectors u and v into the three coordinate planes. We proceed in three steps.
We first look at vectors in the xy-plane. The magnitude of u × v is
so u×v = ±(|u||v| sin )k. The sign is positive if moving u to v through the smaller angle is counterclockwise and negative in the other case. Hence if we let positive in the former and negative in the latter case we get
(We use that sin(-) = - sin .) Now let u be the vector obtained by rotating u counterclockwise by 90°. As |u| = |u| it follows from the definition of the scalar product that
Two possible situations are shown below
We therefore get u × v = (u· v)k.
Now assume that u = u1i + u2j and v = v1i + v2j. Then from the figure below u = -u 2i + u1j.
Using the formula for the scalar product we get
and therefore u × v = (u· v)k = (u 1v2 - u2v1)k as claimed.
Let now u and v be two non-zero and non-parallel vectors in space. Denote by P the parallelogram they span when placed tail-to-tail. By definition of the vector product the area of P is A = |u × v|.
We now project P along the z-axis onto the xy-plane. That projection, P', is spanned by the projections u' and v' of u and v onto the xy-plane.
It will be shown in Step 3 that the area, A', of P' is given by
where is the acute angle between the planes containing P and P'.
Observe that is also the acute angle between the z-axis and w := u × v, which are perpendicular to the respective planes.
Writing w = w1i + w2j + w3k it follows that
as you see from the figures below, representing a cross section along w and the z-axis.
If w3 > 0 the triple u', v', k is right handed, otherwise it is left handed. Hence by the definition of the vector product u'× v' = w3k. If we note that u' = u1i + u2j and v' = v1i + v2j we conclude from Step 1 that w3 = u1v2 - u2v1.
We get the other components w1 and w2 by the cyclic permutation i j k i. If we do so we get w1 = u2v3 - u3v2 and w2 = u3v1 - u1v3, proving the formula for u × v.
We determine how the area of a parallelogram is changed when it is projected onto a plane. Suppose that E1 and E2 are two planes intersecting at a line . If R is a rectangle on E2 with area A and one edge parallel to , then the area of its projection, R', orthogonal to E1 is A cos , where is the acute angle between the two planes:
Now look at an arbitrary parallelogram P on E2 spanned by vectors u and v. We can transform it into a rectangle of the same area such that one edge of that rectangle is parallel to as shown below
The projections of the black, green and red parallelograms into E1 have all the same area as the picture looks very similar in E1. Suppose the area of the original black parallelogram is A2 and that of its projection is A1. Then the area of the red rectangle is A2 and that of its projection is A1. From the previous considerations we know that
with being the acute angle between E1 and E2.