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.
Step 1We first look at vectors in the xyplane. The magnitude of u × v is so u×v = ±(uv 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 = u_{1}i + u_{2}j and v = v_{1}i + v_{2}j. Then from the figure below u^{} = u_{ 2}i + u_{1}j. Using the formula for the scalar product we get and therefore u × v = (u^{}· v)k = (u_{ 1}v_{2}  u_{2}v_{1})k as claimed.
Step 2Let now u and v be two nonzero and nonparallel vectors in space. Denote by P the parallelogram they span when placed tailtotail. By definition of the vector product the area of P is A = u × v. We now project P along the zaxis onto the xyplane. That projection, P', is spanned by the projections u' and v' of u and v onto the xyplane. 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 zaxis and w := u × v, which are perpendicular to the respective planes. Writing w = w_{1}i + w_{2}j + w_{3}k it follows that as you see from the figures below, representing a cross section along w and the zaxis. If w_{3} > 0 the triple u', v', k is right handed, otherwise it is left handed. Hence by the definition of the vector product u'× v' = w_{3}k. If we note that u' = u_{1}i + u_{2}j and v' = v_{1}i + v_{2}j we conclude from Step 1 that w_{3} = u_{1}v_{2}  u_{2}v_{1}. We get the other components w_{1} and w_{2} by the cyclic permutation i j k i. If we do so we get w_{1} = u_{2}v_{3}  u_{3}v_{2} and w_{2} = u_{3}v_{1}  u_{1}v_{3}, proving the formula for u × v.
Step 3We determine how the area of a parallelogram is changed when it is projected onto a plane. Suppose that E_{1} and E_{2} are two planes intersecting at a line . If R is a rectangle on E_{2} with area A and one edge parallel to , then the area of its projection, R', orthogonal to E_{1} is A cos , where is the acute angle between the two planes: Now look at an arbitrary parallelogram P on E_{2} 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 E_{1} have all the same area as the picture looks very similar in E_{1}. Suppose the area of the original black parallelogram is A_{2} and that of its projection is A_{1}. Then the area of the red rectangle is A_{2} and that of its projection is A_{1}. From the previous considerations we know that with being the acute angle between E_{1} and E_{2}.


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