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The vector product

We have discovered how to construct a scalar from the components of two general vectors and . Can we also construct a vector which is not just a linear combination of and ? Consider the following definition:
 (28)

Is a proper vector? Suppose that , . Clearly, . However, if we rotate the basis through about the -axis then , , and . Thus, does not transform like a vector, because its magnitude depends on the choice of axes. So, above definition is a bad one.

Consider, now, the cross product or vector product:

 (29)

Does this rather unlikely combination transform like a vector? Let us try rotating the basis through degrees about the -axis using Eqs. (10)-(12). In the new basis,
 (30)

Thus, the -component of transforms correctly. It can easily be shown that the other components transform correctly as well, and that all components also transform correctly under rotation about the - and -axes. Thus, is a proper vector. Incidentally, is the only simple combination of the components of two vectors which transforms like a vector (which is non-coplanar with and ). The cross product is anticommutative,
 (31)

distributive,
 (32)

but is not associative:
 (33)

The cross product transforms like a vector, which means that it must have a well-defined direction and magnitude. We can show that is perpendicular to both and . Consider . If this is zero then the cross product must be perpendicular to . Now

 (34)

Therefore, is perpendicular to . Likewise, it can be demonstrated that is perpendicular to . The vectors , , and form a right-handed set, like the unit vectors , , and . In fact, . This defines a unique direction for , which is obtained from the right-hand rule (see Fig. 6).

Let us now evaluate the magnitude of . We have

 (35)

Thus,
 (36)

Clearly, for any vector, since is always zero in this case. Also, if then either , , or is parallel (or antiparallel) to .

Consider the parallelogram defined by vectors and (see Fig. 7). The scalar area is . The vector area has the magnitude of the scalar area, and is normal to the plane of the parallelogram, which means that it is perpendicular to both and . Clearly, the vector area is given by
 (37)

with the sense obtained from the right-hand grip rule by rotating on to .

Suppose that a force is applied at position (see Fig. 8). The moment, or torque, about the origin is the product of the magnitude of the force and the length of the lever arm . Thus, the magnitude of the moment is . The direction of the moment is conventionally the direction of the axis through about which the force tries to rotate objects, in the sense determined by the right-hand grip rule. It follows that the vector moment is given by

 (38)

Next: Rotation Up: Vectors Previous: The scalar product
Richard Fitzpatrick 2006-02-02