   Next: Vector Calculus Up: Vectors Previous: The Scalar Product

## 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: (27)

Is a proper vector? Suppose that and . 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, (28)

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,       (29)

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, (30)

distributive, (31)

but is not associative: (32)

Note that can be written in the convenient, and easy to remember, determinant form (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 a 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 .

Suppose that a force is applied at position --see Fig. 7. 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 a right-hand grip rule. It follows that the vector moment is given by (37)    Next: Vector Calculus Up: Vectors Previous: The Scalar Product
Richard Fitzpatrick 2007-07-14