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# 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:
 (1293)

Is a proper vector? Suppose that , . In this case, . However, if we rotate the coordinate axes through about 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:

 (1294)

Does this rather unlikely combination transform like a vector? Let us try rotating the coordinate axes through an angle about using Equations (A.1280)-(A.1282). In the new coordinate system,
 (1295)

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 and . Thus, is a proper vector. Incidentally, is the only simple combination of the components of two vectors that transforms like a vector (which is non-coplanar with and ). The cross product is anticommutative,
 (1296)

distributive,
 (1297)

but is not associative,
 (1298)

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,

 (1299)

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 Figure A.102.

Let us now evaluate the magnitude of . We have

 (1300)

Thus,
 (1301)

where is the angle subtended between and . 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 the vectors and --see Figure A.103. The scalar area of the parallelogram is . By convention, the vector area has the magnitude of the scalar area, and is normal to the plane of the parallelogram, in the sense obtained from a right-hand circulation rule by rotating on to (through an acute angle): i.e., if the fingers of the right-hand circulate in the direction of rotation then the thumb of the right-hand indicates the direction of the vector area. So, the vector area is coming out of the page in Figure A.103. It follows that

 (1302)

Suppose that a force is applied at position --see Figure A.104. The 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 torque is . The direction of the torque is conventionally defined as the direction of the axis through about which the force tries to rotate objects, in the sense determined by a right-hand circulation rule. Hence, the torque is out of the page in Figure A.104. It follows that the vector torque is given by

 (1303)

The angular momentum, , of a particle of linear momentum and position vector is simply defined as the moment of its momentum about the origin: i.e.,

 (1304)

Next: Rotation Up: Vector Algebra and Vector Previous: Scalar Product
Richard Fitzpatrick 2011-03-31