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Previous: The Scalar 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,
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 noncoplanar 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 welldefined 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
Therefore,
is perpendicular to . Likewise, it can
be demonstrated that
is perpendicular to .
The vectors , , and
form a righthanded
set, like the unit vectors , , and . In fact,
. This defines a unique direction for
, which
is obtained from a righthand rulesee Fig. 6.
Figure 6:
The righthand rule for cross products.

Let us now evaluate the magnitude of
. We have
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 righthand grip rule. It follows that the vector moment is
given by

(37) 
Figure 7:
A torque.

Next: Vector Calculus
Up: Vectors
Previous: The Scalar Product
Richard Fitzpatrick
20070714