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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,
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,
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.
Figure A.102:
The right-hand rule for cross products. Here,
is less that
.
 |
Let us now evaluate the magnitude of
. We have
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) |
Figure A.103:
A vector parallelogram.
 |
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) |
Figure A.104:
A torque.
 |
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