The second type of vector multiplication is called the cross
product. Unlike the dot product, which results in a scalar, the
cross product of two vectors is also a vector, with a magnitude or direction.
Magnitude
The magnitude of a cross product is
$$\abs{\vec A\times\vec B}=\abs{\vec A}\abs{\vec B}\sin\theta$$
where \(\theta\) is the angle between them. If vectors point in the
same direction or opposite directions, their cross product is zero;
the cross product is maximal when the vectors are perpendicular.
Notice that if you cross a vector with itself, you get zero:
$$\vec A\times\vec A=0$$
Direction
The cross product \(\vec A\times\vec B\) always points in a direction
which is perpendicular to both \(\vec A\) and \(\vec B\), as well as
the plane both vectors lie on. For instance, if I draw two vectors on
a piece of paper, their cross product must point either into of the
paper or out of the paper. Which it is depends on the arrangement of
the two vectors. If A needs to rotate clockwise to point in the
direction of B, the cross product points into the page; if
counterclockwise, then it points out of the page. Notice that the
order of the vectors matters:
$$\vec A\times\vec B=-\vec B\times\vec A$$
Unlike normal multiplication and the dot product, the cross-product is
noncommutative. You have to be careful, therefore, to
keep your vectors in the right order when calculating.
Another way to figure out the direction is to use a so-called right-hand
rule. Hold your right hand (hence the name) so that your fingers point along the first
vector, and then twist your hand so that your palm faces in the
general direction of the second. Your thumb will then point in the
direction of the cross-product of the two. You're only trying to
distinguish between two directions (in or out of the page, for
instance), so the precise direction your thumb points is not
important.
Cross-Products of Unit Vectors
It is particularly useful to know the cross-products between the
different unit vectors \(\def\^{\hat}\^x,\^y,{\rm and}\,\^z\). For
example, consider
$$\^x\times\^y$$
Because the vectors are perpendicular, the magnitude of this cross product is
$$\abs{\^x\times\^y}=\abs{\^x}\abs{\^y}=(1)(1)=1$$
because the unit vectors have magnitude 1. The cross-product has to be
perpendicular to \(\^x\) and \(\^y\), which means it must point along
the \(z\) axis. Since it has length 1, it must be either \(\^z\) or
\(-\^z\).
Which is it? Actually, it's arbitrary: we could choose either answer,
and mathematics would work exactly the same way. Of course, it's
helpful if we all agree on the same result, and so we've decided that
$$\^x\times\^y=\^z$$
This choice is called a right-handed coordinate system, and
physicists (and physics textbooks) usually assume it. If
we'd made the opposite choice, we would be using a left-handed
coordinate system. If you are asked to choose a coordinate
system for a problem, be sure that you make it a right-handed
coordinate system, or else some of your answers may be incorrect.
We can read off the other unit-vector cross-products from the
coordinate axes, using the right-hand rule. We will get the results
$$\begin{array}
\^x\times\^y=\^z &\^y\times\^z=\^x &\^z\times\^x=\^y\\
\^y\times\^x=-\^z &\^z\times\^y=-\^x &\^x\times\^z=-\^y\\
\^x\times\^x=0 &\^y\times\^y=0 &\^z\times\^z=0 \qquad \hbox{(because
$\vec A\times\vec A$=0)}\\
\end{array}$$
If you look at these results, you may notice a pattern which is
summarized by the triangle shown here on the left. You can use that
as a mnemonic, or you can simply use the right-hand rule and the
coordinate axes.
Cross-Product and Components
If we are given two vectors in terms of their components, we can find
the cross-product in terms of components as well. We start by using
distribution, like we would when multiplying polynomials. For
example,
$$
\begin{eqnarray}
&&({\color{red}3\^x}-{\color{green}4\^y})\times({\color{blue}2\^x}+{\color{purple}\^z})\\
&=&({\color{red}3\^x}\times{\color{blue}2\^x})
+({\color{red}3\^x}\times {\color{purple}\^z})
+({\color{green}-4\^y}\times{\color{blue}2\^x})
+({\color{green}-4\^y}\times{\color{purple}\^z})
\end{eqnarray}$$
Next, we can factor out the scalar bits and multiply them together:
$$=6(\^x\times\^x)+3(\^x\times\^z)-8(\^y\times\^x)-4(\^y\times\^z)$$
and use the results from the previous section:
$$=0+3(-\^y)-8(-\^z)-4(\^x)=\boxed{-4\^x-3\^y+8\^z}$$
Then use the unit vector rules:
$$0-9\^x+16(-\^x)+0=\boxed{-25\^x}$$
Now it's good to double-check your result by seeing if its
direction agrees with the right-hand rule. To the right are the two
vectors in the \(yz\)-plane. For this to be a right-handed
coordinate system, \(\^x\) must point out of the page, and so the
cross-product of these vectors, as given by the right-hand rule,
points into the page.
Loop-Normal Right-Hand Rule
The cross product uses one type of right-hand rule, but there is a
second group of right-hand rules which will come up when we use the
cross product. I call them loop-normal right-hand rules,
because they turn up when we have one quantity which goes in a
loop, which is related to another quantity that is perpendicular (or
normal) to that loop. One example from mechanics is the vector
angular velocity \(\vec\omega\), which points along the axis of a
spinning object. There are two directions \(\vec\omega\) could point
along the axis, and also two directions the object could spin, and
the two quantities should be related (e.g. reversing the direction
of spin should reverse the direction of \(\vec\omega\)). What you
do is take your right hand and curl the fingers in the direction of
the spin; the thumb then points in the direction of \(\vec\omega\).
