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.

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.

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 $$\^x\times\^y=\^z$$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.
Find $$(3\^y+4\^z)\times(4\^y-3\^z)$$

Distribute first:
$$15(\^y\times\^y)-9(\^y\times\^z)+16(\^z\times\^y)-12(\^z\times\^z)$$
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.