Cross Products

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 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: $$\vec A\times\vec A=0$$


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: $$=0+3(-\^y)-8(-\^z)-4(\^x)=\boxed{-4\^x-3\^y+8\^z}$$
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.

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 \(\vec\omega\).