
The vector product of two 3vectors, $\stackrel{\u27f6}{v}$ and $\stackrel{\u27f6}{w}$ , written as $\stackrel{\u27f6}{v}\times \stackrel{\u27f6}{w}$ is the determinant of the 3 by 3 matrix whose first two columns are the components of $\stackrel{\u27f6}{v}$ and $\stackrel{\u27f6}{w}$ and whose third column consists of the basis vectors $\widehat{i},\widehat{j}$ and $\widehat{k}$ .
This definition appears somewhat mysterious. But all that it means is that its components in the direction of the various axes are the cofactors of $\widehat{i},\widehat{j}$ and $\widehat{k}$ here. These are determinants of ordinary two by two matrices.
$\stackrel{\u27f6}{v}\times \stackrel{\u27f6}{w}$ is a vector NOT a number, and is sometimes called the "cross product" of $\stackrel{\u27f6}{v}$ and $\stackrel{\u27f6}{w}$ .
From this definition we can see that the dot product of $\stackrel{\u27f6}{v}\times \stackrel{\u27f6}{w}$ with another vector, $\stackrel{\u27f6}{u}$ , is the determinant of the matrix whose columns (or rows) are the components of these three vectors, in the order $\stackrel{\u27f6}{v},\stackrel{\u27f6}{w},\stackrel{\u27f6}{u}$ which makes its magnitude the volume of the parallelepiped determined by these vectors.
Its other components can be obtained by cyclically shifting among the variables $x,y$ and $z$ changing $x$ to $y$ , $y$ to $z$ and $z$ to $x$ .
The vector product is, from the properties of determinants, linear in both its vector factors. Thus if you multiply $\stackrel{\u27f6}{v}$ by 10, $\stackrel{\u27f6}{v}\times \stackrel{\u27f6}{w}$ gets multiplied by 10, and you also have
Also, $\stackrel{\u27f6}{v}\times \stackrel{\u27f6}{w}$ changes sign if the order of its factors is reversed: $\stackrel{\u27f6}{v}\times \stackrel{\u27f6}{w}=\stackrel{\u27f6}{w}\times \stackrel{\u27f6}{v}$ . These statements follow from similar properties of determinants.
The magnitude of $\stackrel{\u27f6}{v}\times \stackrel{\u27f6}{w}$ is the area of the parallelogram defined by its factors, that is
$\stackrel{\u27f6}{v}\times \stackrel{\u27f6}{w}$ is perpendicular to both its arguments: $\stackrel{\u27f6}{v}\times \stackrel{\u27f6}{w}\xb7\stackrel{\u27f6}{v}=\stackrel{\u27f6}{v}\times \stackrel{\u27f6}{w}\xb7\stackrel{\u27f6}{w}$ . This follows from the statement that the determinant of a matrix with two identical columns, which is what either of these expressions is, is 0.
We can deduce $(\stackrel{\u27f6}{v}\times \stackrel{\u27f6}{w})\xb7(\stackrel{\u27f6}{v}\times \stackrel{\u27f6}{w})=(\stackrel{\u27f6}{v}\xb7\stackrel{\u27f6}{v})(\stackrel{\u27f6}{w}\xb7\stackrel{\u27f6}{w})(\stackrel{\u27f6}{v}\xb7\stackrel{\u27f6}{w})(\stackrel{\u27f6}{w}\xb7\stackrel{\u27f6}{v})$ , since both sides here represent the square of the area of the parallelogram with sides $\stackrel{\u27f6}{v}$ and $\stackrel{\u27f6}{w}$ (see Exercise 3.2 ).
Human beings make numerical errors in evaluating 3 by 3 determinants or vector products roughly one out of every four times they do them, even more so when the vectors components or matrix entries have lots of negative values. It is wise to build a determinant and vector product tool on a spreadsheet and use it to check not replace your own computations. You will then get the right answer every time.
It is actually easier to build such a tool than to do even one cross product by hand. It's just about as easy as doing a dot product. To do it, enter $\stackrel{\u27f6}{v}$ and $\stackrel{\u27f6}{w}$ as two parallel rows, say with ${v}_{x}$ in A2, ${v}_{y}$ in B2, ${v}_{z}$ in C2 and the components of $\stackrel{\u27f6}{w}$ similarly in A3C3. In D2 put =A2 and copy that into D3, E2 and E3. (You are now half done.) Next, in A4 put =B2*C3C2*B3, and copy that into B4 and C4. That's it! Row 4 contains the cross product, $\stackrel{\u27f6}{v}\times \stackrel{\u27f6}{w}$ of $\stackrel{\u27f6}{v}$ with $\stackrel{\u27f6}{w}$ . You can check by verifying that its dot product with either of the first two rows is 0.
Notice that you can now change $\stackrel{\u27f6}{v}$ and $\stackrel{\u27f6}{w}$ and (with luck) row four will then contain the cross product of the new rows 2 and 3.
You can see what vector products look like from the applet in section 4.1 .
Exercises:
4.9 Write out the vector product $\stackrel{\u27f6}{v}\times \stackrel{\u27f6}{w}$ explicitly in terms of the components of its factors.
4.10 Compute the vector product of the two vectors $(1,2,3)$ and $(4,5,6)$ .
4.11 Build a vector product tool as indicated above.
