Last time, we saw that the cross product is a product of two 3d vectors which delivers a vector perpendicular to those two factor vectors.

The cross product is built using three determinants. To determine the x-component of the cross product from the factor vectors (1, 3, 0) and (–2, 0, 0), you find the determinant of the vectors (3, 0) and (0, 0)—the vectors built from the “not-x” components (y- and z-components) of the factors. Repeat this process for the other two components of the cross product, making sure to reverse the sign of the result for the y-component.

But why does this work? How does the cross product make itself perpendicular to the two factor vectors by just using determinants? Below, we’ll still be using magic, but we get a little closer to making our understanding magic free.

Getting the Result We Want

We can actually start with a result we definitely want from the cross product and go from there. (1) The result we want is that when we determine the cross product of a “pure” x-vector (\(\mathtt{1,0,0}\)) and a “pure” y-vector (\(\mathtt{0,1,0}\)), we should get a “pure” z-vector (\(\mathtt{0,0,1}\)). The same goes for other pairings as well. Thus:

\(\begin{bmatrix}\mathtt{1}\\\mathtt{0}\\\mathtt{0}\end{bmatrix} \otimes \begin{bmatrix}\mathtt{0}\\\mathtt{1}\\\mathtt{0}\end{bmatrix} = \begin{bmatrix}\mathtt{0}\\\mathtt{0}\\\mathtt{1}\end{bmatrix} \quad \quad \) \(\begin{bmatrix}\mathtt{1}\\\mathtt{0}\\\mathtt{0}\end{bmatrix} \otimes \begin{bmatrix}\mathtt{0}\\\mathtt{0}\\\mathtt{1}\end{bmatrix} = \begin{bmatrix}\mathtt{0}\\\mathtt{1}\\\mathtt{0}\end{bmatrix} \quad \quad \begin{bmatrix}\mathtt{0}\\\mathtt{1}\\\mathtt{0}\end{bmatrix} \otimes \begin{bmatrix}\mathtt{0}\\\mathtt{0}\\\mathtt{1}\end{bmatrix} = \begin{bmatrix}\mathtt{1}\\\mathtt{0}\\\mathtt{0}\end{bmatrix} \)

A simpler way to write this is to use \(\mathtt{i}\), \(\mathtt{j}\), and \(\mathtt{k}\) to represent the pure x-, y-, and z-vectors, respectively. So, \(\mathtt{i \otimes j = k}\) and so on.

Another thing we want—and here comes some (more) magic—is for (2) the cross product to be antisymmetric, which means that when we change the order of the factors, the cross product’s sign changes but its value does not. So, we want \(\mathtt{i \otimes j = k}\), but then \(\mathtt{j \otimes i = -k}\). And, as before, the same goes for the other pairings as well: \(\mathtt{j \otimes k = i}\), \(\mathtt{k \otimes j = -i}\), \(\mathtt{k \otimes i = j}\), \(\mathtt{i \otimes k = -j}\). This property allows us to use the cross product in order to get a sense of how two vectors are oriented relative to each other in 3d space.

With those two magic beans in hand (and a third and fourth to come in just a second), we can go back to notice that any vector can be written as a linear combination of \(\mathtt{i}\), \(\mathtt{j}\), and \(\mathtt{k}\). The two vectors at the end of the previous post on this topic, for example, (0, 4, 1) and (–2, 0, 0) can be written as \(\mathtt{4j + k}\) and \(\mathtt{-2i}\), respectively.

The cross product, then, of **any** two 3d vectors \(\mathtt{v = (v_x,v_y,v_z)}\) and \(\mathtt{w = (w_x,w_y,w_z)}\) can be written as: \[\mathtt{(v_{x}i+v_{y}j+v_{z}k) \otimes (w_{x}i+w_{y}j+w_{z}k)}\]

For the final bits of magic, we (3) assume that the cross product distributes over addition as we would expect it to, and (4) decide that the cross product of a “pure” vector (i, j, or k) with itself is 0. If that all works out, then we get this: \[\mathtt{v_{x}w_{x}i^2 + v_{x}w_{y}ij + v_{x}w_{z}ik + v_{y}w_{x}ji + v_{y}w_{y}j^2 + v_{y}w_{z}jk + v_{z}w_{x}ki + v_{z}w_{y}kj + v_{z}w_{z}k^2}\]

Then, by applying the ideas in (1) and (4), we simplify to this: \[\mathtt{(v_{y}w_{z} – v_{z}w_{y})i + (-v_{x}w_{z} – v_{z}w_{x})j + (v_{x}w_{y} – v_{y}w_{x})k}\]

And that’s our cross product vector that we saw before. The cross product of the vectors shown in the image above would be the vector (0, –2, 8).