In the following, I’m going to miss out quite a few details which I think are very nicely laid out in Stewart. I will try and add a slightly more pedagogical tone to some of it, and some nice diagrams along the way.

So we saw in the last post that we can write the cross product of two vectors, which itself gives a vector, in terms of the determinant of a 3 by 3 array. We can use this to both find a vector perpendicular to two given vectors (unless they are parallel to one another) and also to find the area of a parallelogram formed by two vectors (the area of which is zero if the vectors are parallel to one another).

The second of these is easy enough to do in two dimensions, but in three dimensions that’s not an easy prospect. Using the cross  (otherwise called the vector) product makes this easy.

What we should keep in mind, is that with both the dot and the cross product, we defined them in terms of the angle in between two vectors, but figured out that using the component form of the vectors, we didn’t even need to know this angle. Better than that, we can use the dot or cross product to calculate the angle between two vectors.

For the dot product we found that we could use it as a test for whether or not two vectors were perpendicular to one another. If they were, then their dot product will give zero. For the cross-product we can use it as a test for whether two vectors are parallel to one another. We can thus use this to calculate whether three points lie in a single straight line. If they do, then the two vectors formed between any two pairs of them will have a vanishing cross product.

Let’s take an example. Below is a plot of three points on a line. The points are P(2, 4, 0), Q(3, 3, 2) and R(4, 2, 4). We can form vectors between them.

For instance ${\vec{PQ}}=\left<1,-1,2\right>$ and ${\vec{PR}}=\left<2,-2,4\right>$. Taking the cross product of these two vectors gives:

$\left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ 1 & -1 & 2 \\ 2 & -2 & 4 \\ \end{array} \right| = \vec{i}\left| \begin{array}{cc} -1 & 2 \\ -2 & 4 \\ \end{array} \right|- \vec{j}\left| \begin{array}{cc} 1 & 2 \\ 2 & 4 \\ \end{array} \right|+\vec{k}\left| \begin{array}{cc} 1 & -1 \\ 2 & -2 \\ \end{array} \right|=\vec{0}$

What we’ve seen then is that we can use the cross product, and the dot product to make some quite strong statements about the relative geometry of various objects – lines and points, and, we will see shortly, planes. After all, a line is simply the intersection of two planes (in 3d) and a point is the intersection of three planes (again, in 3d). We will see that actually everything comes down to the intersections of planes, but the number of planes which are going to intersect are going to be dependent on the number of dimensions we are dealing with.

We will find shortly that we can introduce another set of notation, namely matrices, which are going to encode for us both sets of linear equations, as well as information about planes. These three concepts: matrices, sets of linear equations, and planes are going to become a fascinating playing ground where we will find that while they are all completely equivalent, sometimes one notation is going to be easier to solve a problem, and sometimes another will provide us with the intuition we need.

So we’ve tackled the vector product, and there are plenty of exercises to be done with that. The nice thing about vectors is that it’s very easy to come up with your own examples. Either take two random vectors, or three points, or two points and a vector, and work out what combinations you can put them in to calculate the vector product.

In the case of the dot (also called the scalar) product, the result of this was indeed a scalar. There’s not much we can do with this other than multiply it by a scalar, or a vector, but it doesn’t tell us anything very interesting. On the other hand the cross product gives us a vector, and it turns out that dotting another vector into this, does give us something interesting.

The triple scalar product

Let’s take three vectors, and put them all at the origin. We can form a parallelpiped (like a skewed box) with these three vectors:

The red vector is $\vec{a}=\left<6,0,0\right>$, the green is $\vec{b}=\left<0,6,2\right>$ and the blue is $\vec{c}=\left<2,1,5\right>$. The box is formed by using the three vectors end to end in the appropriate sequences.

Given that the area of the base (let’s say formed by the red and green vectors) is given by $|\vec{b}\times\vec{c}|$, if we multiply this by $|a|$ times the cosine of the angle between the vector $\vec{a}$ and the line perpendicular to the base (ie. the direction of $|\vec{b}\times\vec{c}|$) we will get the volume of the parallelpiped. This is simply given by:

$|\vec{a}.(\vec{b}\times \vec{c})|$

where we take the modulus because this can be negative, but we just want the absolute value.

Although it looks like this might be a bit complicated, in fact it’s quite simple. Let’s stick to generalities and call the components of $\vec{a}=\left$ etc. So we have that $\vec{a}.(\vec{b}\times \vec{c})$ is

$\left.\left|\begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ b_1 &b_2 & b_3 \\ c_1 & c_2 & c_3 \\ \end{array} \right|= a_1\left| \begin{array}{cc} b_2 & b_3 \\ c_2 & c_3 \\ \end{array} \right|- a_2\left| \begin{array}{cc} b_1 & b_3 \\ c_1 & c_3 \\ \end{array} \right|+a_3\left| \begin{array}{cc} b_1 & b_2 \\ c_1 & c_2 \\ \end{array} \right|$

But this is the same as:

$\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 &b_2 & b_3 \\ c_1 & c_2 & c_3 \\ \end{array} \right|$

In the current case we have

$\left| \begin{array}{ccc} 6 & 0 & 0 \\ 0 & 6 & 2 \\ 2 & 1 & 5 \\ \end{array} \right|=6(6\times 5-2\times1)=6(28)=168$

So this is the volume of the above parallelpiped. This again means that we have a geometrical measure of the relative orientation of 3 vectors. A parallelpiped will have zero volume if any of the vectors in it are zero length, or if all the three vectors can be put in the same plane. In general two vectors define a plane for us (unless they are parallel in which case they define an infinite number of different planes), but three vectors, in general define two planes for us. If they are coplanar, then the two planes that are defined are really the same. Thus the triple scalar product is a test for coplanarity. If it is zero, then we know that the three vectors are coplanar.

The triple vector product

We can also take three vectors and cross them into one another. However, this doesn’t have quite such a nice geometric interpretation as the triple scalar product. Of course it gives a vector. In fact this vector can be worked out entirely using the dot product. You can show, using components, or the fundamental definition of the dot and cross products, that:

$\vec{a}\times(\vec{b}\times\vec{c})=(\vec{a}.\vec{c})\vec{b}-(\vec{a}.\vec{b})\vec{c}$.

In the next post we will see much more powerfully how we can use vectors to define shapes for us in any number of dimensions in very powerful ways.

 How clear is this post?