Determinants

The idea of determinants have been about since around the 3rd century when it first appeared in an ancient Chinese book of Mathematics called The Nine Chapters on the Mathematical Art. It was used originally to define certain properties of systems of linear equations, as we will see later in the section on linear algebra, however for now we will simply use it as a particular way to easily calculate the cross product. Let’s take a two by two array of numbers and define the determinant for this. $\left|\begin{array}{cc}a & b \\ c & d \\\end{array}\right|=ad-bc$

The vertical lines on the left and right are the sign that the we are taking a determinant. For now this is just a definition and we will work with it in what follows. Don’t worry too much about where it comes from, but we will see later where it comes from and we will see now why it is useful. The above is for a $2\times 2$ array of numbers. If we have a $3\times 3$ we can write it as follows: $\left| \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \\ \end{array} \right|=a \left| \begin{array}{cc} e & f \\ h & i \\ \end{array} \right|-b\left| \begin{array}{cc} d & f \\ g & i \\ \end{array} \right|+c\left| \begin{array}{cc} d & e \\ g & h \\ \end{array} \right|$

The first term is just the first element in the top row multiplied by the determinant of the bottom two rows, not including the column that $a$ is in. The second term is minus the second term in the top row times the determinant of the bottom two rows, not including the column that $b$ is in. The third term is the third element in the top row times the determinant of the bottom two rows, not including the column that $c$ is in. We could then multiply out the $2\times 2$ determinants using the rule in equation \ref{det}.

Just to see where the pattern is going, let’s look at a $4\times 4$ example. The following defined the determinant of a $4\times 4$ array: $\left| \begin{array}{cccc} a & b & c & d \\ e & f & g & h \\ i & j & k & l \\ m & n & o & p \\ \end{array} \right|=a \left| \begin{array}{ccc} f & g & h \\ j & k & l \\ n & o & p \\ \end{array} \right|-b \left| \begin{array}{ccc} e & g & h \\ i & k & l \\ m & o & p \\ \end{array} \right|+c \left| \begin{array}{ccc} e & f & h \\ i & j & l \\ m & n & p \\ \end{array} \right|-d \left| \begin{array}{ccc} e & f & g \\ i & j & k \\ m & n & o \\ \end{array} \right|$

Notice that the signs between each term alternate between $+$ and $-$ and we define the determinant recursively. Ok, good, now that we have shown how to calculate a determinant, we can give a very simple expression for the cross product in terms of the determinant. For our vectors $\vec{a}$ and $\vec{b}$, we have: $\vec{a}\times\vec{b}=\left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{array} \right|$

Check for yourself that this does indeed give the same expression for the cross product as in equation \ref{cross1} when you expand out the determinant. What we have now is simply a very simple notational form for calculating the cross product of two vectors. We will see in the section on linear algebra a way to define the determinant of any square matrix.

In fact the cross product has a very geometrical significance. The magnitude of the cross product (ie. $\left|\left|\vec{a}\right|\left|\vec{b}\right|\sin\theta\right|$ is precisely the area of the paralellogram given by the two ways of adding the vectors $\vec{a}$ and $\vec{b}$. The cross product of the blue and red arrows will give a new vector whose magnitude is equal to the area of the parallelogram between the vectors shown here.

Properties of the cross product

The fact that the cross product produces a vector which is perpendicular to the two vectors being crossed is extremely useful. In particular it allows us to find vectors perpendicular to planes. Such a vector is known as a normal vector and can be used to define a plane in 3 dimensional space. Any flat surface in 3 dimensions can be defined by a single point on it and a vector perpendicular to it. To calculate the normal vector we can simply take any 3 points on the plane and define two vectors that go between these three points. As long as the points don’t all lie along a straight line we can then take the cross product of the two vectors and the result will be a vector which is perpendicular to the plane. We will see in a bit how this vector can help us to discover relations between points on and off the plane.

Let’s imagine a plane that goes through the three points $P(1,4,6)$, $Q(-2,5,-1)$ and $R(1,-1,1)$. We can form 3 vectors from these three points, but we will only need two.

Let’s form the vectors $\vec{PQ}=(-2-1)\vec{i}+(5-4)\vec{j}+(-1-6)\vec{k}=-3\vec{i}+\vec{j}-7\vec{k}$ and $\vec{PR}=(1-1)\vec{i}+(-1-4)\vec{j}+(1-6)\vec{k}=-5\vec{j}-5\vec{k}$. We can find a vector normal to these: $\vec{PQ}\times\vec{PR}=\left| \begin{array}{ccc} \vec{i} & \vec{j} & \vec{k} \\ -3&1&-7 \\ 0 & -5 & -5 \\ \end{array} \right|=-40\vec{i}-15\vec{j}+15\vec{k}$

How can we check that this vector is perpendicular to the plane? Well, if it is, then it should certainly be perpendicular to any line which lies in the plane, and thus if we take the dot product of any vector that lies in the plane it should give zero. Check for yourselves that when we do the dot product of this vector with the vectors $\vec{PQ}$ and $\vec{PR}$ that it does indeed give zero.

Often we will be most interested not just in a vector which is perpendicular to the plane, but what is known as a unit normal vector, that is a vector of unit magnitude that is perpendicular to the plane. We can thus take the vector we have just found and divide it by its magnitude ( $5 \sqrt{82}$): $\hat{n}=\frac{-40\vec{i}-15\vec{j}+15\vec{k}}{5\sqrt{82}}$

where the hat on the $n$ is sometimes used to denote a unit vector.

In the figure below we show the vectors described above (the perspective has been somewhat warped so all the arrows should really be perpendicular to one another). The cross product of the two red arrows which lie in the plane produces the blue arrow (once normalised to one) which lies perpendicular to the plane.

 How clear is this post?