The vector, or cross product

When we took two vectors previously and found a way to multiply them together using the dot product, we ended up with a scalar. However, there is also a way that we can take two vectors and multiply them together to give a vector, but a vector with very specific properties with respect to the first two. What we will define here will be in three dimensions, and, unlike the dot product, does not generalise easily to other dimensions, (other than 7) though it can in fact be extended.

We are going to define the cross product such that it gives a vector which is perpendicular to the two vectors being crossed. This might sound a bit arbitrary but it shows up in a huge number of different situations in physics in particular and can help us to understand the geometric relation between vectors very simply. Clearly if two vectors are parallel to one another in two dimensions, there won’t be a unique direction which is perpendicular to both vectors. In fact there will be a whole plane of directions perpendicular to the two vectors. We are going to define the cross product such that its magnitude is proportional to the magnitude of the two vectors being crossed, and the $\sin$ of the angle in between them. The direction of the resulting vector will be perpendicular to the two original vectors:

$\vec{a}\times\vec{b}=(\left|\vec{a}\right|\left|\vec{b}\right|\sin\theta)\hat{n}$

where $\hat{n}$ is in the direction perpendicular to $\vec{a}$ and $\vec{b}$. However, this is not uniquely defined. We can think of two vectors (pointing in opposite directions) which are both perpendicular to two other vectors and are unit magnitude. This is shown in the figure below.

The two black arrows are both perpendicular to the red and blue vectors and have magnitude equal to the product of the magnitude of the two vectors times the $\sin$ of the angle in between them. One of these is the cross product. The right hand rule tells us which one.

To disambiguate which of the two perpendicular vectors is the result of the cross product we use the right hand rule. Rather than trying to explain this in words, or in still pictures, I’ll simply point you to a video where this is explained clearly.

In order to understand how to take the cross product, it is informative to look at the cross product of the unit basis vectors. Clearly because their magnitudes are all one and the angle in between them are $\frac{\pi}{2}$, their cross products are going to have unit magnitude, the question is what direction will they be.

For instance, if we want to calculate $\vec{i}\times\vec{j}$ we know that there are two vectors perpendicular to this and of magnitude one. These are $\vec{k}$ and $-\vec{k}$. However, according to the right hand rule we get:

$\vec{i}\times\vec{j}=\vec{k}$

Once you have this, all of the other cross products are the cyclic permutations of this expression. That is, you simply shift each vector to the next position in the expression and take the one at the end and put it at the beginning. Shifting each one of these along one, gets you:

$\vec{k}\times\vec{i}=\vec{j}$

and shifting it along one more gets:

$\vec{j}\times\vec{k}=\vec{i}$

Shifting it along again takes us to the first expression. Swapping around the order of the terms in the cross product changes the result by a sign:

$\vec{j}\times\vec{i}=-\vec{k}$

And the same for the other two above. If we take the cross product of a vector with itself then clearly the angle in between the two vectors is zero and so you are left with the zero vector:

$\vec{i}\times \vec{i}=\vec{0}$

and similarly for $\vec{j}$ and $\vec{k}$ crossed with themselves. The above facts, along with the distributive property of the cross product, will allow us to calculate the cross product of any vectors when given in component form. We can take a product of two vectors: $\vec{a}=a_1\vec{i}+a_2\vec{j}+a_3\vec{k}$ and $\vec{b}=b_1\vec{i}+b_2\vec{j}+b_3\vec{k}$ for arbitrary components $a_i$ and $b_i$ and find the cross product of the two by expanding everything out:

$\vec{a}\times\vec{b}=(a_1\vec{i}+a_2\vec{j}+a_3\vec{k})\times(b_1\vec{i}+b_2\vec{j}+b_3\vec{k})$

$\vec{a}\times\vec{b}=a_1\vec{i}\times b_1\vec{i}+a_1\vec{i}\times b_2\vec{j}+a_1\vec{i}\times b_3\vec{k}+a_2\vec{j}\times b_1\vec{i}...etc.$

$\vec{a}\times\vec{b}=a_1 b_1\vec{0}+a_1 b_2\vec{k}-a_1 b_3\vec{j}-a_2 b_1\vec{k}...etc.$

$\vec{a}\times\vec{b}=\vec{i}\left(a_2 b_3-a_3 b_2\right)+\vec{j}\left(a_3b_1-a_1b_3\right)+\vec{k}\left(a_1b_2-a_2b_1\right)$

Make sure that you can follow the above and fill in the expressions for “etc.”. This looks like a terrible mess but in fact we can come up with a new form of notation that will allow us to remember how to perform this calculation with relative simplicity. First we need to define a determinant of an array.

 How clear is this post?