Vector Cross Product

The cross product is an operation on vectors, but I didn’t include it in the vectors section because I felt like it would be nice to introduce this operation after covering planes, but I didn’t want to introduce planes before having introduced linear systems, so here we are.

In the lesson about planes, we have found the equation of a plane from three given points. The way we did this was by forming two vectors from the three points given - say $\vec{u}$ and $\vec{v}$ - and we created a linear system of equations by saying that the vector we’re looking for has to be orthogonal to both of them. If $\vec{n}$ is the normal vector we are looking for, that would be

The cross product solves exactly this problem. The cross product between two vectors $\vec{u}$ and $\vec{v}$, denoted by the symbol $\vec{u} \times \vec{v}$, returns a vector orthogonal to $\vec{u}$ and $\vec{v}$.

Derivation

We will derive the cross product through a similar process used to find a vector orthogonal to two other vectors as we did in the lesson about planes.

We are given two three-dimensional vectors

We want to find a vector $\vec{n}$ such that

From our lesson about orthogonal vectors, we know that, when two vectors are orthogonal, their dot product equals $0$.

This means that

and

must both be true. We have a system of equations:

We can multiply the top equation by $v_3$ and the bottom equation by $u_3$. That gets us to

We now subtract the second equation from the first to get

In the resulting equation, $n_3 u_3 v_3 - n_3 u_3 v_3$ cancels out, and the rest can be factorized:

Remember, we are given $\vec{u}$ and $\vec{v}$, and we want to find some expression for the components of $\vec{n}$. The system of equations has infinitely many solutions, and we can choose any of the possible solutions. We will intelligently pick our values for $n_1$ and $n_2$ such that the terms cancel out and the equation is solved:

Substituting into our equation, we get

because the terms cancel out.

We now have our expressions for $n_1$ and $n_2$, but we’re missing $n_3$. To find $n_3$, we substitute our chosen values for $n_1$ and $n_2$ into our original linear system of equations and solve it:

As you can see, we get an expression for $n_3$:

We have now found a way to rewrite all components of the normal vector $\vec{n}$ in terms of the components of $\vec{u}$ and $\vec{v}$, such that

That is,

This is not as hard to remember as it may seem at first, because in each component, there is a subtraction taking place. Each operand of the subtraction is a product. The first operand of the product is always a component of $\vec{u}$, the second is always a component of $\vec{v}$.

If you look at the first operand of each subtraction, if the first operand of the product is the $i$th element of $\vec{u}$, the second operand of the product will be the $(i + 1)$th element of $\vec{v}$. If you go past $3$ - the size of the vector - it wraps around back to $1$.

Looking at the second operand of each subtraction, it’s the other way around: the first operand of the product is the $(i + 1)$th element of $\vec{u}$, and the second operand is the $i$th element of $\vec{v}$.

Hopefully, this allows you to memorize this more easily.

Properties of the Resulting Vector

If you have a vector $\vec{n} = \vec{u} \times \vec{v}$, then $\vec{n}$ will tell us some information about how $\vec{u}$ and $\vec{v}$ are related to each other. Also, some interesting facts emerge.

Angle Between the Operands of the Cross Product

For example, the angle between $\vec{u}$ and $\vec{v}$ has an effect on the length of $\vec{n}$.

That is, when $\vec{u}$ and $\vec{v}$ are orthogonal to each other, $\vec{n}$ reaches its maximum length of $\lVert \vec{u} \rVert \lVert \vec{v} \rVert$, whereas when $\vec{u}$ and $\vec{v}$ are parallel, $\vec{n}$ reaches its minimum length of $0$.

In fact, an interesting geometric fact is that the length of $\vec{u} \times \vec{v}$ is the same as the area of the parallelogram spanned by $\vec{u}$ and $\vec{v}$.

Reversing the Operands

Notice how two vectors $\vec{u}$ and $\vec{v}$ always have two possible normal vectors which are orthogonal to both $\vec{u}$ and $\vec{v}$ at the same time. These two possible normal vectors point in exact opposite directions.

But the cross product only gives us one normal vector? How do we find the other? Well, for one, you could just multiply the result by the scalar $-1$ to reverse the direction the normal vector points to and therefore find the second normal vector.

But another, more interesting way, is by using the property of the cross product

Therefore, reversing the order of the operands in the cross product simply finds the other possible normal vector. This also implies that the cross product is not commutative.

Example

Take the vectors

We want to find a vector orthogonal to both of these vectors. That is

We use the formula derived above for the cross product:

We can now say that

We can even check this.

For $\vec{u}$:

Therefore, $\vec{n} \perp \vec{u}$.

For $\vec{v}$:

Therefore, $\vec{n} \perp \vec{v}$.

One Final Remark

Keep in mind that the cross product in this form only exists in three dimensions. There are some more advanced cases in higher dimensions, but that is beyond the scope of this lesson.