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Vector Cross Product

We will prove that:

For any two linearly independent vectors in \(\mathbb{R}^3\) \(\vec{r}\) and \(\vec{v}\), the magnitude of the vector cross product \(\vec{r} \times \vec{v}\) is equal to the area of the parallelogram with \(\vec{r}\) and \(\vec{v}\) as sides.

Additionally, we will show that for any two linearly dependent vectors, the cross product is the null vector and thus the area of the parallelogram described above is zero.

Before we start, it should be mentioned that there are definitely many ways to prove this. This proof is somewhat unconventional.


Consider two linearly independent vectors \(\vec{r}\) and \(\vec{v}\) in any plane in \(\mathbb{R}^3\). For the sake of simplicity, let:

\(\vec{r} = r_x\hat{i} + r_y\hat{j}\)

\(\vec{v} = v_x \hat{i} + v_y \hat{j}\)

Our decision to put \(\vec{r}\) and \(\vec{v}\) in the plane \(z=0\) is completely arbitrary. We could have, for instance, just as easily chosen the plane \(x=5\). This decision is made purely to make the problem more simple.

Define the following:

\(\phi\) is the angle between \(\vec{r}\) and \(\vec{v}\)

\(\theta\) is the angle between \(\vec{v}\) and either \(\hat{i}\) or \(\hat{j}\). (The choice is, again, arbitrary. So that we may be consistent throughout the remainder of this proof, choose \(\hat{i}\))

The area of a parallelogram is given by: \(A= bh\), where \(b\) is the base and \(h\) is the vertical height.

Now, \(\vec{r}\) and \(\vec{v}\) have components that, if expressed in relation to \(\hat{i}\) and \(\hat{j}\), are very cumbersome to work with. What we are going to do is construct a linear map \(A\) such that:

\(A = T(\vec{x}): \mathbb{R}^2 \longrightarrow \mathbb{R}^2\)


\(A\vec{v} = \left[ \begin{array}{c} |\vec{v}|\\ 0 \end{array} \right]\)

In other words, this linear transformation will shift our basis vectors so that one of them is parallel to \(\vec{v}\). Then we may take the cross product in this new basis, where \(\vec{v}\) is defined as above and the components of \(\vec{r}\) are in terms of only one angle.

Now let us construct \(A\):

\(A\vec{v} = \left[ \begin{array}{c} |\vec{v}|\\ 0 \end{array} \right]\)

\(A\vec{r} = \left[ \begin{array}{c} |\vec{r}|\cos(\phi)\\ |\vec{r}|\sin(\phi) \end{array} \right]\)

Letting \(A =\left[ \begin{array}{cc} a_{11} & a_{12}\\ a_{21} & a_{22}\end{array} \right]\)


\(a_{11}v_x +a_{12}v_y = |\vec{v}|\)

\(a_{21}v_x + a_{22}v_y = 0\)

\(a_{11}\cos(\phi +\theta) + a_{12} \sin(\phi + \theta) = \cos(\phi)\)

\(a_{21} \cos(\phi +\theta) + a_{22} \sin(\phi + \theta) = \sin(\phi)\)

Now, we want to find \(A\), so we have a linear system of 4 equations in 4 unknowns, which is solvable. I will leave the details of solving this out, as they are somewhat redundant (just use substitution). The elements of \(A\) are:

\(a_{22} = \frac{\sin(\phi)}{(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))}\)

\(a_{21} = \frac{-\sin(\phi)\tan(\theta)}{(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))}\)

\(a_{12} = \frac{v_x\cos(\phi) - |\vec{v}| \cos(\phi +\theta)}{v_x(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))}\)

\(a_{11} = \frac{ |\vec{v}| \sin(\phi +\theta) - v_x \cos(\phi) \tan(\theta)}{v_x(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))}\)

Now, before we move on, we need to make sure \(\det(A) \neq 0\), as this would imply that \(T(\vec{x})\) is not a one-to-one transformation.

Lemma: \(\det(A) = 0\) if and only if \(\vec{r}\) and \(\vec{v}\) are linearly dependent.


\(A= \left[ \begin{array}{cc} \frac{ |\vec{v}| \sin(\phi +\theta) - v_x \cos(\phi) \tan(\theta)}{v_x(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))} & \frac{v_x\cos(\phi) - |\vec{v}| \cos(\phi +\theta)}{v_x(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))}\\ \frac{-\sin(\phi)\tan(\theta)}{(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))} & \frac{\sin(\phi)}{(\sin(\phi + \theta) - \tan(\theta) \cos(\phi +\theta))}\end{array} \right]\)

Finding \(\det(A)\), it is apparent that:

\(\det(A) = 0\) \(\Rightarrow\) \(0= |\vec{v}|sin(\phi) \sin(\phi+\theta) - |\vec{v}|\cos(\phi +\theta) \sin(\phi) \tan(\theta)\)

\(\Rightarrow\) \(\tan(\theta) \cos(\phi +\theta) = \sin(\phi + \theta)\)

\(\Rightarrow\) \(\tan(\theta) =\tan(\phi + \theta)\) \(\Rightarrow\) \(\phi = 0\)

\(\phi\) is the angle between \(\vec{r}\) and \(\vec{v}\), hence if \(\phi = 0\), then \(\exists\) \(k \in \mathbb{R}\) such that \(\vec{r} = k \vec{v}\). This is the definition of linear dependence and we may conclude that \(\det(A) = 0\) if and only if \(\vec{r}\) and \(\vec{v}\) are linearly dependent.

This proves our lemma.

We will return to this case at the end. For now, let us proceed assuming, as we did at the outset, that \(\vec{r}\) and \(\vec{v}\) are linearly independent.


\(T(\vec{r}) = \left[ \begin{array}{c} |\vec{r}|\cos(\phi)\\ |\vec{r}|\sin(\phi) \end{array} \right]\)

\(T(\vec{v}) = \left[ \begin{array}{c} |\vec{v}|\\ 0 \end{array} \right]\)

Now, taking the cross product in this new basis:

\(T(\vec{r}) \times T(\vec{v}) = \frac{-(|\vec{r}||\vec{v}|\sin(\phi))T(\hat{k})}{|T(\hat{k})|}=-(|\vec{r}||\vec{v}|\sin(\phi))\hat{k}\)

\(\Rightarrow\) \(|T(\vec{r}) \times T(\vec{v})| = |(|\vec{r}||\vec{v}|\sin(\phi))|\)

Now, in our new basis, \(|\vec{v}|\) is the base of the parallelogram. Then:

\(|\vec{v}| = b\)

\(|\vec{r}|\sin(\phi))|\) is the vertical component of \(T(\vec{r})\) in our transformed basis, i.e. the height of the parallelogram. Then:

\(|(|\vec{r}||\vec{v}|\sin(\phi))| = bh =A\)

Thus we have proved the first part of our statement.

Now, return to the case in which \(\vec{r}\) and \(\vec{v}\) are linearly dependent. \(T(\vec{x})\) is no longer valid under this condition, so we cannot use our transformed basis. We can, however, take the cross product in our initial basis:

\(\vec{r} \times \vec{v} = k \vec{v} \times \vec{v} = (kv_xv_y - kv_yv_x)\hat{k} = \vec{0}\)

\(\Rightarrow\) \(|\vec{r} \times \vec{v} | = 0\)

This proves the second part of our statement.


Image Credit: Wikipedia

One thing that is interesting to note is that the transformation we did in this proof is akin to one many physics students do without even thinking twice about it. Consider a block on an inclined ramp. You wish to find the time it will take for the block to reach the bottom of the ramp. Many students use a "tilted axis" to solve this problem. This is essentially what we did above. However, most physicists will never rigorously justify this. This is an example of the fact that math is defined very strictly by axioms, and most often intuition is not enough to solve a problem. In physics, often times, intuition is sufficient, due to the non-axiomatic structure of the discipline.

Note by Ethan Robinett
2 years, 2 months ago

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Nice but very tough to understand Gaurav Agarwal · 7 months, 2 weeks ago

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