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Given vectors $a, b \in \mathbb{R}^3$, the cross (vector) product of $a \times b = c$ is defined as a vector orthogonal to both $a$ and $b$ such that the right hand rule is satisfied, with the additional stipulation that $\|c\|=\|a\|\|b\|\sin(\theta)$ where $\theta$ is the angle between $a,b$. But any vector $c’ \in \mathbb{R}^3$ such that \begin{align*} a \cdot c’ &= 0, \\ b \cdot c’ &=0 \end{align*} satisfies the orthogonality condition. For non-parallel $a,b$, there are infinitely many vectors $c’$ that do this (by the rank theorem). Albeit, these $c’$ lie on the same line. My questions are:

  1. What is the purpose of wanting the cross product to have magnitude equal to the parallelogram spanned by $a$ and $b$?
  2. How does the determinant formulation of the cross product $$ c = \mathrm{det} \begin{bmatrix} \hat{\imath} & \hat{\jmath} & \hat{k} \\ a_x & a_y & a_z \\ b_x & b_y & b_z \end{bmatrix} $$ ensure that $\|c\|=\|a\|\|b\|\sin(\theta)$? I understand how to work backwards, i.e. to show that given this formulation, $\|c\|=\|a\|\|b\|\sin(\theta)$. I am more interested in working forwards: why does this particular formulation give the desired magnitude?
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2 Answers 2

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Q1. What is the purpose of wanting the cross product to have magnitude equal to the parallelogram spanned by $a$ and $b$?

I'm not sure what sort of answer you expect to get to this question. The purpose is that it gives us all of the properties of the cross product that we are familiar with.


Q2. How does the determinant formulation of the cross product ensure that $\|c\|=\|a\|\|b\|\sin(\theta)$? I am more interested in working forwards: why does this particular formulation give the desired magnitude?

I sat down to write an answer and discovered this is fiddlier than I thought. The twist is that we need to know that the cross product follows linearity and the distributive law: $$ \vec{a} \times (\lambda \vec{b} ) = \lambda \, \vec{a} \times \vec{b} $$ $$ \vec{a} \times (\vec{b} +\vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$$ Linearity follows easily from your definition of the cross product but distributivity does not.

If we do know the distributive law then we may write the three Cartesian unit vectors as $\hat{e_i}$ for $i=1,2,3$. Then it follows from your definition of the cross product that (we do not use summation convention because it may lead us to make hidden assumptions): $$ \hat{e_i} \times \hat{e_j} = \sum_k \varepsilon_{ijk} \, \hat{e_k}$$ Then write $$ \vec{a} = \sum_i a_i \hat{e_i} $$ $$ \vec{b} = \sum_j b_j \hat{e_j} $$ $$ \vec{a} \times \vec{b} = (\sum_i a_i \hat{e_i}) \times (\sum_j b_j \hat{e_j}) $$ (now we use the distributive law) $$ = \sum_{i,j} a_i b_j \, \hat{e_i} \times \hat{e_j} $$ $$ = \sum_{i,j,k} a_i b_j \, \varepsilon_{ijk} \, \hat{e_k} $$ But this is exactly the determinant definition which we were looking for.


So how do we prove distributivity? It's kind of obvious if you draw some diagrams, so we can form a 'geometric proof'. It also follows from the definition using determinants or the properties of the scalar triple product - but that's circular.

I can't quite find a purely algebraic proof. Perhaps someone else can.

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Regarding the second question:

The formula for $|a\times b|$ follows from four axioms:

  1. $|a\times b| = |a||b|$ if $a\perp b$.
  2. $a\times a = 0$.
  3. Bilinearity of $\times$.
  4. $a\cdot b = |a||b|\cos\theta$.

This is because $a - \frac{a\cdot b}{|b|^2}b$ is orthogonal to $b$, thus $$\begin{aligned} |a\times b|^2 &= |(a - \frac{a\cdot b}{|b|^2}b)\times b|^2 \\&= |a - \frac{a\cdot b}{|b|^2}b|^2|b|^2 \\&= (|a|^2 - \frac{(a\cdot b)^2}{|b|^2})|b|^2 \\&= |a|^2|b|^2 - |a|^2|b|^2\cos^2\theta \\&= |a|^2|b|^2\sin^2\theta. \end{aligned}$$

(1) is the most difficult to verify for the determinant definition, but once you do the above argument goes through.

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