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Our linear algebra class didn't cover cross products and I require it for a project. I've learned the following methods.

1) $\mathbf{a}\,\times\,\mathbf{b} = ||\mathbf{a}||\,||\mathbf{b}||\,\sin(\theta)\,\mathbf{n}$

2) Determinant method

I don't have a good intuition for why either method works. However, we did learn about orthogonal projections and orthogonal vector spaces so I'm wondering if those methods are in any way related to computing the cross product.

For example, let $\DeclareMathOperator{Span}{Span}W = \Span\{\mathbf{a}, \mathbf{b}\}$. We know that $(Row\, A)^{\perp} = Nul\,A$ so all we need to do is find a vector from $W^{\perp}$ right? (I know that according to methods 1) and 2), the cross product is unique - but does it have to be?)

Anyway, what if we tried finding a vector from $Nul \begin{bmatrix} \mathbf{a}^T \\ \mathbf{b}^T\end{bmatrix}$ where $\mathbf{a}$ and $\mathbf{b}$ are column vectors? In other words, what if we tried finding a vector from the nullspace of a matrix whose rows were the two vectors we are trying to compute the cross product of? Does that say anything about the cross product?

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  • $\begingroup$ The cross product is intrinsically linked to a notion of area. You need some functional which can measure this sort of thing in order to reproduce the formula. $\endgroup$ – Spencer Jan 7 '18 at 11:04
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    $\begingroup$ If you want to understand cross products better, the user @Bye_World has written some answers here that I've really liked. See if any of these are helpful to you at all: 1, 2, 3, ... $\endgroup$ – got it--thanks Jan 8 '18 at 20:03
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    $\begingroup$ 4, 5, and 6. @Carpetfizz $\endgroup$ – got it--thanks Jan 8 '18 at 20:03
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The cross product was introduced by Gibbs as a simplification of quaternion product. Quaternions were considered too hard to teach by Gibbs, so he split the quaternion product in two: cross product and dot product.

Originally, both products were computed always together, since they represent both the orthogonality (symmetric part) and and parallelness (antisymmetric part) of the two vectors, that allowed to define the product inverse.

The quaternion product is defined with the simple rules:

$i j = k$, $k i = j$, $j k = i$, $ i i = j j = k k = -1$

Those rules imply also:

$j i = -k$, $i k = -j$, $k j = -i$

The original meaning of $i$, $j$ and $k$ was of the coordinate vectors, since vectors were numbers you could multiply them together and they produce other vectors. The key property is that the square of any vector is always a scalar. For the vectors $i$, $j$ and $k$ that scalar is -1.

Following the rules of multiplication described above, the product of two vectors produces two quantities, which we now identify with cross product (the vector part) and with dot product (the scalar part).

What Gibbs did was to introduce a mysterious product, which is really half of the quaternion product. Something that most people ignore.

If you study the quaternion math, the cross product calculation will explain by itself.

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  • $\begingroup$ That’s very interesting and I appreciate your answer, however I don’t think this directly answers the question. $\endgroup$ – Carpetfizz Jan 7 '18 at 14:32
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For vectors in $\mathbb{R}^3$ you are right:

the cross product of two vectors lies in the null space of the $2 \times 3$ matrix with the vectors as rows.

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