The cross product between two vectors in $\Bbb{R}^3$ (call them a and b) is denoted a $\times$ b and the result is a vector in $\Bbb{R}^3$ orthogonal to the first two. There are a variety of ways of computing this resultant vector. One way in particular is known from the symbolic determinant involving i j k and the entries of a and b.

I recall a professor awhile back trying to explain why a $\times$ b can be seen from this by appealing to the notion of finding the projection of the parallelogram with sides a and b onto the xy, xz and yz planes (imagine a and b floating in 3-space, so we're finding the projection of a and b onto these planes and finding the new area of the parallelogram after this projection). The goal was to show that the factor scaling the i component of a $\times$ b was the area of the parallelogram formed by a and b after being projected onto the yz plane, and similarly for the j component being scaled by the area of the projected parallelogram of a and b in the xz plane and the k component from the xy plane.

The explanation seemed straightforward at the time, but in trying to remember it I'm a little lost. Is anyone familiar with this notion of finding the areas under projection and these areas being the magnitudes of the i j and k components of a $\times$ b?

I hope that isn't too vague. I know that we had an explanation for the sign change of the j component as well being connected to these smashed projections but I don't recall how that worked either.

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    $\begingroup$ The monolith of words was a little intimidating, so I went ahead and used things called linebreaks to make it more likely that readers would read it all. Whitespace really does miracles for making text more readable! $\endgroup$
    – rschwieb
    Nov 19, 2012 at 0:29
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    $\begingroup$ I made an attempt to answer this here: math.stackexchange.com/questions/596158/cross-products $\endgroup$
    – Agol
    Sep 10, 2014 at 5:21

1 Answer 1


The magnitude of the cross-product $\|\mathbf{a}\times \mathbf{b}\|$ gives the area of the parallelogram formed by vectors $\mathbf{a}$ and $\mathbf{b}$ as edges. I do not know exactly what your professor is talking about so I will take a guess here.

From the Cauchy-Binet Formula, the area of the parallelogram is related to the areas of the projections of the parallelogram onto the $xy,\ xz$ and $yz$ planes. This is a higher dimension analogue of the familiar Pythagorean theorem. If we denote $\|\mathbf{a}\times \mathbf{b}\|_{xy}$ to be the area of the projection onto the $xy$-plane, then the areas are related as $$\|\mathbf{a}\times \mathbf{b}\|^2 = \|\mathbf{a}\times \mathbf{b}\|_{xy}^2 + \|\mathbf{a}\times \mathbf{b}\|_{xz}^2 + \|\mathbf{a}\times \mathbf{b}\|_{yz}^2$$ Now each projection is given by a smaller determinant. The $xy$ projection for example is given by $$\|\mathbf{a}\times \mathbf{b}\|^2_{xy} = \begin{vmatrix}a_1^2 + a_2^2 & a_1b_1 + a_2b_2 \\ a_1b_1 + a_2b_2 & b_1^2 + b_2^2\end{vmatrix}$$ where $a_i$ and $b_i$ are the respective components of $\mathbf{a}$ and $\mathbf{b}$. This is evaluated as $$(a_1^2 + a_2^2)(b_1^2 + b_2^2) - (a_1b_1 + a_2b_2)^2 = a_1^2b_2^2 -2a_1b_1a_2b_2+ a_2^2b_1^2 = (a_1b_2 - a_2b_1)^2$$ This is precisely the (square of the) third component of the cross-product.


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