Why does the magnitude of the cross-product of a and b give the area of a parallelogram spanned by a and b?

Question

I tried looking it up but many websites just state it without proof and without intuition. I'm hoping to learn it a little bit better so that I don't forget how to compute the Jacobian when working with surface integrals where the divergence theorem is not applicable.

If you have a good online reference instead, please feel free to provide it :-)

Thanks,

Answer 1

Choose coordinates so that the two vectors $\vec a, \vec b$ are in the $xy$-plane, with $\vec a$ along the $x$-axis. (Note that as long as you've decided on a unit length, exactly which direction you choose for the coordinate axes doesn't change anything. The vectors and their cross product live in a coordinate-free space, just floating around. We're just imposing coordinates to make concrete calculations simpler.) That means we can set $\vec a = (a_1, 0, 0)$ and $\vec b = (b_1, b_2, 0)$. This gives $$ \vec{a}\times \vec b = (0, 0, a_1b_2) $$ and the length of this vector is $\sqrt{(a_1b_2)^2} = |a_1b_2|$, obviously. But the parallelogram has base $|a_1|$ and height $|b_2|$, which means that the area of the parallelogram is given by the exact same expression.

Answer 2

It is probably because the answer is simple in terms of classical 2D geometry.

$||\vec u\times \vec v||=||\vec u||.||\vec v||.\sin(\vec u,\vec v)$

But the area of the parallelogram defined by $\vec u$ and $\vec v$ is the base multiplied by the height. If you take $\vec u$ as the base, the height is $h=||\vec v||.\sin(\vec u,\vec v)$, hence the result...

Answer 3

Define $A \times_1 B, A \times_2 B$ as the vectors perpendicular to $A, B$, with orientation picked by the right hand rule, and with magnitude $|A||B|sin \theta$, respectively the absolute value of $$ \begin{vmatrix} \bf{i} & \bf{j} & \bf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} $$ A simple geometric proof shows $A \times_1 B$ is linearly dependent in $A, B$. And determinant properties show that $A \times_2 B$ is linearly dependent in $A, B$ also. It remains to observe that $\times_1, \times_2$ coincide on basis vectors $\bf{i}, \bf{j}, \bf{k}$, which means, as expected, that $\times_1 = \times_2$.