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,
$\endgroup$ 14 Answers
$\begingroup$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.
$\endgroup$ 9 $\begingroup$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...
$\endgroup$ 1 $\begingroup$$|A×B|=|A| ×|B|\sin \theta $ where $\theta$ is the angle between A and B Using simple trigonometry then we let the height of the parallelogram be a letter say X
The area of a parallelogram is given by base × height
$\sin\theta=X/|A|$ Which means that $X= |A| \sin\theta $
Therefore the area will be given by the base $|B| × |A| \sin\theta $ That is essentially the magnitude of the cross product of A and B
$\endgroup$ 1 $\begingroup$Define $A \times_1 B, A \times_2 B$ as the vectors perpendicular to $A, B$, with orientation picked by the right hand rule, and with magnitude $|A||B|sin \theta$, respectively the absolute value of$$ \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 $\times_1 = \times_2$.
$\endgroup$
