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

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,

• If you define the cross product by $a \cdot (b \times c) = \det(a,b,c)$ (the best definition of the cross product, IMO), and knowing that determinants represent the volumes of parallelotopes, this just comes down to Cavalieri's principle.
– user137731
Oct 6, 2015 at 3:10

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.

• Hi @Arthur - very cool and simple answer. Just a quick follow-up question - you meant for the vector b to have coordinates (0,$b_2$,0), right? Thanks, Sep 25, 2015 at 22:47
• @LebronJames No, in that case it would be established that $\vec a$ and $\vec b$ are perpendicular, which is not assumed. We know that $a_2=0$ since that's how we chose the coordinate system, but for $\vec b$ we can only tell that it lies in the $xy$-plane. Sep 26, 2015 at 8:12
• Hi @Arthur hmm...isn't the base |a| = |$a_1$|, but then the height is |b| = $\sqrt{b_1^2 + b_2^2}$? Then the product |a||b| doesn't equal $|a_1b_1|$, which is the magnitude of the cross-product. Thanks, Sep 26, 2015 at 8:30
• @LebronJames No, the height of a parallelogram is perpendicular to the base, so $\vec b$ isn't the height. Sep 26, 2015 at 9:03
• Ok, got it now - thanks so much for your help and patience @arthur. Have a great night. Sep 26, 2015 at 10:14

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...

• Hi @Martigan, is your equation a definition ... or can it be derived from basic geometry? If it is a derivation, do you mind just explaining a bit more about where it comes from? Thanks, Sep 25, 2015 at 22:39

$|A×B|=|A| ×|B|\sin \theta$ 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

• @JMP Despite there's an answer which is very similar to this. I think this one made me understood better. The more you think of it is rather trivial. Unless you forget that the area of a paralellogram is the base times the height just as a rectangle. Mar 18, 2020 at 1:14
• How is $\theta$ defined? Should this be |A x B| = |A| |B| |sin(theta)|? Sep 21, 2023 at 17:41

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 $$\times_1 = \times_2$$.