The formula for the magnitude of cross product, $\| u\times v \| = \| u \| \|v \| \sin \theta $

Question

Can someone show me a proof of the magnitude (length) of the cross product: $$\|u \times v \| = \| u \| \|v \| \sin \theta $$

Answer 1

The magnitude of the cross product is the positive area of the parallelogram having $a$ and $b$ as sides:

From here, it is easy to see that when the angle is 0 or 180 degrees, the area of the parallelogram would be 0. Similarly, the area is largest when the angle is 90 degrees.

Answer 2

It depends on the definition of cross product you use. The only definition for which the absolute value is not immediate is $$ (x,y,z)\times(u,v,w)=(yw-zv,zu-xw,xv-yu)\tag{1} $$ Note that the dot product is $$ (x,y,z)\cdot(u,v,w)=xu+yv+zw\tag{2} $$ A rotation is represented by a matrix $R$ so that $R^TR=I$. The dot product is invariant under rotations: $$ \begin{align} (Ra)\cdot(Rb) &=(Ra)^T(Rb)\\ &=a^TR^TRb\\ &=a^Tb\\ &=a\cdot b\tag{3} \end{align} $$ Since lengths are computed using dot products, lengths are preserved by rotations, also. Thus, the side lengths of the triangle formed by $a$, $b$, and $b-a$ are preserved, and likewise, $\theta$, the angle between $a$ and $b$.

If $R$ rotates $a$ to the positive $x$-axis, then $$ \begin{align} a\cdot b &=(|a|,0,0)\cdot(b_x,b_y,b_z)\\ &=|a|b_x\\ &=|a||b|\cos(\theta)\tag{4} \end{align} $$

$\hspace{3.5cm}$

The sum of the square of the length of the cross product and the square of the dot product is $$ \begin{align} &|(x,y,z)\times(u,v,w)|^2+((x,y,z)\cdot(u,v,w))^2\\ &=(yw-zv)^2+(zu-xw)^2+(xv-yu)^2+(xu+yv+zw)^2\\ &=(yw)^2+(zv)^2+(zu)^2+(xw)^2+(xv)^2+(yu)^2 +(xu)^2+(yv)^2+(zw)^2\\ &=x^2(u^2+v^2+w^2)+y^2(u^2+v^2+w^2)+z^2(u^2+v^2+w^2)\\ &=(x^2+y^2+z^2)(u^2+v^2+w^2)\\ &=|(x,y,z)|^2|(u,v,w)|^2\tag{5} \end{align} $$ Equations $(4)$ and $(5)$ yield $$ \begin{align} |(x,y,z)\times(u,v,w)|^2 &=|(x,y,z)|^2|(u,v,w)|^2-((x,y,z)\cdot(u,v,w))^2\\ &=|(x,y,z)|^2|(u,v,w)|^2-|(x,y,z)|^2|(u,v,w)|^2\cos^2(\theta)\\ &=|(x,y,z)|^2|(u,v,w)|^2(1-\cos^2(\theta))\\ &=|(x,y,z)|^2|(u,v,w)|^2\sin^2(\theta)\tag{6} \end{align} $$ Therefore, $$ |a\times b|=|a||b||\sin(\theta)|\tag{7} $$