Prove $|v \times(u \times v)|=\sqrt{((u.u)(v.v)^2-(v.v)(u.v)^2}$

Question

Use the formula for the angle between two vectors $v$ and $u$ to show $$|v \times(u \times v)|=\sqrt{((u.u)(v.v)^2-(v.v)(u.v)^2}$$

I have endlessly used $u.v = |u||v|\cos(\theta)$ but to no avail:

This was my square root after simplifying

$$ \sqrt{\cos^3(\theta)|u|^2|v|^4-\cos^3(\theta)|v|^3|u|^2} $$

But I don't know how to continue further...

Answer 1

Use the angle formula $| u\times v| = |u| \cdot | v| |\sin(\theta )|$

Then we have \begin{eqnarray} | v\times (u\times v)| &&= | v| \cdot | u\times v| \\ &&= | v|^2| u||\sin(\theta)|\\ &&= | v|^2| u|\sqrt{1-\cos^2(\theta)}\\ &&= | v|^2| u|\sqrt{1-\big(\frac{u\cdot v}{ | u| \cdot | v|}\big)^2}\\ &&= \sqrt{(u\cdot u) (v\cdot v)^2 - (u\cdot v)^2(v\cdot v)} \end{eqnarray}

Answer 2

Let $\theta$ the angle between $u$ and $v$. Since $u\times v$ is a vector orthogonal to both of $u$ and $v$ we have \begin{align*} |v\times (u\times v)| &=|v||u\times v|\\ &=|v||u||v||\sin \theta|\\ &=|v||u||v|\sqrt{1-\cos^2\theta}\\ &=|v|\sqrt{|u|^2|v|^2-(u\cdot v)^2}\\ &=\sqrt{|u|^2|v|^4-(u\cdot v)^2|v|^2}\\ &=\sqrt{(u\cdot u)(v\cdot v)^2-(v\cdot v)(u\cdot v)^2}\\ \end{align*}

Answer 3

(Note: the following is not using "the formula for the angle between two vectors $v$ and $u$".)

Using the vector triple product formula...

$$a \times(b \times c) = \langle a , c \rangle \,b - \langle a , b \rangle \,c$$

...for $\,a=v, \,b=u, \,c=v\,$ gives:

$$v \times(u \times v) = \langle v , v \rangle \,u - \langle u , v \rangle \,v$$

Then scalar multiplying both sides by themselves:

$$ \begin{align} |v \times(u \times v)|^2 & = \left\langle \langle v , v \rangle \,u - \langle u , v \rangle \,v \;,\; \langle v , v \rangle \,u - \langle u , v \rangle \,v \right\rangle \\ & = \langle v , v \rangle^2 \,\langle u, u\rangle - 2 \langle u , v \rangle^2 \,\langle v, v\rangle + \langle u , v \rangle^2 \,\langle v, v\rangle \\ & = \langle v , v \rangle^2 \,\langle u, u\rangle - \langle u , v \rangle^2 \,\langle v, v\rangle \end{align} $$