Computing cross product using norm and angle

Question

Sorry for the weird title, if someone finds a better title for my problem be my guest to edit it ;)

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For $\mathbf{v,w} $ in R³ with $\mathbf{||v||=1 ;||w||=4; \theta =\frac{2\pi}{3}}$

Solve the following:

1. $\lVert (3v+w)\times(v-2w)\rVert$
2. $\langle w\times 5v, v-3w\rangle$

I've already calculated

$$\langle v,w\rangle = \frac{-1}{2}||v|| \cdot||w|| =-2$$

Also, I've transformed 1. into

$$\lVert-7(v\times w)\rVert$$

But I'm a bit lost here.

Next I masked $w\times 5v $ as $x$ so as 2. we now have

$$\langle x,v-3w\rangle$$

This all seems to bring me not a bit closer to the solution, can someone suggest a valid approach?

Answer 1

Just as $v \cdot w = |w| |v| \cos \theta$, there is a corresponding formula $|v \times w| = |v| |w| \sin \theta$. Since the angle and the lengths are known, you should be able to compute this magnitude.

Answer 2

Hints: First, $$ \begin{align} (3v+w)\times(v-2w) &=-7v\times w \end{align} $$ and $$ \|-7v\times w\|=7\|v\|\|w\|\left|\sin\left(\frac{2\pi}{3}\right)\right| $$ For $\langle w\times 5v, v-3w\rangle$, note that $w\times v$ is perpendicular to both $v$ and $w$.