I am studying vectors and I have tried to make this exercise:
Given $u, v$ two vectors making an angle of 60 degrees, such that $|| u || = a$ and $|| v || = b$, calculate $|| -2u \times 3v||, || (u+v) \times (u-3v)||$ and $||-2(u\cdot v)u \times (u\cdot v)v||$.
I make this exercise: $|| -2u \times 3v|| = ||-2u|| \;||3v|| \sin 60 = |-2|.|3|.||u|| ||v|| \sin 60 = 6ab\frac{\sqrt{3}}{2} = 3\sqrt{3}ab$.
Another item is
$|| -2(u\cdot v)u \times (u\cdot v)v|| = ||-2(u\cdot v)u|| ||(u\cdot v)v|| \sin 60 = |-2(u\cdot v)|.||u|| .|u\cdot v|||v|| \sin 60 = 2 ||u||^2 ||v||^2 \cos^2 60 ||u||||v||\sin 60 = \frac{\sqrt{3}}{4}a^3b^3 $.
So, I don't have idea how I proceed with the second item.
But I know that $u \times v = - v \times u$, $u \times av = au \times v = a(u \times v)$ and $u \times (v+w) = u \times v + u \times w$. Then, we have $$ (u+v) \times (u-3v) = u \times u + u \times (-3v) + v \times u + v \times (-3v) = u \times u - 3 u \times v + v \times u - 3 v \times v. $$
But I don't see what it is help me. I think that $u \times u = 0$. Then $$ (u+v) \times (u-3v) = - 4 u \times v $$ Therefore, $||(u+v) \times (u-3v)|| = ||- 4 u \times v|| = 4 ||u \times v|| = 4ab\frac{\sqrt{3}}{2}$.