The vectors $u, v$ are such that $||u|| = 4$, $||v|| = 5$ and $(u|v) = −12$. Compute the length of the vector $(3u+2v)×(2v−3u)$.

Question

I'm trying to solve this problem but not sure if I'm doing it correctly. Here's the problem:

The vectors $u, v$ are such that $||u|| = 4$, $||v|| = 5$ and $(u|v) = −12$. Compute the length of the vector $(3u+2v)×(2v−3u)$.

This is how I tried to solve it: $$(3u+2v)×(2v−3u)=3u×2v+3u×3u+2v×2v+2v×(-3u)=2(3u×2v)=12(u×v)$$ and since the length of the equations above is asked for I use the fact that $$|u×v|=||u||||v||sin[u,v].$$ The angle between u and v is obtained by using the given information and using the following definition: $$(u|v)=||u||||v||cos[u,v]$$

$$[u,v]=arccos(-\frac{3}{5})$$ Lastly, I compute the length of the given vector like this: $$12|u×v|=12\times 4\times 5\times sin(arccos(-\frac{3}{5}))=192.$$

Answer 1

Correct. You can also use $|u\times v|^2=|u|^2|v|^2-(u\cdot v)^2$ to avoid working with the trigonometry.