How to use projection of vectors in order to compute length of a cross product?

Question

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

I broke it down into $2u \times (5v-4u) + 3v(5v-4u)$ and then further broke it down into $(2u \times 5v) -(2u \times 4u) + (3v \times 5v) -(3v \times 4u)$.

But what I don't get is, how do I use $(u|v) =−12$ here in order to compute the answer? Is it connected to the formula $u \times v = \|v\| T(u')$?

Answer 1

The cross-product of a vector by itself is always $0$, and $v\times u=-u\times v$. So,$$(2u+3v)\times(5v-4u)=(10+12)(u\times v)=22u\times v.$$On the other hand$$-12=(u|v)=\|u\|\|v\|\cos\theta=20\cos\theta,$$where $\theta$ is the angle between $u$ and $v$. So, $\cos\theta=-\frac35$, and therefore $\sin\theta=\frac45$. Now, use the fact that $\|u\times v\|=|\sin\theta|\|u\|\|v\|$.