Find the area of the parallelogram determined by vectors $\vec m$ and $\vec n$

Question

Given vectors: $$|\vec a|=2\\|\vec b|=4\\\angle(\vec a,\vec b)=\frac{2\pi}{3}\\\vec m=2\vec a+5\vec b\\\vec n=3\vec a-\vec b$$ Find the area of the parallelogram determined by vectors $\vec m$ and $\vec n$.

This is my solution: $$\require{cancel}\begin{align}A&=|(2\vec a+5\vec b)\times(3\vec a-\vec b)|\\&=|\cancel{2\vec a\times3\vec a}-2\vec a\times\vec b+15\vec b\times\vec a-\cancel{5\vec b\times\vec b}|\\&=|-16\cdot\frac{\sqrt 3}{2}-120\cdot\frac{\sqrt 3}{2}|\\&=68\sqrt 3\end{align}$$ But, in my book the solution is $34\sqrt 3$. Where did I make a mistake? I noticed that my solution is twice of the actual result, so maybe I made some stupid mistake in my calculations, but I cannot figure out where.

Answer 1

I think your solution is true. There is a little mistake:

In your proof must be $2\vec{a}\times3\vec{a}=\vec{0}$, but the answer is right.

Your book does not know, how to calculate the area of parallelogram.

I think your book calculated an area of the triangle.