$p,q,r \in \mathbb{R^+}$ $$p^2 +q^2 = 9 $$ $$q^2+qr+r^2=16$$ $$ p^2 + \sqrt{3} pr +r^2 =25$$ Find $2pq + pr+ \sqrt{3} pr$
My attempt:
I tried to interpret this equation as one involving vector dot products by the steps presented below:
Let $\vec{p}$ , $ \vec{q}$ and $ \vec{r}$ be vectors, then the first identity suggests that $ \vec{p} \perp \vec{q}$ as we would see if we compare the formula of dot products:
if, $$ |p+q|^2 = 9 $$
$ \implies p \cdot q = 0 $ and since they are non zero vectors , the vectors must necessarily be perpendicular. Similarly, we find that $ \vec{q} $ and $ \vec{r}$ are angled at $ \frac{\pi}{3}$ degrees and finally $ \vec{p} $ and $ \vec{r}$ are at an angle of $ \frac{\pi}{6}$ degrees.
So, what we want to find is this:
$$ F(p,q,r) = 2 |p| |q| + |p| |r| + \sqrt{3} |p| |r|$$
One last thing which I thought of doing is writing $ \vec{r}$ in the basis of $ \vec{p}$ and $ \vec{q}$ since they are orthogonal, but, I'm worried that $ \vec{r}$ may not be in the plane defined by $ \vec{p}$ and $ \vec{q}$. Anyways I have shown the decomposition below:
$$ \vec{r} = \frac{ \vec{r} \cdot \vec{p} }{ |p| } \vec{p} + \frac{ \vec{r} \cdot \vec{q} }{|q| } \vec{q}$$
Not sure how to proceed next/ if it is even solvable by this method. Appreciate any help given.
