### Coordinates and vectors

Let $\hat{e}_3=a\hat{e}_1+b\hat{e}_2$. Since $\hat{e}_3\cdot\hat{e}_3=1$ we have
$a^2+b^2+2ab\hat{e}_1\cdot\hat{e}_2=1$
Let $\alpha$ be the angle between $\hat{e}_1$ and $\hat{e}_3$, and $\beta$ be the angle between $\hat{e}_2$ and $\hat{e}_3$. We have
$\cos\alpha=\hat{e}_1\cdot\hat{e}_3=a+b\hat{e}_1\cdot\hat{e}_2\\\cos\beta=\hat{e}_2\cdot\hat{e}_3=b+a\hat{e}_1\cdot\hat{e}_2$
Since $\alpha=2\beta$ we have $\cos\alpha=\cos(2\beta)=2\cos^2\beta-1$ so that
$\hat{e}_1\cdot\hat{e}_3=a+b\hat{e}_1\cdot\hat{e}_2=2(\hat{e}_2\cdot\hat{e}_3=b+a\hat{e}_1\cdot\hat{e}_2)^2-1$
Since $\hat{e}_1\cdot\hat{e}_2$ is known we get two equations for $a$ and $b$ which have to be solved. Please solve.