For $\vec{x}\in\mathbb{R}^3-\{\vec{0}\}$, let for every $i=1,2,3$ $\alpha_i =\cos^{-1}\langle\hat{\vec{x}}|\underline{e_i}\rangle$, the angle que between the $\vec{x}$ with the $ith$ vector in the base canonique of $\mathbb{R}^3$ and with $\hat{\vec{x}}=\frac{\vec{x}}{\|\vec{x}\|}$. Show that $$\cos^2\alpha_1 + \cos^2\alpha_2 + \cos^2\alpha_3 = 1.$$
Notes:
This exercise is about director cosine, my problem is I do not know how to write the solution with the notation $\langle\hat{\vec{x}}|\underline{e_i}\rangle$ which is in this link (It is translated into English) http://delta.cs.cinvestav.mx/~gmorales/Biberstein/fvd/node32.html
I use $\underline{e_i}$ instead of $e_i$ with the down arrow.
$\hat{x} = \frac{x}{|| x||}$.
As $\{ e_i \} $ form a basis for $\mathbb{R}^3$,
$$ \hat{x} = \sum_i \langle \hat{x} | e_i \rangle e_i = \sum_i \cos (\alpha_i )e_i$$
Lastly, $||\hat{x}|| = \frac{|| x ||}{||x||} = 1$
Therefore,
$$1 = \langle \hat{x} | \hat{x} \rangle = \sum_i \sum_j \cos \alpha_i \cos \alpha_j \langle e_i | e_j \rangle = \sum_i \cos^2 \alpha_i$$
Where the last equality is from the fact that the canonical basis is orthonormal.