2.7. Vector $\vec{a}$ forms equal angles with the coordinate axes. Find its coordinates if $|\bar{a}|=\sqrt{3}$.
The problem is solved. The direction cosines satisfy condition (2.29), and since the angles formed by vector $\vec{a}$ with the coordinate axes are equal, i.e., $\alpha=\beta=\gamma$, taking this into account, (2.29) in the considered case will take the form: $3 \cos ^{2} \alpha=1$. From this, it follows that

$$
\cos \alpha=\cos \beta=\cos \gamma= \pm \frac{1}{\sqrt{3}}
$$

Knowing the direction cosines and the magnitude of vector $|\vec{a}|$, we can find its coordinates. In the first case, when

$$
\begin{aligned}
& \cos \alpha=\cos \beta=\cos \gamma=\frac{1}{\sqrt{3}} \\
& x=|\vec{a}| \cdot \cos \alpha=\sqrt{3} \cdot \frac{1}{\sqrt{3}}=1 \\
& y=|\vec{a}| \cdot \cos \beta=\sqrt{3} \cdot \frac{1}{\sqrt{3}}=1 \\
& z=|\bar{a}| \cdot \cos \gamma=\sqrt{3} \cdot \frac{1}{\sqrt{3}}=1
\end{aligned}
$$

Thus, $\vec{a}=\{1,1,1\}$. Similarly, in the case when

$$
\cos \alpha=\cos \beta=\cos \gamma=-\frac{1}{\sqrt{3}}
$$

we get that $\vec{a}=\{-1,-1,-1\}$.