Problem 841

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 $\vec{a}=\{-1,-1,-1\}$.