Find points on the sphere with center at the origin and…

Find points on the sphere with center at the origin and radius $R=2\sqrt{3}$, whose radius vector makes equal angles with the positive directions of the coordinate axes. What are those angles?

Step 1: Setup

\[ x^2 + y^2 + z^2 = R^2 = 12, \qquad \alpha = \beta = \gamma \]

Step 2: The angles

\[ 3\cos^2\alpha = 1 \;\Longrightarrow\; \cos\alpha = \pm\frac{1}{\sqrt{3}} \]

Equal angles mean equal cosines, so all three cosines are $+\frac{1}{\sqrt3}$ or all three are $-\frac{1}{\sqrt3}$: \[\alpha=\beta=\gamma = \arccos\frac{1}{\sqrt{3}} \approx 54^\circ44' \quad\text{or}\quad \alpha=\beta=\gamma = \arccos\left(-\frac{1}{\sqrt{3}}\right) \approx 125^\circ16'\]

Step 3: The points

\[ x = y = z = R\cos\alpha = 2\sqrt{3}\cdot\left(\pm\frac{1}{\sqrt{3}}\right) = \pm 2 \]

\[M_1(2,\,2,\,2) \qquad \text{and} \qquad M_2(-2,\,-2,\,-2)\]

$M_1$ makes equal acute angles $\approx 54^\circ44'$ with the positive axes; $M_2$ makes equal obtuse angles $\approx 125^\circ16'$.

Step 4: Verification

\[ |\overrightarrow{OM_{1,2}}| = \sqrt{4+4+4} = 2\sqrt{3} = R \;\checkmark, \qquad \cos\alpha = \frac{2}{2\sqrt{3}} = \frac{1}{\sqrt{3}} \;\checkmark \]

Step 5: Geometric meaning

$M_1$ and $M_2$ are opposite vertices of the cube $[-2,2]^3$; $M_1M_2$ is a space diagonal through $O$. The angle $54^\circ44'$ is the angle between a space diagonal of a cube and its edges: for edge $a$, the diagonal is $a\sqrt3$ and $\cos\alpha = \frac{a}{a\sqrt3}=\frac{1}{\sqrt3}$.