Vector r makes angles α=30^° and β=60^° with coordinate axes…

Vector $\vec{r}$ makes angles $\alpha=30^\circ$ and $\beta=60^\circ$ with coordinate axes $Ox$ and $Oy$ respectively. Find the angle $\gamma$ that the vector makes with the $Oz$ axis.

Step 1: Fundamental identity

By Problem 29, $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$.

Step 2: Compute

\[ \cos^2 30^\circ = \frac{3}{4}, \qquad \cos^2 60^\circ = \frac{1}{4} \]

\[ \frac{3}{4} + \frac{1}{4} + \cos^2\gamma = 1 \;\Longrightarrow\; \cos^2\gamma = 0 \;\Longrightarrow\; \cos\gamma = 0 \]

\[\gamma = 90^\circ\]

The vector is perpendicular to $Oz$: it lies in (a plane parallel to) the $xOy$ plane.

Step 3: Components

\[ x = r\cos30^\circ = \frac{\sqrt{3}}{2}r, \qquad y = r\cos60^\circ = \frac{1}{2}r, \qquad z = r\cos90^\circ = 0 \]

\[ \vec{r} = r\left(\frac{\sqrt{3}}{2}\,\vec{i}+\frac{1}{2}\,\vec{j}\right), \qquad \hat{r} = \frac{\sqrt{3}}{2}\,\vec{i}+\frac{1}{2}\,\vec{j} \]

Step 4: Interpretation and a general note

Since $z=0$, $\vec r$ lies in the $xOy$ plane. There, because $Ox\perp Oy$, a vector making $30^\circ$ with $Ox$ (in the first quadrant) makes $90^\circ-30^\circ=60^\circ$ with $Oy$ — consistent with the data.

In general the two given angles must satisfy $\cos^2\alpha+\cos^2\beta \le 1$, otherwise no vector exists. Here the sum equals exactly $1$, which forces $\cos\gamma=0$, so $\gamma=90^\circ$ is the only solution.