Determine the position vector of point M given that it makes…

Determine the position vector of point $M$ given that it makes an angle of $45^\circ$ with the $Ox$ axis, an angle of $60^\circ$ with the $Oy$ axis, its magnitude is $6$, and its $z$-coordinate is negative.

Explanation of Method

Position vector from two angles, the magnitude and the sign of the $z$-coordinate.

Step 1: Data

\[ \alpha = 45^\circ, \qquad \beta = 60^\circ, \qquad r = 6, \qquad z \lt 0 \]

Step 2: Find $\cos\gamma$

\[ \cos^2\gamma = 1 - \cos^2 45^\circ - \cos^2 60^\circ = 1 - \frac{1}{2} - \frac{1}{4} = \frac{1}{4} \;\Longrightarrow\; \cos\gamma = \pm\frac{1}{2} \]

Since $z = r\cos\gamma \lt 0$ and $r \gt 0$, we need $\cos\gamma \lt 0$: $\cos\gamma = -\frac{1}{2} \;\Longrightarrow\; \gamma = 120^\circ$.

Step 3: Coordinates of $M$

\[ x = 6\cdot\frac{\sqrt{2}}{2} = 3\sqrt{2}, \qquad y = 6\cdot\frac{1}{2} = 3, \qquad z = 6\cdot\left(-\frac{1}{2}\right) = -3 \]

\[\overrightarrow{OM} = 3\sqrt{2}\,\vec{i} + 3\,\vec{j} - 3\,\vec{k}, \qquad M(3\sqrt{2},\,3,\,-3)\]

Step 4: Verification

\[ |\overrightarrow{OM}| = \sqrt{18+9+9} = 6 \;\checkmark, \qquad \frac{1}{2}+\frac{1}{4}+\frac{1}{4} = 1 \;\checkmark \]

Unit vector: $\hat{r} = \frac{\sqrt{2}}{2}\,\vec{i}+\frac{1}{2}\,\vec{j}-\frac{1}{2}\,\vec{k}$.

$\alpha$ $\beta$ $\gamma$ $O$ $M(3\sqrt2,3,-3)$ $M'$ $3\sqrt2\,\vec i$ $3\vec j$ $-3\vec k$

$\overrightarrow{OM}=3\sqrt2\,\vec i+3\vec j-3\vec k,\ r=6$
$d_{xy}=3\sqrt3$
$\alpha=45^\circ$
$\beta=60^\circ$
$\gamma=120^\circ$

$M$ lies below the $xOy$ plane ($z\lt 0$), so the angle with $Oz$ is obtuse.

Result

Axis	Angle	$\cos$	Coordinate
$Ox$	$45^\circ$	$\frac{\sqrt{2}}{2}$	$x = 3\sqrt{2}$
$Oy$	$60^\circ$	$\frac{1}{2}$	$y = 3$
$Oz$	$120^\circ$	$-\frac{1}{2}$	$z = -3$