If α , β and γ are the angles that vector overrightarrowOM…

If $\alpha$, $\beta$ and $\gamma$ are the angles that vector $\overrightarrow{OM}$ makes with the coordinate axes, prove that $\cos\alpha = \frac{x}{\sqrt{x^2+y^2+z^2}}, \quad \cos\beta = \frac{y}{\sqrt{x^2+y^2+z^2}}, \quad \cos\gamma = \frac{z}{\sqrt{x^2+y^2+z^2}}$ and that $\cos^2\alpha+\cos^2\beta+\cos^2\gamma = 1$, where $x$, $y$, $z$ are the coordinates of point $M$.

Step 1: Setup

By Problem 27, $\overrightarrow{OM} = x\vec{i}+y\vec{j}+z\vec{k}$, with length $r = \sqrt{x^2+y^2+z^2}$ ($M\neq O$). The angles are

\[ \alpha = \angle(\overrightarrow{OM}, Ox), \qquad \beta = \angle(\overrightarrow{OM}, Oy), \qquad \gamma = \angle(\overrightarrow{OM}, Oz), \qquad \alpha,\beta,\gamma\in[0^\circ,180^\circ]. \]

Step 2: The direction cosines

$\cos\alpha$

Using the dot product with the unit vector $\vec i$ of the $x$-axis: \[\cos\alpha = \frac{\overrightarrow{OM}\cdot\vec{i}}{|\overrightarrow{OM}|\,|\vec{i}|} = \frac{x(\vec i\cdot\vec i) + y(\vec j\cdot\vec i) + z(\vec k\cdot\vec i)}{r} = \frac{x}{r}\]

$\cos\beta$ and $\cos\gamma$

In the same way, with $\vec j$ and $\vec k$: \[\cos\beta = \frac{\overrightarrow{OM}\cdot\vec{j}}{r} = \frac{y}{r}, \qquad \cos\gamma = \frac{\overrightarrow{OM}\cdot\vec{k}}{r} = \frac{z}{r}.\]

\[\cos\alpha = \frac{x}{\sqrt{x^2+y^2+z^2}}, \quad \cos\beta = \frac{y}{\sqrt{x^2+y^2+z^2}}, \quad \cos\gamma = \frac{z}{\sqrt{x^2+y^2+z^2}}\]

Geometrically: the foot of the perpendicular from $M$ to the $x$-axis is $M_x(x,0,0)$; in the right triangle $OM_xM$ the adjacent leg is $x$ (signed) and the hypotenuse is $r$. The same holds for the $y$- and $z$-axes.

Step 3: The fundamental identity

Direct computation

\[ \cos^2\alpha+\cos^2\beta+\cos^2\gamma = \frac{x^2+y^2+z^2}{r^2} = \frac{r^2}{r^2} = 1 \]

\[\cos^2\alpha+\cos^2\beta+\cos^2\gamma = 1\]

Via the unit vector

The unit vector along $\overrightarrow{OM}$ is

\[ \hat{r} = \frac{\overrightarrow{OM}}{r} = \cos\alpha\,\vec{i}+\cos\beta\,\vec{j}+\cos\gamma\,\vec{k}, \] and $|\hat{r}|^2 = 1$ gives the identity directly. $\blacksquare$

Step 4: Remarks

• The direction cosines are the components of the unit vector $\hat r$ along $\overrightarrow{OM}$.
• $\overrightarrow{OM} = r(\cos\alpha\,\vec{i}+\cos\beta\,\vec{j}+\cos\gamma\,\vec{k})$: magnitude times direction.
• The three angles are not independent: two of them determine the third up to sign, so a direction in space has only two degrees of freedom.

Step 5: Numerical example

For $M(2, 3, 6)$: $r = \sqrt{4+9+36} = 7$, and

\[ \cos\alpha = \tfrac{2}{7},\ \ \cos\beta = \tfrac{3}{7},\ \ \cos\gamma = \tfrac{6}{7}, \qquad \tfrac{4+9+36}{49} = 1 \;\checkmark \]

\[ \alpha \approx 73.4^\circ, \qquad \beta \approx 64.6^\circ, \qquad \gamma \approx 31.0^\circ \]