Relate a direction to the positive coordinate axes

Relate a direction to the positive coordinate axes

For a nonzero vector, compare its direction with each positive coordinate axis. Divide each signed component by the vector length, then take inverse cosine if an angle is requested. A negative component gives an obtuse angle to that positive axis.

The positive x-, y- and z-axes are represented by the unit columns

$$\mathbf i=\begin{pmatrix}1\\0\\0\end{pmatrix},\qquad\mathbf j=\begin{pmatrix}0\\1\\0\end{pmatrix},\qquad\mathbf k=\begin{pmatrix}0\\0\\1\end{pmatrix}.$$

An angle to an axis is therefore another angle between vectors. For a nonzero $\mathbf v=(v_1,v_2,v_3)$ making angles $\theta,\phi,\omega$ with the positive axes,

$$\cos\theta=\frac{\mathbf v\cdot\mathbf i}{|\mathbf v||\mathbf i|}=\frac{v_1}{|\mathbf v|},\qquad \cos\phi=\frac{v_2}{|\mathbf v|},\qquad\cos\omega=\frac{v_3}{|\mathbf v|}.$$

These are directional cosines; their squares sum to one. Example 14 works through all three unit-column substitutions for $\overrightarrow{OA}=(1,4,7)$, giving $1/\sqrt{66},4/\sqrt{66},7/\sqrt{66}$ and the squared sum $(1+16+49)/66=1$. Keep signs when a component is negative.

Check why the squared cosines sum to one

Appendix A gives the general proof. Write $\mathbf a=a_x\mathbf i+a_y\mathbf j+a_z\mathbf k\ne\mathbf0$. For each positive coordinate axis,

$$\begin{aligned} \mathbf a\cdot\mathbf i&=|\mathbf a||\mathbf i|\cos\theta_x,&\cos\theta_x&=\frac{\mathbf a\cdot\mathbf i}{|\mathbf a||\mathbf i|}=\frac{a_x}{|\mathbf a|},\\ \mathbf a\cdot\mathbf j&=|\mathbf a||\mathbf j|\cos\theta_y,&\cos\theta_y&=\frac{\mathbf a\cdot\mathbf j}{|\mathbf a||\mathbf j|}=\frac{a_y}{|\mathbf a|},\\ \mathbf a\cdot\mathbf k&=|\mathbf a||\mathbf k|\cos\theta_z,&\cos\theta_z&=\frac{\mathbf a\cdot\mathbf k}{|\mathbf a||\mathbf k|}=\frac{a_z}{|\mathbf a|}. \end{aligned}$$

Consequently,

$$\begin{aligned} \cos^2\theta_x+\cos^2\theta_y+\cos^2\theta_z &=\frac{a_x^2}{|\mathbf a|^2}+\frac{a_y^2}{|\mathbf a|^2}+\frac{a_z^2}{|\mathbf a|^2}\\ &=\frac{a_x^2+a_y^2+a_z^2}{|\mathbf a|^2}\\ &=\frac{a_x^2+a_y^2+a_z^2}{(\sqrt{a_x^2+a_y^2+a_z^2})^2}=1. \end{aligned}$$

A zero vector has no direction angles, so this division requires $|\mathbf a|>0$. Angles are measured to the positive axes, not to whichever orientation gives the smaller angle.