11b.4. Direction cosines

Recall the dot product formula $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}| \, |\mathbf{b}| \cos \theta$, which is often used to find angles between two vectors $\mathbf{a}$ and $\mathbf{b}$. If we want to find the angle $\mathbf{a}$ makes with the $x$-axis, then we can use the direction vector of the $x$-axis $\mathbf{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$.

Let $\mathbf{a} = \begin{pmatrix} x \\ y \\ z \end{pmatrix}$ and $\theta_x$ be the angle $\mathbf{a}$ makes with the $x$-axis. The dot product formula then gives us $$ \begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} = \left | \begin{pmatrix} x \\ y \\ z \end{pmatrix} \right | \, \left | \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \right | \cos \theta_x$$ Simplifying gives us $\cos \theta_x = \displaystyle \frac{x}{\sqrt{x^2+y^2+z^2}}$. We call $\cos \theta_x$ the direction cosine of $\mathbf{a}$ with respect to the $x$-axis and denote it by $\alpha$.

If we repeat the process for the $y$- and $z$-axis, denoting the angles by $\theta_y$ and $\theta_z$ and the direction cosines by $\beta$ and $\gamma$, we get:

$ \begin{align} \alpha &= \cos \theta_x = \frac{x}{\sqrt{x^2+y^2+z^2}} \\ \beta &= \cos \theta_y = \frac{y}{\sqrt{x^2+y^2+z^2}} \\ \gamma &= \cos \theta_z = \frac{z}{\sqrt{x^2+y^2+z^2}} \end{align} $

It can be easily checked that the three direction cosines satisfy the formula

$ \alpha^2 + \beta^2 + \gamma^2 = 1$.

This means that, once we have two of the direction cosines, we can determine the last one (up to a $\pm$) and can thus deduce the original vector itself. This is illustrated in the second example below.

Solution to examples

1. It is given that $\mathbf{a} = -\mathbf{i} + 2 \mathbf{j} + 3 \mathbf{k}$. Find the cosines of the angle $\mathbf{a}$ makes with the $x$, $y$ and $z$ axis respectively.

Solution:
Direction cosine with respect to the $x$-axis, $\alpha = \frac{-1}{\sqrt{14}} $
Direction cosine with respect to the $y$-axis, $\beta = \frac{2}{\sqrt{14}} $
Direction cosine with respect to the $z$-axis, $\gamma = \frac{3}{\sqrt{14}} $ .

2. It is given that the direction cosines of a vector $\mathbf{b}$ with respect to the $x$ -and $y$- axes are $\frac{2}{\sqrt{6}}$ and $\frac{-1}{\sqrt{6}}$ respectively. Find two possible vectors $\mathbf{b}$.

Solution: Since $\alpha^2+\beta^2 + \gamma^2 = 1,$
$\frac{4}{6} + \frac{1}{6} + \gamma^2 = 1$
Hence $\gamma = \pm \frac{1}{\sqrt{6}}$.
Thus $\mathbf{b} = \begin{pmatrix} 2 \\ -1 \\ 1 \end{pmatrix}$ or $\mathbf{b} = \begin{pmatrix} 2 \\ -1 \\ -1 \end{pmatrix}$