# Thornton & Marion (Fifth Edition), Chapter 01, Exercise Problem 02 Solution

## Chapter 01. Matrices, Vectors, and Vector Calculus

### Problem 02. Trigonometric properties of direction cosines

##### Solution
For Equation (1.10): $\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$

Proof)

Using Figure 1-4 (a) in the book, let the length of a point P on the line $(\alpha, \beta, \gamma)$ from the origin be $l$, then

$l^2\cos^2\alpha + l^2\cos^2\beta + l^2\cos^2\gamma = l^2$
Therefore, it becomes
$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$

For Equation (1.11): $\cos\alpha\cos\alpha' + \cos\beta\cos\beta' + \cos\gamma\cos\gamma' = \cos\theta$

Proof)

In Figure 1-4 (b) in the book, let the length of a point P on the line $(\alpha, \beta, \gamma)$ from the origin be $l$ and the length of a point P' on the line $(\alpha', \beta', \gamma')$ be $l'$.

$l = \sqrt{l^2\cos^2\alpha + l^2\cos^2\beta + l^2\cos^2\gamma}$
$l' = \sqrt{l'^2\cos^2\alpha' + l'^2\cos^2\beta' + l'^2\cos^2\gamma'}$
Using the law of cosines,
$\overline{PP'}^2 = l^2 + l'^2 - 2 l l' \cos\theta$
Since $\begin{array}{rcl} \overline{PP'}^2 & = & (l\cos\alpha - l'\cos\alpha')^2 + (l\cos\beta - l'\cos\beta')^2 + (l\cos\gamma - l'\cos\gamma')^2 \\ & = & l^2(\cos^2\alpha + \cos^2\beta + \cos^2\gamma) + l'^2(\cos^2\alpha' + \cos^2\beta' + \cos^2\gamma') - 2 l l' (\cos\alpha\cos\alpha' + \cos\beta\cos\beta' + \cos\gamma\cos\gamma') \\ & = & l^2 + l'^2 - 2 l l' (\cos\alpha\cos\alpha' + \cos\beta\cos\beta' + \cos\gamma\cos\gamma') \end{array}$ the equation becomes
$l^2 + l'^2 - 2 l l' (\cos\alpha\cos\alpha' + \cos\beta\cos\beta' + \cos\gamma\cos\gamma') = l^2 + l'^2 - 2 l l' \cos\theta$
Therefore,
$\cos\alpha\cos\alpha' + \cos\beta\cos\beta' + \cos\gamma\cos\gamma' = \cos\theta$
##### Reference
https://en.wikipedia.org/wiki/Law_of_cosines