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

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