Thornton & Marion (5th Edition), Chapter 01, Problem 02 Solution

Thornton & Marion, Classical Dynamics of Particles and Systems, 5th Edition

Chapter 1. Matrices, Vectors, and Vector Calculus

Problem 02. Trigonometric properties of direction cosines

Solution

For Equation (1.10): $latex \\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 $latex (\\alpha, \\beta, \\gamma)$ from the origin be $latex l$, then

$latex l^2\\cos^2\\alpha + l^2\\cos^2\\beta + l^2\\cos^2\\gamma = l^2 $
 

Therefore, it becomes

$latex \\cos^2\\alpha + \\cos^2\\beta + \\cos^2\\gamma = 1 $
 

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

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

$latex l = \\sqrt{l^2\\cos^2\\alpha + l^2\\cos^2\\beta + l^2\\cos^2\\gamma}$ 
$latex l\’ = \\sqrt{l\’^2\\cos^2\\alpha\’ + l\’^2\\cos^2\\beta\’ + l\’^2\\cos^2\\gamma\’}$
 

Using the law of cosines,

$latex \\overline{PP\’}^2 = l^2 + l\’^2 – 2 l l\’ \\cos\\theta$
 

Since $latex \\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

$latex l^2 + l\’^2 – 2 l l\’ (\\cos\\alpha\\cos\\alpha\’ + \\cos\\beta\\cos\\beta\’ + \\cos\\gamma\\cos\\gamma\’) = l^2 + l\’^2 – 2 l l\’ \\cos\\theta$
 

Therefore,

$latex \\cos\\alpha\\cos\\alpha\’ + \\cos\\beta\\cos\\beta\’ + \\cos\\gamma\\cos\\gamma\’ = \\cos\\theta$

Reference

https://en.wikipedia.org/wiki/Law_of_cosines

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