Find the cosines of the direction angles of the vector \(\Vect{x} = (-1,1,1)\) in \(\RNrSpc{3}\).
We have \(| \Vect{x} | = \sqrt{3}\), and so
\[\cos \omega_1 = - \dfrac{1}{\sqrt{3}} \qquad \cos \omega_2 = \dfrac{1}{\sqrt{3}} \qquad \cos \omega_3 = \dfrac{1}{\sqrt{3}}\]We get some corroboration for the validity of these computations from the ‘sum of cosines formula’:
\[\cos^2 \omega_1 + \cos^2 \omega_2 + \cos^2 \omega_3\ =\ \left( - \dfrac{1}{3}\right)^2 + \left( \dfrac{1}{3}\right)^2 + \left( \dfrac{1}{3}\right)^2 \ =\ 1\]as expected.
In this particular case we can use a lucky short cut to solve this problem: Inspection of the diagram below
tells us that, up to a sign, all direction angles are equal. Therefore
\[\cos^2 \omega_1 = \cos^2 \omega_2 = \cos^2 \omega_3\]and, therefore, \(3\cdot \cos^2 \omega_i =1\); i.e. \(| \cos \omega_i | = \frac{1}{\sqrt{3}}\).