Answer
The bond angle is $\alpha = 1.91$ or $\alpha = 109.47^\circ $
Work Step by Step
Let the point $C = \left( {\frac{1}{2},\frac{1}{2},\frac{1}{2}} \right)$ denote the location of the carbon atom and let ${H_1} = \left( {0,0,0} \right)$ and ${H_2} = \left( {1,1,0} \right)$ denote the locations of the two hydrogen atoms as is shown in Figure 17. We write
1. the vector ${\bf{u}} = \overrightarrow {C{H_1}} = \left( {0,0,0} \right) - \left( {\frac{1}{2},\frac{1}{2},\frac{1}{2}} \right) = \left( { - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}} \right)$
2. the vector ${\bf{v}} = \overrightarrow {C{H_2}} = \left( {1,1,0} \right) - \left( {\frac{1}{2},\frac{1}{2},\frac{1}{2}} \right) = \left( {\frac{1}{2},\frac{1}{2}, - \frac{1}{2}} \right)$
By Eq. (1) of Theorem 2, the angle $\alpha$ between the vectors ${\bf{u}} = \overrightarrow {C{H_1}} $ and ${\bf{v}} = \overrightarrow {C{H_2}} $ is given by
$\cos \alpha = \frac{{{\bf{u}}\cdot{\bf{v}}}}{{||{\bf{u}}||||{\bf{v}}||}} = \frac{{\left( { - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}} \right)\cdot\left( {\frac{1}{2},\frac{1}{2}, - \frac{1}{2}} \right)}}{{||\left( { - \frac{1}{2}, - \frac{1}{2}, - \frac{1}{2}} \right)||||\left( {\frac{1}{2},\frac{1}{2}, - \frac{1}{2}} \right)||}}$
$\cos \alpha = \frac{{ - \frac{1}{4}}}{{\sqrt {\frac{3}{4}} \sqrt {\frac{3}{4}} }} = - \frac{1}{3}$
$\alpha = 1.91$ ${\ \ \ }$ or ${\ \ \ }$ $\alpha = 109.47^\circ $
