Chapter 13 - Vector Geometry - 13.3 Dot Product and the Angle Between Two Vectors - Exercises - Page 668: 79

Geometrically, the vectors ${{\bf{e}}_{\bf{v}}}$, ${{\bf{e}}_{\bf{w}}}$ and ${\bf{u}}$ construct a parallelogram of equal sides. Therefore, the vector ${\bf{u}}$ bisects the angle between vectors ${\bf{v}}$ and ${\bf{w}}$. Hence, the angle between ${\bf{u}}$ and ${\bf{v}}$ is equal to the angle between ${\bf{u}}$ and ${\bf{w}}$.

Notice that ${{\bf{e}}_{\bf{v}}}$ and ${{\bf{e}}_{\bf{w}}}$ are the unit vectors in the direction of ${\bf{v}}$ and ${\bf{w}}$, respectively. So, ${{\bf{e}}_{\bf{v}}} = \frac{{\bf{v}}}{{||{\bf{v}}||}}$ ${\ \ }$ and ${\ \ }$ ${{\bf{e}}_{\bf{w}}} = \frac{{\bf{w}}}{{||{\bf{w}}||}}$ Or ${\bf{v}} = ||{\bf{v}}||{{\bf{e}}_{\bf{v}}}$ ${\ \ }$ and ${\ \ }$ ${\bf{w}} = ||{\bf{w}}||{{\bf{e}}_{\bf{w}}}$ Let $\alpha$ denote the angle between ${\bf{u}}$ and ${\bf{v}}$ and $\beta$ denote the angle between ${\bf{u}}$ and ${\bf{w}}$. By Eq. (1) of Theorem 2, $\cos \alpha = \frac{{{\bf{u}}\cdot{\bf{v}}}}{{||{\bf{u}}||||{\bf{v}}||}}$, ${\ \ \ }$ $\cos \beta = \frac{{{\bf{u}}\cdot{\bf{w}}}}{{||{\bf{u}}||||{\bf{w}}||}}$ Substituting ${\bf{v}} = ||{\bf{v}}||{{\bf{e}}_{\bf{v}}}$ and ${\bf{w}} = ||{\bf{w}}||{{\bf{e}}_{\bf{w}}}$ in the equations above gives $\cos \alpha = \frac{{{\bf{u}}\cdot{\bf{v}}}}{{||{\bf{u}}||||{\bf{v}}||}} = \frac{{{\bf{u}}\cdot||{\bf{v}}||{{\bf{e}}_{\bf{v}}}}}{{||{\bf{u}}||||{\bf{v}}||}} = \frac{{{\bf{u}}\cdot{{\bf{e}}_{\bf{v}}}}}{{||{\bf{u}}||}}$ $\cos \beta = \frac{{{\bf{u}}\cdot{\bf{w}}}}{{||{\bf{u}}||||{\bf{w}}||}} = \frac{{{\bf{u}}\cdot||{\bf{w}}||{{\bf{e}}_{\bf{w}}}}}{{||{\bf{u}}||||{\bf{w}}||}} = \frac{{{\bf{u}}\cdot{{\bf{e}}_{\bf{w}}}}}{{||{\bf{u}}||}}$ Since ${{\bf{e}}_{\bf{v}}}\cdot{{\bf{e}}_{\bf{v}}} = 1$, ${{\bf{e}}_{\bf{w}}}\cdot{{\bf{e}}_{\bf{w}}} = 1$ and ${\bf{u}} = {{\bf{e}}_{\bf{v}}} + {{\bf{e}}_{\bf{w}}}$, we get $\cos \alpha = \frac{{{\bf{u}}\cdot{{\bf{e}}_{\bf{v}}}}}{{||{\bf{u}}||}} = \frac{{\left( {{{\bf{e}}_{\bf{v}}} + {{\bf{e}}_{\bf{w}}}} \right)\cdot{{\bf{e}}_{\bf{v}}}}}{{||{\bf{u}}||}} = \frac{{{{\bf{e}}_{\bf{v}}}\cdot{{\bf{e}}_{\bf{v}}} + {{\bf{e}}_{\bf{w}}}\cdot{{\bf{e}}_{\bf{v}}}}}{{||{\bf{u}}||}}$ $\cos \alpha = \frac{{1 + {{\bf{e}}_{\bf{w}}}\cdot{{\bf{e}}_{\bf{v}}}}}{{||{\bf{u}}||}}$ $\cos \beta = \frac{{{\bf{u}}\cdot{{\bf{e}}_{\bf{w}}}}}{{||{\bf{u}}||}} = \frac{{\left( {{{\bf{e}}_{\bf{v}}} + {{\bf{e}}_{\bf{w}}}} \right)\cdot{{\bf{e}}_{\bf{w}}}}}{{||{\bf{u}}||}} = \frac{{{{\bf{e}}_{\bf{v}}}\cdot{{\bf{e}}_{\bf{w}}} + {{\bf{e}}_{\bf{w}}}\cdot{{\bf{e}}_{\bf{w}}}}}{{||{\bf{u}}||}}$ $\cos \beta = \frac{{1 + {{\bf{e}}_{\bf{v}}}\cdot{{\bf{e}}_{\bf{w}}}}}{{||{\bf{u}}||}}$ Since ${{\bf{e}}_{\bf{w}}}\cdot{{\bf{e}}_{\bf{v}}} = {{\bf{e}}_{\bf{v}}}\cdot{{\bf{e}}_{\bf{w}}}$, so $\cos \alpha = \cos \beta $. Therefore $\alpha = \beta $. Hence, the angle between ${\bf{u}}$ and ${\bf{v}}$ is equal to the angle between ${\bf{u}}$ and ${\bf{w}}$. Geometrically, since ${{\bf{e}}_{\bf{v}}}$ and ${{\bf{e}}_{\bf{w}}}$ are unit vectors, the three vectors ${{\bf{e}}_{\bf{v}}}$, ${{\bf{e}}_{\bf{w}}}$ and ${\bf{u}}$ construct a parallelogram of equal sides as is shown in the figure attached. Therefore, the vector ${\bf{u}}$ bisects the angle between vectors ${\bf{v}}$ and ${\bf{w}}$. Hence, the angle between ${\bf{u}}$ and ${\bf{v}}$ is equal to the angle between ${\bf{u}}$ and ${\bf{w}}$.