## Calculus 8th Edition

$c$ bisects the angle between $a$ and $b$.
Consider $\alpha$ be the angle between $c$ and $a$ then $cos \alpha =\dfrac{c \cdot a}{|c| |a|}$ $=\dfrac{(|a|b+|b|a) \cdot a}{|c| |a|}$ $=\dfrac{|a|b \cdot a+|b|a \cdot a}{|c| |a|}$ $=\dfrac{|a|b \cdot a+|b||a|^2}{|c| |a|}$ $=\dfrac{b \cdot a+|b||a|}{|c|}$ Now, consider $\beta$ be the angle between $c$ and $b$ then $cos \alpha =\dfrac{c \cdot b}{|c| |b|}$ $=\dfrac{(|a|b+|b|a) \cdot a}{|c| |b|}$ $=\dfrac{|a|b \cdot b+|b|a \cdot b}{|c| |b|}$ $=\dfrac{|a||b|^2+|b|a \cdot b}{|c| |b|}$ $=\dfrac{|a||b|+a \cdot b}{|c|}$ As we can see from above calculations $cos \alpha =cos \beta$ , thus, $\alpha = \beta$ This shows that $c$ bisects the angle between $a$ and $b$.