Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - 12.3 The Dot Product - 12.3 Exercises - Page 854: 58


$c$ bisects the angle between $a$ and $b$.

Work Step by Step

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 $b$.
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