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: 45

Answer

The vector $orth_a{b}=(b-proj_ab)$ is orthogonal to $a$.

Work Step by Step

By definition: $b-proj_ab=b-\frac{ab}{|a|^2}a$ Let us take $orth_a{b}.{a}=(b-proj_ab).a$ $orth_a{b}.{a}=(b-\frac{ab}{|a|^2}a).a$ $orth_a{b}.{a}=b \cdot a-\frac{ab}{|a|^2}a \cdot a$ $orth_a{b}.{a}=b \cdot a-\frac{ab}{|a|^2}|a|^2 $ $orth_a{b}.{a}=b \cdot a- a \cdot b $ $orth_a{b}.{a}=0$ Because the dot product of $orth_a{b}$ and $a$ is $0$, thus the two vectors are orthogonal.
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