## Calculus 8th Edition

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