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.
