If v and w are two mutually perpendicular unit vectors and u = av + bw, where a and b are non-zero real numbers, then the angle between u and w is
(a) ${{\cos }^{-1}}\left( \dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)$
(b) ${{\cos }^{-1}}\left( \dfrac{a}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)$
(c) ${{\cos }^{-1}}\left( b \right)$
(d) ${{\cos }^{-1}}\left( a \right)$

Hint: To calculate the angle between vectors u and w, we must find the dot product of these two vectors. We must remember that the dot product of perpendicular vectors is 0, and hence, we can simplify the equation to find the angle between these two vectors.

Complete step by step answer:
We are given that v and w are two mutually perpendicular unit vectors. So, we can write
$\begin{align}
  & \left| v \right|=1 \\
 & \left| w \right|=1 \\
\end{align}$
It is also given that $u=av+bw$.
We can thus write that the magnitude of vector u is given by
$\left| u \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}$
We also know that the dot product of any two vectors p and q is equal to $p\cdot q=\left| p \right|\left| q \right|\cos \phi $, where $\phi $ is the angle between vectors p and q.
We need to find the angle between u and w. Assuming $\theta $ to be the angle between these two vectors, we can write
$u\cdot w=\left| u \right|\left| w \right|\cos \theta $
Substituting the values of $\left| u \right|$ and $\left| w \right|$, we get
$u\cdot w=\sqrt{{{a}^{2}}+{{b}^{2}}}\cos \theta $
We can also substitute $u=av+bw$ on the left hand side of above equation. Thus, we get
$\left( av+bw \right)\cdot w=\sqrt{{{a}^{2}}+{{b}^{2}}}\cos \theta $
Hence, we can also write
$av\cdot w+bw\cdot w=\sqrt{{{a}^{2}}+{{b}^{2}}}\cos \theta ...\left( i \right)$
We know that v and w are mutually perpendicular unit vectors. Thus, we can write $v\cdot w=\left| v \right|\left| w \right|\cos {{90}^{\circ }}=0$ and $w\cdot w=\left| w \right|\left| w \right|\cos {{0}^{\circ }}=1$.
Thus, equation (i) becomes
$0+b=\sqrt{{{a}^{2}}+{{b}^{2}}}\cos \theta $
On dividing both sides of the above equation by $\sqrt{{{a}^{2}}+{{b}^{2}}}$, we get
$\dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}}=\cos \theta $
Thus, we can write
$\theta ={{\cos }^{-1}}\left( \dfrac{b}{\sqrt{{{a}^{2}}+{{b}^{2}}}} \right)$

So, the correct answer is “Option a”.

Note: We can see here that the vectors v and w are mutually perpendicular unit vectors. So, the magnitude of vector u is $\left| u \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}$, because the term $2ab\cos \theta $ becomes 0 since the angle between these vectors is a right angle.