Let O be the origin and A, B be two points. If $\overrightarrow{p},\overrightarrow{q}$ are vectors represented by $\overrightarrow{OA}$ and $\overrightarrow{OB}$ and their magnitudes are p, q. Then, the unit vector bisection the angle AOB is?
(a) $\dfrac{\dfrac{\overrightarrow{p}}{p}+\dfrac{\overrightarrow{q}}{q}}{\left| \dfrac{\overrightarrow{p}}{p} \right|+\left| \dfrac{\overrightarrow{q}}{q} \right|}$
(b) $\dfrac{\dfrac{\overrightarrow{p}}{p}+\dfrac{\overrightarrow{q}}{q}}{\left| \dfrac{\overrightarrow{p}}{p} \right|-\left| \dfrac{\overrightarrow{q}}{q} \right|}$
(c) $\dfrac{\dfrac{\overrightarrow{p}}{p}+\dfrac{\overrightarrow{q}}{q}}{\left| \dfrac{\overrightarrow{p}}{p}+\dfrac{\overrightarrow{q}}{q} \right|}$
(d) $\dfrac{\overrightarrow{p}+\overrightarrow{q}}{2}$

Hint: In this problem, first of all we will find the unit vectors in the direction of $\overrightarrow{p}$ and in the direction of $\overrightarrow{q}$. The vector bisecting the angle AOB is directed from O that is from origin to the midpoint of the vector $\overrightarrow{AB}$.

Complete step-by-step solution -

We know that a unit vector is a vector whose magnitude is 1. Unit vectors are often chosen to form the basis of a vector space. Every vector in the space may be written as a linear combination of unit vectors.
Any unit vector in the direction of vector $\overrightarrow{a}$ is given as the vector divided by its magnitude, that is $\dfrac{\overrightarrow{a}}{\left| \overrightarrow{a} \right|}$.
In this problem, the two vectors that we have are $\overrightarrow{p}$ and $\overrightarrow{q}$.
Unit vector in the direction of $\overrightarrow{p}$ will be = $\dfrac{\overrightarrow{p}}{\left| \overrightarrow{p} \right|}$
And, unit vector in the direction of $\overrightarrow{q}$ will be = $\dfrac{\overrightarrow{q}}{\left| \overrightarrow{q} \right|}$
Now, we have to find the unit vector bisecting the angle AOB.
The vector which bisects the angle AOB will be directed from the origin to the mid-point of the vector $\overrightarrow{AB}$.
So, the vector bisecting the angle AOB is = $\left( \dfrac{\dfrac{\overrightarrow{p}}{\left| \overrightarrow{p} \right|}+\dfrac{\overrightarrow{q}}{\left| \overrightarrow{q} \right|}}{2} \right)$
The magnitude of this vector is = $\left| \left( \dfrac{\dfrac{\overrightarrow{p}}{\left| \overrightarrow{p} \right|}+\dfrac{\overrightarrow{q}}{\left| \overrightarrow{q} \right|}}{2} \right) \right|=\dfrac{1}{2}\left| \dfrac{\overrightarrow{p}}{\left| \overrightarrow{p} \right|}+\dfrac{\overrightarrow{q}}{\left| \overrightarrow{q} \right|} \right|$
Therefore, the unit vector bisecting the angle AOB is =$\dfrac{\left( \dfrac{\dfrac{\overrightarrow{p}}{\left| \overrightarrow{p} \right|}+\dfrac{\overrightarrow{q}}{\left| \overrightarrow{q} \right|}}{2} \right)}{\left| \dfrac{\dfrac{\overrightarrow{p}}{\left| \overrightarrow{p} \right|}+\dfrac{\overrightarrow{q}}{\left| \overrightarrow{q} \right|}}{2} \right|}=\dfrac{\dfrac{\overrightarrow{p}}{\left| \overrightarrow{p} \right|}+\dfrac{\overrightarrow{q}}{\left| \overrightarrow{q} \right|}}{\left| \dfrac{\overrightarrow{p}}{\left| \overrightarrow{p} \right|}+\dfrac{\overrightarrow{q}}{\left| \overrightarrow{q} \right|} \right|}$
Hence, option (c) is the correct answer.

Note: Students should note here that a unit vector represents direction. The magnitude of a unit vector is always 1 unit. So, it is always given as the vector divided by its magnitude.