Answer
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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.
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.
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