question_answer

If \[\vec{a}\] and \[\vec{b}\] are two unit vectors and \[\theta \] is the angle between them, then the unit vector along the angular bisector of \[\vec{a}\] and \[\vec{b}\] will be given by

A) \[\frac{\vec{a}-\vec{b}}{2\cos (\theta /2)}\]

B) \[\frac{\vec{a}+\vec{b}}{2\cos (\theta /2)}\]

C) \[\frac{\vec{a}-\vec{b}}{\cos (\theta /2)}\]

D) none of these

Correct Answer: B

Solution :

[b] Vector in the direction of angular bisector of \[\vec{a}\] and \[\vec{b}\] is\[\frac{\vec{a}+\vec{b}}{2}\]. Unit vector in this direction is\[\frac{\vec{a}+\vec{b}}{\left| \vec{a}+\vec{b} \right|}\]. From the figure, position vector of E is \[\frac{\vec{a}+\vec{b}}{2}\] Now in triangle \[AEB,AE=AB\cos \frac{\theta }{2}\] \[\Rightarrow \left| \frac{\vec{a}+\vec{b}}{2} \right|=\cos \frac{\theta }{2}\] Hence, unit vector along the bisector is\[\frac{\vec{a}+\vec{b}}{2\cos (\theta /2)}\].