question_answer

Three unit vectors a, b and c are such that \[\hat{a}+\hat{b}=\hat{c},\] then \[|\hat{a}-\hat{b}+\hat{c}|\]is equal to

A) \[1\]                      

B)        \[\sqrt{2}\]                    

C) \[\sqrt{3}\]                    

D)        \[2\]

Correct Answer: D

Solution :

[d] Since \[\hat{a}+\hat{b}=\hat{c},\] vector \[\hat{a}+\hat{b}\] is along angle bisector of vectors \[\hat{a}\] and \[\hat{b}\]. Angle between \[\hat{a}\] and \[\hat{b}\] is \[\frac{2\pi }{3}\] and \[\vec{c}\] is along the angle bisector, \[\hat{a}-\hat{b}\] is along another angle bisector and hence perpendicular to \[\hat{c}\]. Also, \[|\hat{a}-\hat{b}{{|}^{2}}=1+1-2\times 1\times \cos \frac{2\pi }{3}=3\] \[\therefore \,\,\,\,\,\,|\hat{a}-\hat{b}+\hat{c}{{|}^{2}}=\,\,|\hat{a}-\hat{b}{{|}^{2}}+{{\hat{c}}^{2}}+2|\hat{a}-\hat{b}||\hat{c}|\,\cos 90{}^\circ \]   \[=3+1=4\] \[\therefore \,\,\,\,\,\,|\hat{a}-\hat{b}+\hat{c}|=2\]