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, \[|\hat{a}-\hat{b}{{|}^{2}}=1+1-2\times 1\times \cos \frac{2\pi }{3}=3\] \[\therefore \,\,\,\,\,\,|\hat{a}-\hat{b}+\hat{c}{{|}^{2}}=\,\,|\hat{a}-\hat{b}{{|}^{2}}+{{\hat{c}}^{2}}+2|\hat{a}-\hat{b}||\hat{c}|\,\cos 90{}^\circ \] \[=3+1=4\] \[\therefore \,\,\,\,\,\,|\hat{a}-\hat{b}+\hat{c}|=2\]You need to login to perform this action.
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