A) \[\pi \]
B) \[\frac{5\pi }{6}\]
C) \[\frac{2\pi }{3}\]
D) \[\frac{3\pi }{4}\]
Correct Answer: A
Solution :
Given, \[|a|=7\] and \[|b|=11\] The angle between vectors \[a+b\] and \[a-b\] is \[\cos \theta =\frac{(a+b)\,.\,(a-b)}{|a+b|\,.\,|a-b|}=\frac{|a{{|}^{2}}-|b{{|}^{2}}}{18.4}\] \[|\because |a+b|\le |a|+|b||a-b|\ge |a|-|b|\] \[=\frac{49-121}{18.4}=\frac{-72}{72}=-1\] \[\Rightarrow \] \[\cos \,\theta =cos\,\pi \] \[\Rightarrow \] \[\theta =\pi \]
You need to login to perform this action.
You will be redirected in
3 sec
