A) \[{{30}^{o}}\]
B) \[{{45}^{o}}\]
C) \[{{60}^{o}}\]
D) \[{{90}^{o}}\]
Correct Answer: C
Solution :
Since, vectors \[a+2b\] and \[5a-4b\] are perpendicular to each other. \[\therefore \] \[(a+2b).(5a-4b)=0\] \[\Rightarrow \] \[5|a{{|}^{2}}+6a.b-8|b{{|}^{2}}=0\] \[\Rightarrow \] \[5{{(1)}^{2}}+6a.b-8{{(1)}^{2}}=0\] \[(\because \,\,|a|=|b|=1,\,\,given)\] \[\Rightarrow \] \[6a.\,b=3\] \[\Rightarrow \] \[|a||b|\,\cos \theta =\frac{3}{6}=\frac{1}{2}\] \[\Rightarrow \] \[1\times 1\times \cos \theta =\frac{1}{2}\] \[\Rightarrow \] \[\cos \theta =\frac{1}{2}\] \[\Rightarrow \] \[\theta ={{60}^{o}}\]
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