Question

Let a and b be two unit vectors such that $ a+2b $ and $ 5\text{ }a-4b $ are perpendicular to each other, then the angle between a and b is

• A. $ {{30}^{o}} $
• B. $ {{45}^{o}} $
• C. $ {{60}^{o}} $
• D. $ {{90}^{o}} $

Answer 1

Answer: (C)

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}} $