Let \(\vec{\text{a}}\) and \(\vec{\text{b}}\) are two unit vectors such that \(\vec{\text{a}}+2 \vec{\text{b}}\) and \(5\vec{\text{a}}−4\vec{\text{b}}\) are perpendicular. What is the angle between \(\vec{\text{a}}\) and \(\vec{\text{b}}\) ?

Concept:

• The angle between two vectors  \(\vec{\text{a}}\) and \(\vec{\text{b}}\) is given by, \(\cos θ = {\vec{\text{a}}.\vec{\text{b}} \over |\vec{\text{a}}||\vec{\text{b}}|}\) 
• If \(\vec{\text{a}}\) and \(\vec{\text{b}}\) are perpendicular vectors, then \(\vec{\text{a}}\).\(\vec{\text{b}}\) = 0 
• For any two vectors \(\vec{\text{a}}\) and \(\vec{\text{b}}\), \(\vec{\text{a}}.\vec{\text{b}} = \vec{\text{b}}.\vec{\text{a}}\) __(i)

Calculation:

Given:  \(\vec{\text{a}}\) and \(\vec{\text{b}}\) are two unit vectors such that \(\vec{\text{a}}+2 \vec{\text{b}}\) and \(5\vec{\text{a}}−4\vec{\text{b}}\) are perpendicular.

As  \(\vec{\text{a}}\) and \(\vec{\text{b}}\) are two unit vectors

⇒  \(|\vec{\text{a}}| = |\vec{\text{b}}| = 1\)___(ii)

And  \(\vec{\text{a}}+2 \vec{\text{b}}\) and \(5\vec{\text{a}}−4\vec{\text{b}}\) are perpendicular

⇒ (\(\vec{\text{a}}+2 \vec{\text{b}}\)).(\(5\vec{\text{a}}−4\vec{\text{b}}\)) = 0

⇒ \(5\vec{\text{a}}.\vec{\text{a}}-4 \vec{\text{a}}.\vec{\text{b}} + 10\vec{\text{b}}.\vec{\text{a}}-8\vec{\text{b}}.\vec{\text{b}}\) = 0

⇒ \(5|\vec{\text{a}}|^2-4 \vec{\text{a}}.\vec{\text{b}} + 10\vec{\text{a}}.\vec{\text{b}}-8|\vec{\text{b}}|^2\) = 0 {from (i)}

⇒ \(5(1) + 6\vec{\text{a}}.\vec{\text{b}}-8(1)\) = 0 {from (ii)}

⇒ \(\vec{\text{a}}.\vec{\text{b}}= {1 \over 2}\) __(iii)

Now angle between  \(\vec{\text{a}}\) and \(\vec{\text{b}}\) is given by, 

\(\cos θ = {\vec{\text{a}}.\vec{\text{b}} \over |\vec{\text{a}}||\vec{\text{b}}|}\)

⇒ \(\cos θ = {{1 \over 2} \over 1.1} = {1 \over 2}\) (from (ii) and (iii))

⇒ θ = \(\frac{\pi}{3}\)

∴ The correct option is (3).