Question

Let $\hat a$ and $\hat b$ be two unit vectors. If the vectors $\vec c = \hat a + 2\hat b$ and $\vec d = 5\hat a - 4\hat b$ are perpendicular to each other, then the angle between $\hat a$ and $\hat b$ is
$\left( a \right)\dfrac{\pi }{6}$
$\left( b \right)\dfrac{\pi }{2}$
$\left( c \right)\dfrac{\pi }{3}$
$\left( d \right)\dfrac{\pi }{4}$

Answer 1

Hint: In this particular question use the concept that if two vectors are perpendicular to each other than their dot product should be equal to zero, and use the concept that the modulus of the unit vector is always one, so use these concepts to reach the solution of the question.

Complete step-by-step answer:
Given data:
Let $\hat a$ and $\hat b$ be two unit vectors.
So if $\hat a$ and $\hat b$ be two unit vectors, then the modulus of unit vectors are always one, so we have,
$ \Rightarrow \left| {\hat a} \right| = \left| {\hat b} \right| = 1$............... (1)
Now it is also given that the vectors $\vec c = \hat a + 2\hat b$ and $\vec d = 5\hat a - 4\hat b$ are perpendicular to each other.
Now as we know that if two vectors are perpendicular to each other then their dot product should be equal to zero.
$ \Rightarrow \vec c.\vec d = 0$
$ \Rightarrow \left( {\hat a + 2\hat b} \right).\left( {5\hat a - 4\hat b} \right) = 0$
Now simplify it we have,
$ \Rightarrow \left( {5\hat a.\hat a + 6\hat a.\hat b - 8\hat b.\hat b} \right) = 0$
Now as we know that $\hat a.\hat a = {\left| {\hat a} \right|^2},\hat b.\hat b = {\left| {\hat b} \right|^2}$ so use this property in the above equation we have,
$ \Rightarrow \left( {5{{\left| {\hat a} \right|}^2} + 6\hat a.\hat b - 8{{\left| {\hat b} \right|}^2}} \right) = 0$
Now from equation (1) we have,
$ \Rightarrow \left( {5\left( 1 \right) + 6\hat a.\hat b - 8\left( 1 \right)} \right) = 0$
$ \Rightarrow 6\hat a.\hat b = 3$
$ \Rightarrow \hat a.\hat b = \dfrac{3}{6} = \dfrac{1}{2}$.................. (2)
Now as we know that the dot product of any two vectors is given as,
$ \Rightarrow \hat a.\hat b = \left| {\hat a} \right|\left| {\hat b} \right|\cos \theta $, where $\theta $ is the angle between the vectors a, and b.
Now from equations (1) and (2) we have,
$ \Rightarrow \dfrac{1}{2} = \left( {1 \times 1} \right)\cos \theta $
$ \Rightarrow \cos \theta = \dfrac{1}{2} = \cos \left( {\dfrac{\pi }{3}} \right)$
$ \Rightarrow \theta = \dfrac{\pi }{3}$
So this is the required answer.
Hence option (c) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that the dot product of any two vectors is given as, $\hat a.\hat b = \left| {\hat a} \right|\left| {\hat b} \right|\cos \theta $, where $\theta $ is the angle between the vectors a, and b, so simply substitute the values in this equation as above we will get the required angle between the unit vectors a, and b.