Question

If vector $ \hat a + 2\hat b $ is perpendicular to vector $ 5\hat a - 4\hat b $ then find the angle between $ \hat a $ and $ \hat b $ where $ \hat a $ and $ \hat b $ are unit vectors.

Answer 1

Hint: From the given vectors $ \hat a $ and $ \hat b $ are unit vectors. The magnitude of any unit vector is $ 1 $ and hence the magnitude of $ \hat a $ and $ \hat b $ i.e. $ \left| {\hat a} \right| $ and $ \left| {\hat b} \right| $ is one. If two vectors are perpendicular to each other, then their dot product would be zero.

Complete step-by-step answer:
Given to us are two vectors $ \hat a + 2\hat b $ and $ 5\hat a - 4\hat b $ which are perpendicular to each other.
 $ \hat a $ and $ \hat b $ are unit vectors and hence their magnitude would be one. This can be written as $ \left| {\hat a} \right| = 1 $ and $ \left| {\hat b} \right| = 1 $
Let us assume that the angle between $ \hat a $ and $ \hat b $ to be $ \theta $
Now, from the given information the vector $ \hat a + 2\hat b $ is perpendicular to the vector $ 5\hat a - 4\hat b $ which means that their dot product is equal to zero. We can write this as follows.
 $ \left( {\hat a + 2\hat b} \right).\left( {5\hat a - 4\hat b} \right) = 0 $
On solving this we get $ 5\hat a \cdot \hat a - 4\hat a \cdot \hat b + 10\hat a \cdot \hat b - 8\hat b \cdot \hat b = 0 $
On further solving, we get $ 6\hat a \cdot \hat b + 5\hat a \cdot \hat a - 8\hat b \cdot \hat b = 0 $
Now we know that the dot product of a vector with itself gives the value one. Hence $ \hat a \cdot \hat a = 1 $ and $ \hat b \cdot \hat b = 1 $
Now the above equation becomes, $ 6\hat a \cdot \hat b + 5 - 8 = 0 \Rightarrow 6\hat a \cdot \hat b = 3 $
From this we get $ \hat a \cdot \hat b = \dfrac{1}{2} $
Now we use the formula $ \hat a \cdot \hat b = \left| {\hat a} \right|\left| {\hat b} \right|\cos \theta $ to solve the above equation.
Now the equation becomes $ \left| {\hat a} \right|\left| {\hat b} \right|\cos \theta = \dfrac{1}{2} $
We have already calculated the values of $ \left| {\hat a} \right| $ and $ \left| {\hat b} \right| $ as one. We substitute these values to get $ \cos \theta = \dfrac{1}{2} $
From this we get $ \theta = 60^\circ $
Therefore the angle between $ \hat a $ and $ \hat b $ is $ 60^\circ $
So, the correct answer is “$ 60^\circ $ ”.

Note: The dot product of any two vectors A and B is given as $ A.B = \left| A \right|\left| B \right|\cos \theta $ where $ \left| A \right| $ and $ \left| B \right| $ are the magnitudes of vectors A and B respectively. The angle between the vector A and B is $ \theta $ . If the vectors are perpendicular to each other then the angle between them would be $ 90^\circ $ so the value of $ \cos \theta $ would be zero and hence $ A \cdot B = 0 $