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 $A \cdot B = 0$