Question

# What is the angle between $\left( {\widehat i + 2\widehat j + 2\widehat k} \right)$ and $\widehat i$ A) ${0^ \circ }$ B) $\dfrac{\pi }{6}$ C) $\dfrac{\pi }{3}$ D) None of these

Hint: In this question, we can use the dot product to find the angle between the vectors. In dot products, we can calculate the magnitude of each vector separately. After that we can find the dot product of the vectors. We can find the angle by using the formula$\overline A \bullet \overline B = \left| {\overline A } \right|\left| {\overline B } \right|\cos \theta$and substituting the all calculated values in it.

Complete step by step solution:
First, we can let
$\overline A = \widehat i + 2\widehat j + 2\widehat k$
And $\overline B = \widehat i$
We know that the dot product of two vectors is given as-
$\overline A \bullet \overline B = \left| {\overline A } \right|\left| {\overline B } \right|\cos \theta$ (i)
Where $\left| {\overline A } \right| =$ magnitude of $\overline A$ , $\left| {\overline B } \right| =$ magnitude of $\overline B$and $\theta =$ angle between the two vectors.
Now, let us find the magnitudes of the two vectors separately.
So, the magnitude of $\left| {\overline A } \right|$is
$\left| {\overline A } \right| = \sqrt {{1^2} + {2^2} + {2^2}} \\ \Rightarrow \left| {\overline A } \right| = \sqrt {1 + 4 + 4} \\ \Rightarrow \left| {\overline A } \right| = \sqrt 9 \\ \Rightarrow \left| {\overline A } \right| = 3 \\$
And the magnitude of $\left| {\overline B } \right|$is
$\left| {\overline B } \right| = \sqrt {{1^2}} \\ \Rightarrow \left| {\overline B } \right| = \sqrt 1 \\ \Rightarrow \left| {\overline B } \right| = 1 \\$
Now, let us find the dot product of both vectors, which will be-
$\overline A \bullet \overline B = \left( {\widehat i + 2\widehat j + 2\widehat k} \right) \bullet \left( {\widehat i} \right) \\ \Rightarrow \overline A \bullet \overline B = 1 \\$
Now, the angle $\theta$will be written as the following equation by using equation (i), we get-
$\theta = {\cos ^{ - 1}}\left( {\dfrac{{\overline A \bullet \overline B }}{{\left| {\overline A } \right|\left| {\overline B } \right|}}} \right)$
Substituting all the values, we get-
$\theta = {\cos ^{ - 1}}\left( {\dfrac{{\overline A \bullet \overline B }}{{\left| {\overline A } \right|\left| {\overline B } \right|}}} \right) \\ \Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{1}{{\left( 1 \right)\left( 3 \right)}}} \right) \\ \Rightarrow \theta = {\cos ^{ - 1}}\left( {\dfrac{1}{3}} \right) \\ \Rightarrow \theta = {\cos ^{ - 1}}\left( {\cos {{70.53}^ \circ }} \right) \\ \Rightarrow \theta = {70.53^ \circ } \\$

Hence, The angle between the vectors is ${70.53^ \circ }$.

Note: We have to remember the rules of dot product. In dot product, we know that $\widehat i \bullet \widehat i = 1,\widehat j \bullet \widehat j = 1,\widehat k \bullet \widehat k = 1$ . The other products will be zero in dot product. We have to remember that while finding the magnitude all the coefficients will be squared.
We have to remember that we can find the angle between the two vectors by using dot product. In a vector product, we have unit vectors also in the equation, which gives a difficult calculation. Sometimes, we don’t have the value of the unit vector. But in dot products, we only need to find the magnitudes of the vectors separately and the dot product of the vectors.