Question

Find the cosine and sine angle between the vectors $\overrightarrow a = 2\widehat i + \widehat j + 3\widehat k$ and $\overrightarrow b = 4\widehat i - 2\widehat j + 2\widehat k$.

Answer 1

Hint: Here, we will use the formula of the dot product of two vectors. Then substituting the given vectors in the formula, we will be able to find the cosine angle between the given vectors. We will then substitute the cosine angle in the formula depicting the relationship between cosine and sine to find both the required angles between the given two vectors.

Formula Used:
We will use the following formulas:
1.$\overrightarrow a \cdot \overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta $
2.${\sin ^2}\theta + {\cos ^2}\theta = 1$

Complete step-by-step answer:
Given vectors are:
$\overrightarrow a = 2\widehat i + \widehat j + 3\widehat k$
$\overrightarrow b = 4\widehat i - 2\widehat j + 2\widehat k$
Now, by dot product formula, we know that,
$\overrightarrow a \cdot \overrightarrow b = \left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta $
This can also be written as:
$ \Rightarrow \cos \theta = \dfrac{{\overrightarrow a \cdot \overrightarrow b }}{{\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}$
Now, substituting the vectors, we get,
$ \Rightarrow \cos \theta = \dfrac{{\left( {2\widehat i + \widehat j + 3\widehat k} \right) \cdot \left( {4\widehat i - 2\widehat j + 2\widehat k} \right)}}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 3 \right)}^2}} \cdot \sqrt {{{\left( 4 \right)}^2} + {{\left( { - 2} \right)}^2} + {{\left( 2 \right)}^2}} }}$
$ \Rightarrow \cos \theta = \dfrac{{8 - 2 + 6}}{{\sqrt {4 + 1 + 9} \cdot \sqrt {16 + 4 + 4} }}$
Solving this further, we get,
$ \Rightarrow \cos \theta = \dfrac{{12}}{{\sqrt {14} \cdot \sqrt {24} }} = \dfrac{{12}}{{\sqrt {2 \times 7} \cdot \sqrt {2 \times 2 \times 2 \times 3} }}$
$ \Rightarrow \cos \theta = \dfrac{{12}}{{4\sqrt {21} }} = \dfrac{3}{{\sqrt {3 \times 7} }} = \dfrac{{\sqrt 3 }}{{\sqrt 7 }}$
Therefore, the cosine angle between the given vectors is $\cos \theta = \dfrac{{\sqrt 3 }}{{\sqrt 7 }}$
Now, we know that: ${\sin ^2}\theta + {\cos ^2}\theta = 1$
Hence, $\sin \theta = \sqrt {1 - {{\cos }^2}\theta } $
Substituting the value of $\cos \theta $, we get,
$\Rightarrow \sin \theta = \sqrt {1 - {{\left( {\dfrac{{\sqrt 3 }}{{\sqrt 7 }}} \right)}^2}} = \sqrt {1 - \dfrac{3}{7}} = \sqrt {\dfrac{4}{7}} $
Therefore,
$\Rightarrow \sin \theta = \dfrac{2}{{\sqrt 7 }}$
Hence, the cosine and sine angle between the vectors $\overrightarrow a = 2\widehat i + \widehat j + 3\widehat k$ and $\overrightarrow b = 4\widehat i - 2\widehat j + 2\widehat k$ are $\dfrac{{\sqrt 3 }}{{\sqrt 7 }}$ and $\dfrac{2}{{\sqrt 7 }}$ respectively.
Thus, this is the required answer.

Note: A scalar is a quantity that has a magnitude whereas; a vector is a mathematical quantity that has both magnitude and direction. A line of a given length and pointing along a given direction, such as an arrow, is a typical representation of a vector. Also, for further knowledge, position vector is also known as location vector, it is a straight line having one end fixed and the other end attached to a moving point, it is used to describe the position of a certain point, which turns out to be its respective coordinates.