Question

Find the angle between the vectors $\vec{a}=2\hat{i}+\hat{j}-3\hat{k}$ and $\vec{b}=3\hat{i}-2\hat{j}-\hat{k}$

Answer 1

Hint: In this question, we are given two vectors and we need to find the angle between them. We should therefore first understand the method to find the angle between vectors using dot product between them and their magnitudes and then use it for the given vectors to obtain the answer to this question.

Complete step-by-step answer:

The dot product between two vectors $\vec{a}={{a}_{x}}\hat{x}+{{a}_{y}}\hat{y}+{{a}_{z}}\hat{z}$ and $\vec{b}={{b}_{x}}\hat{x}+{{b}_{y}}\hat{y}+{{b}_{z}}\hat{z}$ is defined as
$\vec{a}.\vec{b}={{a}_{x}}{{b}_{x}}+{{a}_{y}}{{b}_{y}}+{{a}_{z}}{{b}_{z}}..........................(1.1)$
Also, geometrically, the dot product between $\vec{a}$ and $\vec{b}$ is defined as
$\vec{a}.\vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta ............................(1.2)$
Therefore, comparing the RHS of the equations (1.1) and (1.2), we can write
$\begin{align}
  & {{a}_{x}}{{b}_{x}}+{{a}_{y}}{{b}_{y}}+{{a}_{z}}{{b}_{z}}=\vec{a}.\vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta \\
 & \Rightarrow \cos \theta =\dfrac{{{a}_{x}}{{b}_{x}}+{{a}_{y}}{{b}_{y}}+{{a}_{z}}{{b}_{z}}}{\left| {\vec{a}} \right|\left| {\vec{b}} \right|}..........................(1.3) \\
\end{align}$
where $\theta $ is the angle between $\vec{a}$ and $\vec{b}$ and $\left| {\vec{a}} \right|\text{ and }\left| {\vec{b}} \right|$ are given by
$\begin{align}
  & \left| {\vec{a}} \right|=\sqrt{{{a}_{x}}^{2}+{{a}_{y}}^{2}+{{a}_{z}}^{2}} \\
 & \left| {\vec{b}} \right|=\sqrt{{{b}_{x}}^{2}+{{b}_{y}}^{2}+{{b}_{z}}^{2}}..........................(1.4) \\
\end{align}$
To solve this question, we can take $\vec{a}=2\hat{i}+\hat{j}-3\hat{k}$ and $\vec{b}=3\hat{i}-2\hat{j}-\hat{k}$ and use it in equation (1.4) to obtain
$\begin{align}
  & \left| {\vec{a}} \right|=\sqrt{{{2}^{2}}+{{1}^{2}}+{{\left( -3 \right)}^{2}}}=\sqrt{14} \\
 & \left| {\vec{b}} \right|=\sqrt{{{3}^{2}}+{{\left( -2 \right)}^{2}}+{{\left( -1 \right)}^{2}}}=\sqrt{14}..........................(1.5) \\
\end{align}$
We can then use these values in equation (1.3) to obtain
$\cos \theta =\dfrac{2\times 3+1\times -2+-3\times -1}{\sqrt{14}\sqrt{14}}=\dfrac{7}{14}=\dfrac{1}{2}..........................(1.6)$
Also, we know that the cosine of $\dfrac{\pi }{3}$ is given by
$\cos \left( \dfrac{\pi }{3} \right)=\dfrac{1}{2}$
Therefore, using the above equation and equation (1.6), we get
\[\cos \left( \dfrac{\pi }{3} \right)=\cos \theta \]
Also, we know that
$\begin{align}
  & \cos \left( \theta \right)=\cos \left( \alpha \right) \\
 & \Rightarrow \theta =2n\pi \pm \alpha \\
\end{align}$
Where n is an integer. Using this in equation (1.7), we get
$\theta =2n\pi \pm \dfrac{\pi }{3}....................(1.9)$
Now, increasing an angle by $2n\pi $ gives the same angle, and also as we are only asked to find the angle between the two vectors and not its orientation, we can omit the sign of the angle. Therefore, we get the angle between the vectors as
$\theta =\dfrac{\pi }{3}$
This is the required answer to this question.

Note: We should note that as $\pi ={{180}^{\circ }}$ , therefore $\dfrac{\pi }{3}={{60}^{\circ }}$ and therefore, the angle can also be expressed as ${{60}^{\circ }}$ . However, expressing the angle in radian or degrees does not matter as long as we are expressing the correct angle in either of the ways i.e. in radians or in degrees.