Question

How do you find the angle between the vectors $u=\left\langle 1,0 \right\rangle $ and $v=\left\langle 0,-2 \right\rangle $? \[\]

Answer 1

Hint: We recall the component wise representation of vectors in the plane in terms of unit orthogonal vectors $\hat{i},\hat{j}$ . We find the angle between two vectors $\overrightarrow{a},\overrightarrow{b}$ as $\theta ={{\cos }^{-1}}\left( \dfrac{\overrightarrow{a}\cdot \overrightarrow{b}}{ab} \right)$ where $\overrightarrow{a}\cdot \overrightarrow{b}$ represent the dot product of the vector and $a,b$ are magnitudes of the vectors $\overrightarrow{a},\overrightarrow{b}$ respectively.

Complete answer:
We know that the dot product of two vectors $\vec{a}$ and $\vec{b}$ is denoted as $\vec{a}\cdot \vec{b}$ and is given by $\vec{a}\cdot \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta $ where $\theta $ is the angle between the vectors $\vec{a}$ and $\vec{b}$/\[\]
We also know that $\hat{i}$ and $\hat{j}$ are unit vectors (vectors with magnitude 1) along $x$ and $y$axes respectively. So the magnitude of these vectors are $\left| {\hat{i}} \right|=\left| {\hat{j}} \right|=1$. The vectors just like their axes are perpendicular to each other which means any angle among $\hat{i}$ and $\hat{j}$is ${{90}^{\circ }}.$ So $\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=1\cdot 1\cdot \cos {{0}^{\circ }}=1$ and $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{i}=1\cdot 1\cdot \cos {{90}^{\circ }}=0$.\[\]
We know that all the vectors in terms of the coordinate components if $\overrightarrow{a}$ has ${{a}_{x}}$ and ${{a}_{y}}$ as the vector components along $x$ and $y-$axes respectively then we can represent $\overrightarrow{a}$ as $\overrightarrow{a}=\left\langle {{a}_{x}},{{a}_{y}} \right\rangle ={{a}_{x}}\hat{i}+{{a}_{y}}\hat{j}$ and then we can we can find the magnitude of $\overrightarrow{a}$ as $a=\left| \overrightarrow{a} \right|=\sqrt{a_{x}^{2}+a_{y}^{2}}$. If there is another vector $\overrightarrow{b}=\left\langle {{b}_{x}},{{b}_{y}} \right\rangle ={{b}_{x}}\hat{i}+{{b}_{y}}\hat{j}$ then dot product between $\overrightarrow{a},\overrightarrow{b}$ is given by
\[\begin{align}
  & \vec{a}\cdot \vec{b}={{a}_{x}}{{b}_{x}}+{{a}_{y}}{{b}_{y}} \\
 & \vec{a}\cdot \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \theta =ab\cos \theta \\
\end{align}\]
So the angle between two vectors is given by
\[\begin{align}
  & \overrightarrow{a}\cdot \overrightarrow{b}=ab\cos \theta \\
 & \Rightarrow \cos \theta =\dfrac{\overrightarrow{a}\cdot \overrightarrow{b}}{ab} \\
 & \Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{\overrightarrow{a}\cdot \overrightarrow{b}}{ab} \right) \\
 & \Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{{{a}_{x}}{{b}_{x}}+{{a}_{y}}{{b}_{y}}}{\sqrt{a_{x}^{2}+a_{y}^{2}}\sqrt{b_{x}^{2}+b_{y}^{2}}} \right) \\
\end{align}\]
We are given in the question two vectors $u=\left\langle 1,0 \right\rangle $ and $v=\left\langle 0,-2 \right\rangle $. So the $x$ and $y$component of vector $\overrightarrow{u}$ is ${{u}_{x}}=1,{{u}_{y}}=0$ respectively. The $x$ and $y$component of vector $\overrightarrow{u}$ is ${{u}_{x}}=0,{{u}_{y}}=-2$ respectively. Then the angles between two vectors is
\[\begin{align}
  & \theta ={{\cos }^{-1}}\left( \dfrac{{{u}_{x}}{{v}_{x}}+{{u}_{y}}{{v}_{y}}}{\sqrt{u_{x}^{2}+u_{y}^{2}}\sqrt{v_{x}^{2}+v_{y}^{2}}} \right) \\
 & \Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{1\left( 0 \right)+0\left( -2 \right)}{\sqrt{{{1}^{2}}+{{0}^{2}}}\sqrt{{{0}^{2}}+{{\left( -2 \right)}^{2}}}} \right) \\
 & \Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{0}{1\cdot 2} \right)={{\cos }^{-1}}\left( 0 \right)=\dfrac{\pi }{2} \\
\end{align}\]

Note:
We note that we measure the angle $\theta \in \left[ 0,2\pi \right)$ using dot product between two vectors as the smaller angle between them by a smaller amount of rotation from one vector to another vector. If we rotate in the opposite sense and then the greater angle is $2\pi -\theta $. So here in this problem the smaller between $\overrightarrow{u},\overrightarrow{v}$ is $\theta =\dfrac{\pi }{2}$ and the greater angle is $2\pi -\theta =2\pi -\dfrac{\pi }{2}=\dfrac{3\pi }{2}$. We can also use cross product $\vec{a}\times \vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\sin \theta \hat{n}$ to find the angle .