Question

How do you find the angle between the vectors $u=\langle 3,2\rangle$ and $v=\langle 4,0\rangle$?

Answer 1

Consider the vectors u and v lying in x-y plane and write the vectors as $\vec{u}=3\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,$ and $\vec{v}=4\overset{\wedge }{\mathop{i}}\,+0\overset{\wedge }{\mathop{j}}\,$. Take the dot product of the two vectors by assuming that the angle between them is $\theta$. To find $\overset{\to }{\mathop{u}}\,.\overset{\to }{\mathop{v}}\,$, multiply the coefficients of $\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,$ of vector u with the corresponding coefficients of $\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,$ of vector v. Use the formula $\overset{\to }{\mathop{u}}\,.\overset{\to }{\mathop{v}}\,=uv\cos \theta$and find the magnitude of the vector given as $\overset{\to }{\mathop{a}}\,=x\overset{\wedge }{\mathop{i}}\,+y\overset{\wedge }{\mathop{j}}\,$ by using the formula: $a=\sqrt{{{x}^{2}}+{{y}^{2}}}$ and hence find the value of $\theta$.

Complete step by step solution:
Here, we have been provided with two vectors, $u=\langle 3,2\rangle$ and $v=\langle 4,0\rangle$, and we are asked to find the angle between the two vectors. Here, we will use the formula of the dot product of two vectors.
Here we are assuming that the angle is lying in x-y plane, so we can write the given vectors as $\vec{u}=3\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\,$ and $\vec{v}=4\overset{\wedge }{\mathop{i}}\,+0\overset{\wedge }{\mathop{j}}\,$.
Now, the dot product of two vectors is defined as the product of the magnitude of two vectors and the cosine of the angle between them. For example: - let us consider two vectors, $\overset{\to }{\mathop{x}}\,$ and $\overset{\to }{\mathop{y}}\,$ and the angle between them is $\theta$. So, the dot product of these vectors is given as: -

$\Rightarrow \overrightarrow{x}.\overrightarrow{y}=xy\cos \theta$
Here, x denotes the magnitude of $\overrightarrow{x}$ and similarly y denotes the magnitude of $\overrightarrow{y}$. Now, one thing you may note is that when the two vectors are perpendicular, i.e., $\theta ={{90}^{\circ }}$, then the dot product of two vectors will become zero. This is because $\cos {{90}^{\circ }}=0$.
Now, we know that $\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,$ are perpendicular to each other as they represent unit vectors along the x, y axis respectively. So, we have,
$\Rightarrow \overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{j}}\,=\overset{\wedge }{\mathop{j}}\,.\overset{\wedge }{\mathop{i}}\,=0$
That means we have to take the product of $\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,$ of one vector with the corresponding $\overset{\wedge }{\mathop{i}}\,,\overset{\wedge }{\mathop{j}}\,$ of the second vector. Here, $\overset{\wedge }{\mathop{i}}\,.\overset{\wedge }{\mathop{i}}\,=\overset{\wedge }{\mathop{j}}\,.\overset{\wedge }{\mathop{j}}\,=1$ because in these two cases $\theta ={{0}^{\circ }}$ and we know that $\cos {{0}^{\circ }}=1$.
Let us come to the question, considering the dot product of vectors u and v by assuming that the angle between them is $\theta$, we get,

& \Rightarrow \overset{\to }{\mathop{u}}\,.\overset{\to }{\mathop{v}}\,=\left( 3\overset{\wedge }{\mathop{i}}\,+2\overset{\wedge }{\mathop{j}}\, \right).\left( 4\overset{\wedge }{\mathop{i}}\,+0\overset{\wedge }{\mathop{j}}\, \right) \\\ & \Rightarrow uv\cos \theta =\left( 3\times 4 \right)+\left( 2\times 0 \right) \\\ & \Rightarrow uv\cos \theta =12 \\\ \end{aligned}$$ Using the formula for finding the magnitude of $$\overset{\to }{\mathop{a}}\,=x\overset{\wedge }{\mathop{i}}\,+y\overset{\wedge }{\mathop{j}}\,$$ given as: $a=\sqrt{{{x}^{2}}+{{y}^{2}}}$, we get, $$\begin{aligned} & \Rightarrow \sqrt{{{3}^{2}}+{{2}^{2}}}\times \sqrt{{{4}^{2}}}\times \cos \theta =12 \\\ & \Rightarrow \sqrt{13}\times 4\times \cos \theta =12 \\\ & \Rightarrow 4\sqrt{13}\cos \theta =12 \\\ & \Rightarrow \cos \theta =\dfrac{12}{4\sqrt{13}} \\\ & \Rightarrow \cos \theta =\dfrac{3}{\sqrt{13}} \\\ & \Rightarrow \theta ={{\cos }^{-1}}\left( \dfrac{3}{\sqrt{13}} \right) \\\ \end{aligned}$$ **Note:** One may note that we cannot find the value of the angle $\theta $ obtained above without using the cosine table or calculator. Here we have assumed that the vectors are lying in x-y plane that is why we have taken the unit vectors as $$\overset{\wedge }{\mathop{i}}\,$$ and $$\overset{\wedge }{\mathop{j}}\,$$. You can also assume them to be lying in the y-z plane or x-z plane. In these cases the value of the angle will not change but only the unit vectors will change.