Question

How do you find the component form of $v$ given its magnitude $7/2$ and the angle it makes with the positive $x$ -axis is $\theta ={{150}^{\circ }}$ ?

Answer 1

For answering this question we need to find the component form of $v$ given its magnitude $7/2$ and the angle it makes with the positive $x$ -axis is $\theta ={{150}^{\circ }}$. The general form of any vector $v$ is given as ${{v}_{x}}+{{v}_{y}}$ then its magnitude is $\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}$ and the angle it makes with the positive $x$ -axis is $\theta$ which is known as argument and given as $\theta ={{\tan }^{-1}}\left( \dfrac{{{v}_{y}}}{{{v}_{x}}} \right)$ .

Complete step by step answer:
Now considering from the question we have been asked to find the component form of $v$ given its magnitude $7/2$ and the angle it makes with the positive $x$ -axis is $\theta ={{150}^{\circ }}$.
From the basic components we know that the general form of any vector $v$ is given as ${{v}_{x}}+{{v}_{y}}$ then its magnitude is $\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}$ and the angle it makes with the positive $x$ -axis is $\theta$ which is known as argument and given as $\theta ={{\tan }^{-1}}\left( \dfrac{{{v}_{y}}}{{{v}_{x}}} \right)$ where ${{v}_{x}}$ is the $x$ component and ${{v}_{x}}$ is the $x$ component hence it is known as component form.
From the question we have magnitude as $7/2$ and the argument as $\theta ={{150}^{\circ }}$.
Therefore we can say that $\sqrt{{{v}_{x}}^{2}+{{v}_{y}}^{2}}=7/2$ which can be simplified as
$\begin{aligned} & {{v}_{x}}^{2}+{{v}_{y}}^{2}={{\left( 7/2 \right)}^{2}} \\\ & \Rightarrow {{v}_{x}}^{2}+{{v}_{y}}^{2}=\dfrac{49}{4} \\\ \end{aligned}$
We can also say that $\theta ={{\tan }^{-1}}\left( \dfrac{{{v}_{y}}}{{{v}_{x}}} \right)\Rightarrow \tan \theta =\left( \dfrac{{{v}_{y}}}{{{v}_{x}}} \right)$ as we have $\theta ={{150}^{\circ }}$ we will have $\tan {{150}^{\circ }}=\dfrac{-1}{\sqrt{3}}$ which implies that
$\begin{aligned} & \Rightarrow \left( \dfrac{{{v}_{y}}}{{{v}_{x}}} \right)=\dfrac{-1}{\sqrt{3}} \\\ & \Rightarrow {{v}_{y}}=-\dfrac{{{v}_{x}}}{\sqrt{3}} \\\ & \Rightarrow -\sqrt{3}{{v}_{y}}={{v}_{x}} \\\ \end{aligned}$
So by using this value in ${{v}_{x}}^{2}+{{v}_{y}}^{2}=\dfrac{49}{4}$ we will have
$\begin{aligned} & {{\left( -\sqrt{3}{{v}_{y}} \right)}^{2}}+{{v}_{y}}^{2}=\dfrac{49}{4} \\\ & \Rightarrow 4{{v}_{y}}^{2}=\dfrac{49}{4} \\\ & \Rightarrow 2{{v}_{y}}=\dfrac{7}{2} \\\ & \Rightarrow {{v}_{y}}=\dfrac{7}{4} \\\ \end{aligned}$
By using this we will have ${{v}_{x}}=-\sqrt{3}\left( \dfrac{7}{4} \right)\Rightarrow \dfrac{-7\sqrt{3}}{4}$ .
Hence we can conclude that the component form of $v$ given its magnitude $7/2$ and the angle it makes with the positive $x$ -axis is $\theta ={{150}^{\circ }}$ will be given as $\dfrac{-7\sqrt{3}}{4}+\dfrac{7}{4}$ .

Note: Here we should make sure that the sign of the value of the trigonometric ratio is correct because generally it changes with the change in the quadrant. We have four quadrants in the first all ratios will be positive, in the second one only sine and cosecant ratios will be positive, in the third one only tangent and cotangent ratios will be positive and in the fourth one only cosine and secant ratios will be positive. This question can also be answered using the formulae $\vec{v}=\left| v \right|\cos \theta +\left| v \right|\sin \theta$ which is done as follows $\vec{v}=\left( \dfrac{7}{2} \right)\left( \dfrac{-\sqrt{3}}{2} \right)+\left( \dfrac{7}{2} \right)\left( \dfrac{1}{2} \right)\Rightarrow v=\dfrac{-7\sqrt{3}}{4}+\dfrac{7}{4}$ the same answer.