# Vector Physics: Explanation, Addition, Subtraction, Angles, Resultant, Decomposition, and Problems with Solutions

By a Guy Who Teaches Physics for Fun
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## What is a vector in physics?

Vector Physics – A vector is a quantity that has both magnitude(number) and direction. Vectors are written mathematically using the notation $\hat{i}$, $\hat{j}$, and $\hat{k}$ and can be drawn as an arrow.
In daily life during the pandemic, you regularly have to check your temperature before entering any public place. Your temperature is normal if it measures 36-37 degrees celsius. In physics, when we measured a quantity and only numbers show up, such as body temperature, the quantity is called a scalar quantity. But when a quantity is a combination of direction and a number such as velocity then the quantity is a vector.
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### • Symbol of vector

How do we write the symbol of a vector? The symbol of a vector could be any letter with the presence of an arrow above the letter such as $\vec{P}$, $\vec{x}$, $\vec{v}$, and $\vec{F}$. Vectors can also be written in bold letters such as P, x, v, and F.
The magnitude of a vector is a number and its symbol is any letter without an arrow on it such as P, x, v, and F. In some books, the magnitude of a vector can also be symbolized as $\left |\mathbf{F} \right |$ or $\left \|\mathbf{F} \right \|$.
The direction of a vector is described as the angle between the vector and the positive x-axis. The angle is symbolize as $\theta$.

### • How to describe a vector us based on its visual

How do we describe a vector? Look at the image below.
The length of the vector is the magnitude of the vector. The image above has a scaling of one square equal to one unit.
The image shows 4 vectors, namely $\vec{A}$, $\vec{B}$, $\vec{C}$, and $\ vec{D}$. Each of these vectors has a magnitude and a direction. Let’s describe them one by one.
• The vector $\overrightarrow{A}$ has a magnitude of 5 units and its direction is to the right($\theta=0^o$).
• Vector $\overrightarrow{B}$ has a magnitude of 3 units and has an upward direction($\theta=90^o$).
• The vector $\overrightarrow{C}$ has a magnitude of 6 units and its direction is 30 degrees from the positive x-axis($\theta=30^o$).
• Vector $\overrightarrow{D}$ has a magnitude of 4 units and has its direction is $-120$ degrees from positive x-axis($\theta=-120^o$).

### • Vector Decomposition

A 2D vector can be decomposed into 2 components, namely the x component and y component.
Take a look at the vector $\vec{V}$ below. This vector has a magnitude of V and has its angle is $\theta$.
In the image above there are 3 arrows. The blue arrow is a $\overrightarrow{V}$. The red arrow ($V_x$) is the x component of $\vec{V}$. The green arrow ($V_y$) is the y component of $\vec{V}$.
The magnitude of the x component of $\overrightarrow{V}$ can be calculated using the following formula.
$V_{x}=V \cos \theta$
We can calculate the magnitude of the y component of $\overrightarrow{V}$ by using the following formula.
$V_{y}=V \sin \theta$
You could use the following problems to practice how to decompose vectors.
Problem 1
The vector $\vec{E}$ has a magnitude of 5 units and its angle is 37 degrees. Determine the vector x component($E_x$) and the y component($E_y$).
Solution
Let’s first sketch the vector $\vec{E}$.
You can slide the image.
Next, we sketch and calculate $E_x$ and $E_y$ using the previously described formula. Please slide the image above to see how it is done.

### • How do we express a vector mathematically

We can write a 2D vector using the $\hat{i}$ and $\hat{j}$ notation as follows. Look at the image below.
$\hat{i}$ is the notation given when a vector has a component that points to the right or to the left. If $\hat{i}$ is positive, it means that the vector has a component that points to the right, while if $\hat{i}$ is negative, it means that the vector has a component that points to the left. In the image above, the vector has an x component which is 7 squares to the right, which means we write $7 \hat{i}$.
$\hat{j}$ is the notation given when a vector has a component that points upward or downward. If $\hat{j}$ is positive, this means that the vector has a component that points upward. Meanwhile, if $\hat{j}$ is negative, it means that the vector has a component pointing downwards. For the case above, the vector has a y component which is 4 squares to the top which means we write it as 4 $\hat{j}$.
If we have a vector written using the $\hat{i}$ and $\hat{j}$ notation, we can find the magnitude of the vector using the following formula.
\begin{aligned} V=\sqrt{{V_x}^2 + {V_y}^2} \end{aligned}
The vector $\overrightarrow{V}$ shown on the image above has a magnitude that we can calculate.
\begin{aligned} V=&\sqrt{{V_x}^2 + {V_y}^2}\\ =&\sqrt{{7}^2 + {4}^2}\\ \approx&\;8 \; \text{unit} \end{aligned}
And we can also find the angle $\theta$ of  \overrightarrow{V} by using the following equation.
\begin{aligned} \theta=&\tan^{-1}\frac{V_y}{V_x} \end{aligned}
We could apply the equation to find the angle$\theta$ of $\overrightarrow{V}$.
\begin{aligned} \theta=&\;\tan^{-1}\frac{V_y}{V_x}\\ =&\;\tan^{-1}\frac{4}{7}\\ =&\;29.7^o\\ \end{aligned}
Problem 2
A vector $\overrightarrow{F}$ has a magnitude of 10 units and its angle $\theta$ is 53 degrees. Express the vector $\vec{F}$ using the notation $\hat{i}$ and $\hat{j}$.
Solution
First, let’s sketch the vector $\vec{F}$ on a Cartesian plane along with its components on the x and y axis.
Next, we calculate the vector components as follows.
\begin{aligned} F_{x}=&F \cos \theta\\ =&10 \cos 53^o\\ =&6 \end{aligned}
\begin{aligned} F_{y}=&\;F \cos \theta\\ =&\;10 \sin 53^o\\ =&\;8 \end{aligned}
Finally, we write the vector $\vec{F}$ using the $\hat{i}$ and $\hat{j}$ notation as follows.
\begin{aligned} \vec{F}=&\;F_x \hat{i} + F_y \hat{j}\\ =&\;6 \hat{i} + 8 \hat{j}\\ \end{aligned}
Problem 3
A vector $\overrightarrow{G}$ has the following expression.
\begin{aligned} \overrightarrow{G}=6\sqrt{3}\hat{i}+6\hat{j} \end{aligned}
Sketch the vector, determine the magnitude of the vector, and determine the angle of the vector.
Solution
let’s sketch the vector
Next we calculate the magnitude of the vector.
\begin{aligned} G=&\sqrt{{G_x}^2+{G_y}^2}\\ =&\sqrt{({6\sqrt{3}})^2+6^2}\\ =&\;12\; \text{unit} \end{aligned}
Finally, we determine the angle of the vector.
\begin{aligned} \theta=&\;\tan^{-1}\frac{G_y}{G_x}\\ =&\;\tan^{-1}\frac{6}{6\sqrt{3}}\\ =&\;30^o \end{aligned}

### • Negating a Vector

Vector can be converted into negative form. How do we do that? we reverse the vector 180 degrees without changing magnitude. Let’s use the image below to understand better.
The picture shows vector $\overrightarrow{A}$ which magnitude’s is 5 units and its angle is $\theta= 37^o$. The negative form of $\overrightarrow{A}$ is $\overrightarrow{-A}$. The angle of $\overrightarrow{-A}$ is $\theta= 37^o + 180^o=217^o$ and its magnitude is 5 units.
The vector $\overrightarrow{A}$ can be written using the $\hat{i}$ and $\hat{j}$ notation as follows.
\begin{aligned} \overrightarrow{A}=4\hat{i} + 3\hat{j} \end{aligned}
Mathematically, negating $\overrightarrow{A}$ could be achieved by multiplying $\overrightarrow{A}$ by $-1$ as follows.
\begin{aligned} -\overrightarrow{A}=&(-1)(4\hat{i} + 3\hat{j})\\ =&-4\hat{i} – 3\hat{j} \end{aligned}

We can do addition or subtraction to vectors. Mathematically, we operate something like $\overrightarrow{A}+\overrightarrow{B}$ or $\overrightarrow{A}-\overrightarrow{B}$. The outcome of doing addition or subtraction to vectors is something we call as “resultant”. There are actually 2 ways to find the resultant, by sketching and calculation.

### •Triangle method(connecting tail with head)

Let’s say, we want to operate $\overrightarrow{A} + \overrightarrow{B}$ then what we do is to connect the head of $\overrightarrow{A}$ with the tail of $\overrightarrow{B}$. The resultant of the operation is sketched by connecting the tail of $\overrightarrow{A}$ with the head of $\overrightarrow{B}$ . Take a look at the slides below.
Next, if we want to do $\overrightarrow{A}-\overrightarrow{B}$ operation, we first need to negate the vector $\overrightarrow{B}$ so that the operation becomes $\overrightarrow {A}-\overrightarrow{B}=\overrightarrow{A}+(-\overrightarrow{B})$. Then we do the same as before. You can look at the second slide above to see how to do vector substraction by sketching.

### • Parallelogram method(connecting the tail with tail)

We can also perform vector addition and subtraction by using the parallelogram method. This is done by connecting the both tail of $\overrightarrow{A}$ and $\overrightarrow{B}$ then creating a parallelogram. The result of the operation is the diagonal line of the parallelogram. You can see how we do this by looking at the image slides below.

### • Calculate the resultant by math

Yeah it’s nice to sketch vectors but how do we calculate the magnitude of a resultant? First, we need understand that the resultant itself is also a vector which can be expressed mathematically as follows.
\begin{aligned} \overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B} \end{aligned}
In the case of subtraction we write
\begin{aligned} \overrightarrow{R}=\overrightarrow{A}-\overrightarrow{B} \end{aligned}
There are 2 ways to find $\overrightarrow{R}$, 1) using the angles and 2) using the unit vector $\hat{i}$ and $\hat{j}$.

#### • Using the angles

Look at the image above. You can see vector $\overrightarrow{A}$, vector $\overrightarrow{B}$, and vector $\overrightarrow{R}$ which is the resultant of $\overrightarrow{A}+\overrightarrow{B}$. All vectors are connected to each other at the origin. And there are also $\alpha$, $\beta$, and $\theta$ as you can see in the picture. We can find the magnitude of R using the following equation.
\begin{aligned} R=&\; \sqrt{A^2+B^2+2AB\cos \theta} \end{aligned}
We can calculate the angle $\theta$ by using the following formula.
\begin{aligned} \frac{R}{\sin \theta}=\frac{A}{\sin \beta}=\frac{B}{\sin \alpha} \end{aligned}

#### • Using the unit vector $\hat{i}$ dan $\hat{j}$

The advantage of this method is that you can use this method for addition and subtraction of 2 vectors or more.
How do we do this? First, we decompose every vector in operation and then write it using the notation $\hat{i}$ and $\hat{j}$ as follows.
\begin{aligned} \overrightarrow{R}=&\overrightarrow{A}+\overrightarrow{B}\\ =&({A_x}\hat{i}+{A_y}\hat{j})+({B_x}\hat{i}+{B_y}\hat{j})\\ =&({A_x}+{B_x})\hat{i}+({A_y}+{B_y})\hat{j} \end{aligned}
Next, we can find the resultant R with the following equation.
\begin{aligned} R=&\;\sqrt{({A_x}+{B_x})^2+({A_y}+{B_y})^2} \end{aligned}
##### Substraction
\begin{aligned} \overrightarrow{R}=&\overrightarrow{A}-\overrightarrow{B}\\ =&({A_x}\hat{i}+{A_y}\hat{j})-({B_x}\hat{i}+{B_y}\hat{j})\\ =&({A_x}-{B_x})\hat{i}+({A_y}-{B_y})\hat{j} \end{aligned}
Next, we can find the resultant R with the following equation.
\begin{aligned} R=&\;\sqrt{({A_x}-{B_x})^2+({A_y}-{B_y})^2} \end{aligned}
Let’s study the following problems to practice.
Problem 4
Given vector $\overrightarrow{A}=7\hat{i}+4\hat{j}$ and vector $\overrightarrow{B}=4\hat{i}-8\hat{ j}$. Determine the resultant vector $\overrightarrow{R}$ from the operation $\overrightarrow{A}+ \overrightarrow{B}$, the magnitude of the vector R, the angle of the resultant vector from the positive x-axis ($\theta$), and sketch everything on the Cartesian plane.
Solution
We can find the resultant vector $\overrightarrow{R}$ from the operation $\overrightarrow{A}$ and $\overrightarrow{B}$ as follows.
\begin{aligned} \overrightarrow{R}=&\;\overrightarrow{A}+\overrightarrow{B}\\ =&\;({7}\hat{i}+{4}\hat{j})+({4}\hat{i}-{8}\hat{j})\\ =&\;({7}+{4})\hat{i}+({4}-{8})\hat{j}\\ =&\;11\hat{i}-4\hat{j} \end{aligned}
Next, we calculate the magnitude of $\overrightarrow{R}$.
\begin{aligned} R=&\;\sqrt{11^2+(-4^2)}\\ =&\;11.7 \;\text{unit} \end{aligned}
We could also find the angle $\theta$
\begin{aligned} \theta=&\;\tan^{-1}\left [\frac{-4}{11} \right ]\\ \approx&\; -20^o \end{aligned}
The image below is how exactly we sketch all the vectors.

## Physics Problems Using Vector Operations

Some examples of problems below are physics problems that require vector operations to solve. It’s as everyone says, practice makes perfect.

### • Coulomb's Law

Look at the image above. Point charge C experiences coulomb’s force due to point charges A and B. It is know that $F_{CA}=2\sqrt{13} \;N$ and $F_{CB}=5\; N$. The angle between $\overrightarrow{F_{CA}}$ and $\overrightarrow{F_{CB}}$ is $109.44^o$. Determine the resultant force point charge C experiences.
Solution
We could find the resultant force point charge C experience by calculating the resultant magnitude of $\overrightarrow{F_{CA}}+\overrightarrow{F_{CA}}$. We do it as follows.
\begin{aligned} R=&\;\sqrt{{F_{CA}}^2+{F_{CB}}^2+2F_{CA}F_{CB}\cos{\theta}}\\ =&\;\sqrt{52+25+(2)(2\sqrt{13})(5)\cos{109.44^o}}\\ =&\;7.280\; N \end{aligned}

### • Decomposing a force vector

Look at the image above. A person is pulling a bag with a force of 6 Newtons at an angle of 56 degrees. Determine the force component on the x-axis and y-axis.
Solution
We find the component of the force on the x-axis$(F_x)$ in the following way.
\begin{aligned} F_x=&\;F \cos{\theta}\\ =&\;6 \cos{56^o}\\ =&\;3.36 \;\text{N} \end{aligned}
Next, we find the y component of the force$(F_y)$ as follows.
\begin{aligned} F_y=&\;F \cos{\theta}\\ =&\;6 \sin{56^o}\\ =&\;4.97 \;\text{N} \end{aligned}

### • Electric field by point charges

Look at the image above. Point O experiences electric field of 3 point charges(A,B, and C). Each points exerts electric field on point O and they are $\overrightarrow{E_A}=(4\hat{i}+3\hat{j})\;\frac{N}{C }$, $\overrightarrow{E_B}=(-6\hat{j})\;\frac{N}{C}$, and $\overrightarrow{E_C}=( -3\hat{i})\;\frac{N}{C}$. Determine the resultant electric field at point O and it’s angle from the positive x-axis.
Solution
Let’s symbolize the resultant electric field at point O $\overrightarrow{E_R}$. We could find $\overrightarrow{E_R}$ by operating $\overrightarrow{E_A}+\overrightarrow{E_B}+\overrightarrow{E_C}$.
\begin{aligned} \overrightarrow{E_R}=&\;\overrightarrow{E_A}+\overrightarrow{E_B}+\overrightarrow{E_C} \\ =&\;4\hat{i}+3\hat{j}-3\hat{i}-6\hat{j}\\ =&\;(1\hat{i}-3\hat{j}) \;\frac{N}{C} \end{aligned}
Next, we could find the magnitude of $\overrightarrow{E_R}$ as follows.
\begin{aligned} {E_R}=&\;\sqrt{{E_{Rx}}^2+{E_{Ry}}^2} \\ =&\;\sqrt{1^2+(-3)^2} \\ =&\;\sqrt{10} \;\frac{N}{C} \end{aligned}
The angle between the resultant vector with the x-axis is calculated as follows.
\begin{aligned} \theta=&\;\tan^{-1}\left [\frac{E_{Ry}}{E_{Rx}} \right ] \\ =&\;\tan^{-1}\left [\frac{-3}{1} \right ] \\ =&\;-71.57^\circ \end{aligned}
Practice makes perfect, right? Go practice here 13 Problems with Solutions to Practice Vector Physics
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