## Vector Addition

In vector addition you always need to consider both the magnitude and the direction of vectors. Consider two vectors $\vec{A}$ and $\vec{B}$ in Figure 1 which we add together to get the resultant vector. Note that the vector $\vec{B}$ shown in dotted line in Figure 2 whose tail is at the head of $\vec{A}$ is equal to the original $\vec{B}$ and now the sum of $\vec{A}$ and $\vec{B}$ is the resultant vector whose tail is at the starting point, that is at the tail of $\vec{A}$ and head is at the ending point, that is at the head of $\vec{B}$.

In Figure 3 you can see that the sum of $\vec{A}$ and $\vec{B}$ is the resultant vector straight from the tail of $\vec{A}$ to the head of$\vec{B}$. In other words when you move along the direction of $\vec{A}$ through the magnitude $A$ and then along the direction of $\vec{B}$ through the magnitude $B$, your resultant magnitude and direction of your movement will be that shown by the resultant vector $\vec{R}=\vec{A}+\vec{B}$. Now vector subtraction is the same as vector addition. If you subtract $\vec{A}$ and $\vec{B}$, like $\vec{A}-\vec{B}$, this is the same as $\vec{R}=\vec{A}+(-\vec{B})$. Now you know that the negative of a vector has the same magnitude but has opposite direction. So $-\vec{B}$ has the same magnitude as that of $\vec{B}$ but has opposite direction and the resultant of the vectors will be that as shown in Figure 4.

## Components of a Vector

Here we discuss about resolving a vector into its components. We find the x and y-components of the vector $\vec{A}$ in Figure 5. The x-component of the vector is the component along x-axis and y-component along y-axis.

Notice in Figure 5 that $\vec{A}$ makes an angle $\phi$ with x-axis. Now we can easily find the value of ${{A}_{x}}$ and${{A}_{y}}$ to be, ${{A}_{x}}=A\cos \phi $ and ${{A}_{y}}=A\sin \phi $. So the x and y-components of $\vec{A}$ can be written in vector form as,

\[{{\vec{A}}_{x}}=(A\cos \phi )\hat{i}\]

\[{{\vec{A}}_{y}}=(A\sin \phi )\hat{j}\]

Note that the unit vectors $\hat{i}$ and $\hat{j}$ give the direction along positive x-axis and positive y-axis respectively. $\vec{A}$ can be written as $\vec{A}={{A}_{x}}\hat{i}+{{A}_{y}}\hat{j}=(A\cos \phi )\hat{i}+(A\sin \phi )\hat{j}$. Here we also can find the magnitude and direction of $\vec{A}$ as,

\[A=\sqrt{{{A}_{x}}^{2}+{{A}_{y}}^{2}} \tag{1} \label{1}\]

\[\phi =\arctan \left( \frac{{{A}_{y}}}{{{A}_{x}}} \right) \tag{2} \label{2}\]

Here the angle $\phi$ gives the direction for $\vec{A}$ and you obviously know $\tan \phi =\frac{{{A}_{y}}}{{{A}_{x}}}$.

## Addition of Vectors Using Components

Addition of vectors using components is the easiest way to add vectors. In Figure 6 there are two vectors. The tail of $\vec{B}$ is at the head of $\vec{A}$. First we find the x and y-components of $\vec{A}$ and $\vec{B}$, then we add all x-components to find the total x-component of the resultant vector and all y-components to find the total y-component of the resultant vector.

Now the total x and y-components of the resultant vector $\vec{R}$ are ${{R}_{x}}={{A}_{x}}+{{B}_{x}}$ and ${{R}_{y}}={{A}_{y}}+{{B}_{y}}$. Now we can find the magnitude and direction of the resultant vector easily. Suppose $\vec{R}$ makes an angle $\phi$ with x-axis.

\[R=\sqrt{{{R}_{x}}^{2}+{{R}_{y}}^{2}}\]

\[\phi =\arctan \frac{{{R}_{y}}}{{{R}_{x}}}\]

Now $\vec{R}$ is $\vec{R}={{R}_{x}}\hat{i}+{{R}_{y}}\hat{j}$. Obviously the unit vectors make it easy to represent a vector as well as its components.

## Parallelogram Addition of Vectors

The name parallelogram addition arrived as we arrange the vectors in the form of parallelogram by making the opposite vectors equal. In Figure 7. the vectors $\vec{A}$ and $\vec{B}$ shown in dotted line are equal to the original vectors (these vectors are acting as the opposite sides of the parallelogram) and help to form a parallelogram. You obviously know that the resultant vector $\vec{R}$ is the sum of $\vec{A}$ and $\vec{B}$ and here we find the magnitude and direction of the resultant vector.

In Figure 7. produce the line $\text{OM}$ up to $\text{N}$ and draw $\text{PN}\bot \text{ON}$. Let the angle $\angle \text{PMN=}\alpha $ and the angle made by the resultant vector $\vec{R}$ with $\vec{A}$ be $\theta$. Now $\sin \alpha =\frac{\text{PN}}{B}$ which gives $\text{PN}=B\sin \alpha $ and $\cos \alpha =\frac{\text{MN}}{B}$ which gives $\text{MN=}B\cos \alpha $, so $\text{ON}=\text{OM}+\text{MN}=A+B\cos \alpha $. Since $\vartriangle \text{OPN}$ is a right-angled triangle, the magnitude of $\vec{R}$ is,

\[\begin{align*} R&=\sqrt{\text{O}{{\text{N}}^{2}}+\text{P}{{\text{N}}^{2}}} \\ & =\sqrt{{{(A+Bcos\alpha )}^{2}}+(Bsin\alpha )} \\ & =\sqrt{{{A}^{2}}+{{B}^{2}}+2AB\cos \alpha } \tag{3} \label{3} \end{align*}\]

When $\alpha ={{90}^{\circ }}$, it means when the vectors are perpendicular Eq. \eqref{3} becomes $R=\sqrt{{{A}^{2}}+{{B}^{2}}}$ as it should be. Again, $\tan \theta =\frac{\text{PN}}{\text{ON}}$, so,

\[\begin{align*} \tan \theta&=\frac{B\sin \alpha }{A+B\cos \alpha } \\ \text{and,}\quad \quad \theta& =\arctan \left( \frac{B\sin \alpha }{A+B\cos \alpha } \right) \tag{4} \label{4} \end{align*}\]

You may have known that the angle $\theta$ indicates the direction of the resultant vector. When $\alpha ={{0}^{\circ }}$, $\theta ={{0}^{\circ }}$ which is we should get.