# Addition And Subtraction Of Vectors

We already learnt about the vectors and their notations in the previous physics article. Now we will learn about their addition and subtraction.

The addition is done based on Triangle law. Let us see what triangle rule is:

Suppose there are two vectors: $\overrightarrow{a}$ and $\overrightarrow{b}$

Now, draw a line $AB$ representing $\overrightarrow{a}$ with $A$ as the tail and $B$ as the head. Draw another line $BC$ representing ($\overrightarrow{b}$) with $B$ as the tail and $C$ as the head. Now join the line $AC$ with $A$ as the tail and $C$ as the head. The line $AC$ represents the resultant sum of the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$

The line AC represents $\overrightarrow{a}$ + $\overrightarrow{b}$

The magnitude of $\overrightarrow{a}$ + $\overrightarrow{b}$ is:

$\sqrt{a^2~+~b^2~+~2ab~cos~\theta}$

Where,

$a$ = magnitude of vector $\overrightarrow{a}$

$b$ = magnitude of vector $\overrightarrow{b}$

$\theta$ = angle between $\overrightarrow{a}$ and $\overrightarrow{b}$

Let the resultant make an angle of $\phi$ with $\overrightarrow{a}$, then:

$tan\phi$ = $\frac{b~sin~\theta}{a~+~b~cos~\theta}$

Let us understand this by the means of an example. Suppose there are two having equal magnitude $A$, and they make an angle $θ$ with each other. Now, to find the magnitude and direction of the resultant, we will use the formulas mentioned above.

Let the magnitude of the resultant vector be $B$

$B$ = $\sqrt{A^2~+~A^2~+~2AA~cos~\theta}$ = $2~A~cos~\frac{θ}{2}$

Let’s say that the resultant vector makes an angle $Ɵ$ with the first vector

$tan~\phi$ = $\frac{A~sin~θ}{A~+~A~cos~θ}$ = $tan~\frac{θ}{2}$

Or,

$Ɵ$ = $\frac{θ}{2}$

### Vector Subtraction:

Subtraction of two vectors is similar to addition. Suppose $\overrightarrow{a}$ is to be subtracted from $\overrightarrow{b}$.

$\overrightarrow{a}$$\overrightarrow{b}$ can be said as the addition of the vectors $\overrightarrow{a}$ and (-$\overrightarrow{b}$). Thus, the formula for addition can be applied as:

$\overrightarrow{a}$$\overrightarrow{b}$ = $\sqrt{a^2~+~b^2~-~2ab~cos~\theta}$

(-$\overrightarrow{b}$) is nothing but $\overrightarrow{b}$ reversed in direction.