Vector Addition and Subtraction with Analytical Method

Calculating a Resultant with Components

The graphical method shows the resultant direction, but its numerical accuracy depends on the drawing. The analytical method calculates components and therefore gives a reproducible numerical result.

The method is to resolve every vector into vector components along the $x$ and $y$ axes, add the components along each axis, and then rebuild the resultant.

R_x=\sum F_x

In the analytical method, add the vector components on each axis. The arrow lengths alone do not give the sum.

Steps for Vector Addition

Suppose several forces $\vec{F}_1$ , $\vec{F}_2$ , and $\vec{F}_3$ act on an object. Each force is first split into horizontal and vertical components.

After all components are known, the resultant components are:

R_x=F_{1x}+F_{2x}+F_{3x}+\cdots

The resultant magnitude is:

R=\sqrt{R_x^2+R_y^2}

Use the two-argument arctangent for the direction of the resultant relative to the positive $x$ axis:

\theta=\operatorname{atan2}(R_y,R_x)

Unlike $\tan^{-1}(R_y/R_x)$ , $\operatorname{atan2}$ uses the signs of both components to place the direction in the correct quadrant.

Two Stage Displacement

A person walks in two stages. The first stage is equivalent to $80\text{ m}$ east and $60\text{ m}$ north. The second stage is equivalent to $120\text{ m}$ east and $90\text{ m}$ north.

Displacement vector	$x$ component	$y$ component
$\vec{s}_1$	$80\text{ m}$	$60\text{ m}$
$\vec{s}_2$	$120\text{ m}$	$90\text{ m}$
$\vec{R}$	$200\text{ m}$	$150\text{ m}$

The resultant components are:

R_x=80+120=200\text{ m}

The total displacement magnitude is:

\begin{aligned} R&=\sqrt{200^2+150^2} \\ &=\sqrt{40000+22500} \\ &=250\text{ m} \end{aligned}

The resultant direction is:

\tan\theta=\frac{150}{200}=0.75

$\theta\approx36.9^\circ$ above east. The total displacement is $250\text{ m}$ at $36.9^\circ$ north of east.

Subtraction with Components

Vector subtraction is also done component by component. If $\vec{C}=\vec{A}-\vec{B}$ , then:

C_x=A_x-B_x

For example, let $\vec{A}=7\hat{i}+4\hat{j}$ and $\vec{B}=2\hat{i}+6\hat{j}$ .

\begin{aligned} \vec{C}&=(7-2)\hat{i}+(4-6)\hat{j} \\ &=5\hat{i}-2\hat{j} \end{aligned}

The negative sign in $-2\hat{j}$ means the resulting $y$ component points downward.

So subtraction does not need a new rule. Keep the same axes, subtract matching components, then read the signs to recover the direction of the result.