# 2.3 Vector addition and subtraction: analytical methods  (Page 2/6)

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Then $A=10.3$ blocks and $\theta =29.1º$ , so that

${A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta =\left(\text{10.3 blocks}\right)\left(\text{cos}\phantom{\rule{0.25em}{0ex}}29.1º\right)=\text{9.0 blocks}$
${A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta =\left(\text{10.3 blocks}\right)\left(\text{sin}\phantom{\rule{0.25em}{0ex}}29.1º\right)=\text{5.0 blocks}\text{.}$

## Calculating a resultant vector

If the perpendicular components ${\mathbf{A}}_{x}$ and ${\mathbf{A}}_{y}$ of a vector $\mathbf{A}$ are known, then $\mathbf{A}$ can also be found analytically. To find the magnitude $A$ and direction $\theta$ of a vector from its perpendicular components ${\mathbf{A}}_{x}$ and ${\mathbf{A}}_{y}$ , we use the following relationships:

$A=\sqrt{{A}_{{x}^{2}}+{A}_{{y}^{2}}}$
$\theta ={\text{tan}}^{-1}\left({A}_{y}/{A}_{x}\right)\text{.}$

Note that the equation $A=\sqrt{{A}_{x}^{2}+{A}_{y}^{2}}$ is just the Pythagorean theorem relating the legs of a right triangle to the length of the hypotenuse. For example, if ${A}_{x}$ and ${A}_{y}$ are 9 and 5 blocks, respectively, then $A=\sqrt{{9}^{2}{\text{+5}}^{2}}\text{=10}\text{.}3$ blocks, again consistent with the example of the person walking in a city. Finally, the direction is $\theta ={\text{tan}}^{–1}\left(\text{5/9}\right)=29.1º$ , as before.

## Determining vectors and vector components with analytical methods

Equations ${A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ and ${A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ are used to find the perpendicular components of a vector—that is, to go from $A$ and $\theta$ to ${A}_{x}$ and ${A}_{y}$ . Equations $A=\sqrt{{A}_{x}^{2}+{A}_{y}^{2}}$ and $\theta ={\text{tan}}^{\text{–1}}\left({A}_{y}/{A}_{x}\right)$ are used to find a vector from its perpendicular components—that is, to go from ${A}_{x}$ and ${A}_{y}$ to $A$ and $\theta$ . Both processes are crucial to analytical methods of vector addition and subtraction.

## Adding vectors using analytical methods

To see how to add vectors using perpendicular components, consider [link] , in which the vectors $\mathbf{A}$ and $\mathbf{B}$ are added to produce the resultant $\mathbf{R}$ .

If $\mathbf{A}$ and $\mathbf{B}$ represent two legs of a walk (two displacements), then $\mathbf{R}$ is the total displacement. The person taking the walk ends up at the tip of $\mathbf{R}.$ There are many ways to arrive at the same point. In particular, the person could have walked first in the x -direction and then in the y -direction. Those paths are the x - and y -components of the resultant, ${\mathbf{R}}_{x}$ and ${\mathbf{R}}_{y}$ . If we know ${\mathbf{\text{R}}}_{x}$ and ${\mathbf{R}}_{y}$ , we can find $R$ and $\theta$ using the equations $A=\sqrt{{{A}_{x}}^{2}+{{A}_{y}}^{2}}$ and $\theta ={\text{tan}}^{–1}\left({A}_{y}/{A}_{x}\right)$ . When you use the analytical method of vector addition, you can determine the components or the magnitude and direction of a vector.

Step 1. Identify the x- and y-axes that will be used in the problem. Then, find the components of each vector to be added along the chosen perpendicular axes . Use the equations ${A}_{x}=A\phantom{\rule{0.25em}{0ex}}\text{cos}\phantom{\rule{0.25em}{0ex}}\theta$ and ${A}_{y}=A\phantom{\rule{0.25em}{0ex}}\text{sin}\phantom{\rule{0.25em}{0ex}}\theta$ to find the components. In [link] , these components are ${A}_{x}$ , ${A}_{y}$ , ${B}_{x}$ , and ${B}_{y}$ . The angles that vectors $\mathbf{A}$ and $\mathbf{B}$ make with the x -axis are ${\theta }_{\text{A}}$ and ${\theta }_{\text{B}}$ , respectively.

Step 2. Find the components of the resultant along each axis by adding the components of the individual vectors along that axis . That is, as shown in [link] ,

${R}_{x}={A}_{x}+{B}_{x}$

and

${R}_{y}={A}_{y}+{B}_{y}\text{.}$

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