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having trouble with this one :confused:

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- Aug 9th 2006, 06:36 PMbigstarzHelp with Vectors plz
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having trouble with this one :confused: - Aug 9th 2006, 07:33 PMThePerfectHacker
The vector sum of all these vectors would be your answer.

But you need to express every vector in the form, $\displaystyle <a,b>$---> The*components*of your vector.

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In order to solve this problem we need to develope a formula. Let $\displaystyle \theta$ be the angle from the x-axis (meaning measure conterclockwise).

Then if $\displaystyle (x,y)$ are the endpoints of the vector. The length is $\displaystyle l=\sqrt{x^2+y^2}$. And$\displaystyle \sin \theta = \frac{y}{\sqrt{x^2+y^2}}$

In order to solve this problem. You need to solve for both $\displaystyle x,y$

More simply,

$\displaystyle \sqrt{x^2+y^2}=l$

$\displaystyle \sqrt{x^2+y^2}\sin \theta =y$

Now, divide the second equation by the first to get, (it is nonzero)

$\displaystyle \sin \theta = \frac{y}{l}$

Thus,

$\displaystyle l\sin \theta = y$

Similary,

$\displaystyle l\cos \theta = x$

As a result we have converstion formula:

*If $\displaystyle x,y$ the the endpoints of a vector having length $\displaystyle l$ and angle meausured from the x-axis $\displaystyle \theta$ then,*

$\displaystyle \left\{ \begin{array}{c} l\sin \theta =y\\l\cos \theta = x \end{array}\right\}$

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Now construct a chart in terms of $\displaystyle \theta$, i.e. counterclockwise angle.

$\displaystyle \left\{ \begin{array}{cc}\mbox{Leg #} & \theta\\

1&135^o\\2&345^o\\3&215^o\\4&90^o\\5&0^o\\6&264^ o $

Therefore you new table is,

$\displaystyle \left\{ \begin{array}{ccc}\mbox{Leg #}&\mbox{length}&\theta\\

1&22.5&135^o\\

2&35.4&345^o\\

3&18.6&215^o\\

4&15.5&90^o\\

5&31.8&0^o\\

6&44.8&264^o $

Now use the converstions formula posted above to find the coordinates.

So you have,

$\displaystyle \left\{ \begin{array}{cc}x&y\\

15.91&-15.91\\

-9.16&34.19\\

-10.67&-15.24\\

15.5&0\\

0&31.8\\

-44.55&-4.68$

Now by the rule of vectors you can add their components thus,

$\displaystyle \left\{ \begin{array}{cc}\sum x&\sum y\\

-32.97&30.16

$

Now compute the length,

$\displaystyle \sqrt{(32.97)^2+(30.16)^2}\approx 44.68\mbox{ km}$ - Aug 9th 2006, 09:06 PMearbothQuote:

Originally Posted by**bigstarz**

in addition to TPHs result: Calculate the direction between starting point to ending point by the components of the resulting vector [-32.97 , 30.16]:

$\displaystyle \sin(\theta)=\frac{-32.97}{30.16}$. Thus $\displaystyle \theta\approx 113.8^o$

Greetings

EB