# Help with Vectors plz

• Aug 9th 2006, 06:36 PM
bigstarz
Help with Vectors plz
http://img158.imageshack.us/img158/3...titled1uc3.gif

having trouble with this one :confused:
• Aug 9th 2006, 07:33 PM
ThePerfectHacker
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 PM
earboth
Quote:

Originally Posted by bigstarz
[IMG]...having trouble with this one :confused:

Hello,

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