# Thread: Help with Vectors plz

1. ## Help with Vectors plz

having trouble with this one

2. 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.
---
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,
$\displaystyle \left\{ \begin{array}{c} l\sin \theta =y\\l\cos \theta = x \end{array}\right\}$
---
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,
$\displaystyle \sqrt{(32.97)^2+(30.16)^2}\approx 44.68\mbox{ km}$

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