Results 1 to 3 of 3

Math Help - Help with Vectors plz

  1. #1
    Newbie
    Joined
    Jul 2006
    Posts
    7

    Help with Vectors plz



    having trouble with this one
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    The vector sum of all these vectors would be your answer.
    But you need to express every vector in the form, <a,b>---> The components of your vector.
    ---
    In order to solve this problem we need to develope a formula. Let \theta be the angle from the x-axis (meaning measure conterclockwise).
    Then if (x,y) are the endpoints of the vector. The length is l=\sqrt{x^2+y^2}. And \sin \theta = \frac{y}{\sqrt{x^2+y^2}}
    In order to solve this problem. You need to solve for both x,y
    More simply,
    \sqrt{x^2+y^2}=l
    \sqrt{x^2+y^2}\sin \theta =y
    Now, divide the second equation by the first to get, (it is nonzero)
    \sin \theta = \frac{y}{l}
    Thus,
    l\sin \theta = y
    Similary,
    l\cos \theta = x
    As a result we have converstion formula:
    If x,y the the endpoints of a vector having length l and angle meausured from the x-axis \theta then,
    \left\{ \begin{array}{c} l\sin \theta =y\\l\cos \theta = x \end{array}\right\}
    ---
    Now construct a chart in terms of \theta, i.e. counterclockwise angle.
    \left\{ \begin{array}{cc}\mbox{Leg #} & \theta\\<br />
1&135^o\\2&345^o\\3&215^o\\4&90^o\\5&0^o\\6&264^  o
    Therefore you new table is,
    \left\{ \begin{array}{ccc}\mbox{Leg #}&\mbox{length}&\theta\\<br />
1&22.5&135^o\\<br />
2&35.4&345^o\\<br />
3&18.6&215^o\\<br />
4&15.5&90^o\\<br />
5&31.8&0^o\\<br />
6&44.8&264^o
    Now use the converstions formula posted above to find the coordinates.
    So you have,
    \left\{ \begin{array}{cc}x&y\\<br />
15.91&-15.91\\<br />
-9.16&34.19\\<br />
-10.67&-15.24\\<br />
15.5&0\\<br />
0&31.8\\<br />
-44.55&-4.68
    Now by the rule of vectors you can add their components thus,
    \left\{ \begin{array}{cc}\sum x&\sum y\\<br />
-32.97&30.16<br />

    Now compute the length,
    \sqrt{(32.97)^2+(30.16)^2}\approx 44.68\mbox{ km}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,804
    Thanks
    115
    Quote 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]:
    \sin(\theta)=\frac{-32.97}{30.16}. Thus \theta\approx 113.8^o

    Greetings

    EB
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: November 15th 2011, 05:10 PM
  2. Replies: 3
    Last Post: June 30th 2011, 08:05 PM
  3. Replies: 2
    Last Post: June 18th 2011, 10:31 AM
  4. [SOLVED] Vectors: Finding coefficients to scalars with given vectors.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: January 23rd 2011, 12:47 AM
  5. Replies: 4
    Last Post: May 10th 2009, 06:03 PM

Search Tags


/mathhelpforum @mathhelpforum