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HELP Intersection of two lines (VECTORS)

  1. Aug 8, 2011 #1
    HELP!!! Intersection of two lines (VECTORS)

    1. The problem statement, all variables and given/known data

    Find the common point of the lines r=i+j+k+x(j-3k) and r=i+y(k-j)

    2. Relevant equations



    3. The attempt at a solution

    If the lines intersect then there are numbers x and y such that

    i+j+k+x(j-3k)=i+y(k-j)

    The two lines above does intersect at a point. But my question is how did they find that point!!!!!!!!. Im guessing you must need to find x and y to find the point BUT HOW!!!!

    kinds regards

    I will send the right answer after somebody attempts the above question which i need help on
     
  2. jcsd
  3. Aug 8, 2011 #2
    Re: HELP!!! Intersection of two lines (VECTORS)

    I think it could be helpful to see a way to solve if you put the lines in vectorial notation:
    [itex] r: \left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\1\\1\end{array}\right]+x\cdot\left[\begin{array}{c}0\\1\\-3\end{array}\right][/itex]
    [itex]s: \left[\begin{array}{c}x\\y\\z\end{array}\right]=\left[\begin{array}{c}1\\0\\0\end{array}\right]+y\cdot\left[\begin{array}{c}0\\-1\\1\end{array}\right][/itex]

    Now, it should be easy to find the intersection...

    (Just check if I wrote correctly the vectors)
     
    Last edited: Aug 8, 2011
  4. Aug 8, 2011 #3
    Re: HELP!!! Intersection of two lines (VECTORS)

    I can see what you mean. I wrote the vectorial notation below. Since we are writing it in this way we will have to change the x and y before parallel vector into other letter e.g. a and b. So i wrote the Vectorial notation as it should be below

    r:⎡⎣xyz⎤⎦=⎡⎣111⎤⎦+a⋅⎡⎣01−3⎤⎦
    r:⎡⎣xyz⎤⎦=⎡⎣100⎤⎦+b⋅⎡⎣0−11⎤⎦
    This does help

    From here i would write

    x=1+a(0) (from this we get x=1)
    x=1+b(0)

    y=1+a(1)=1+a
    y=0+b(-1)=-b

    z=1+a(-3)=1-3a
    z=0+b(1)=b

    We need to find a or b to help find y and z. I have chosen to find a. I did simultaneous equations for both y to get 1-3a=-1-a. Which rearranges to give 1+1=3a-a which gives a=1

    I put a in above equation to get y. e.g. y=1+1(1)=2

    and put a in above equation to get z e.g. z=1-3(1)=-2

    Which give me the coordinates (1,2,-2)

    which is the correct answer

    Thanks DiracRules
     
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