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Find the vector, parametric, symmetric equations of a line

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the vector, parametric and symmetric equations of a line that intersect both line 1 and line 2 at 90°.


    L1:
    x = 4 + 2t
    y = 8 + 3t
    z = -1 − 4t

    L2:
    x = 7 - 6t
    y = 2+ t
    z = -1 + 2t

    2. Relevant equations

    vector, parametric, symmetric equations of line in R3 and cross product equation.

    3. The attempt at a solution

    I obtained the direction vector for the line (L3) that intersects L1 and L2. It is [1,2,2].


    And I let the point of intersection between L3 and L1 be:
    [x1,y1,z1]=[4,8-1]+t[2,3,-4], tεℝ

    And let the point of intersection between L3 and L2 be:
    [x2,y2,z2] = [7,2,-1]+s[-6,1,2], sεℝ

    And now to find the scalar multiple of direction vector of the L3 that intersects point 1 and point 2.
    So, [x1,y1,z1]+n[1,2,2]=[x2, y2, z2], nεℝ

    And when I solve I obtain s=-1, n=-1,..
    But then I saw a different approach in which the individual used,

    [x1,y1,z1]-[x2, y2, z2]=n[1,2,2], nεℝ

    And this yields, s=1, n=1, t=-1.

    So my question is which method is accurate/correct, or does it not really matter a great deal?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 9, 2013 #2

    SammyS

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    Take your equation,  [x1,y1,z1] + n[1,2,2] = [x2, y2, z2]  and rearrange it to

    [STRIKE][x1,y1,z1] - n[x2, y2, z2] = -n[1,2,2] .[/STRIKE]

    [x1,y1,z1] - [x2, y2, z2] = -n[1,2,2] .

    This is equivalent to the other individual's result !

    Edited post to fix typo.
     
    Last edited: Mar 9, 2013
  4. Mar 9, 2013 #3
    But there not equivalent. If you rearrange, [x1,y1,z1]+n[1,2,2]=[x2, y2, z2]
    We obtain:[x1,y1,z1]-[x2, y2, z2]=-n[1,2,2] or [x2, y2, z2]-[x1,y1,z1]=n[1,2,2], which are not equivalent to [x1,y1,z1]-[x2, y2, z2]=n[1,2,2]---subtraction is not commutative.
    And plus, if they were equivalent how would the s,t and n values obtained vary??
    So once again does it not matter which method you use, or is there something I'm missing?
     
  5. Mar 9, 2013 #4

    SammyS

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    Your n is equal to -n of your 'friend' , but that just makes the equations exactly equivalent, if you have the same t as he/she.
     
  6. Mar 9, 2013 #5
    Okay I see what your saying. So then it really doesn't matter which equation I use.
    Thanks.
     
  7. Mar 9, 2013 #6
    One other question, when you substitute the s=-1 value into the equation
    [x2,y2,z2] = [7,2,-1]+s[-6,1,2] the answer varies when you input the other s value.
    But would it still be right??
     
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