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Yet 'nother vector problem!

  1. Feb 26, 2008 #1
    1. The problem statement, all variables and given/known data
    following equations represent lines P nad Q in 3 dimensional space:
    P: 2-x=y+1=(z-3)/2
    Q: 5-x=y-2=(z+1)/2
    (a) Show that P & Q are two different lines which are parallel
    (b) Find the distance between the two lines, measured along a line perpendicular to P & Q.


    2. Relevant equations



    3. The attempt at a solution
    (a) I attempted to turn the above equations into vector equations. I came out with the following:
    v(P) = (2,-1,3) + t(-1,1,2)
    v(Q) = (5,2,-1) + s(-1,1,2)
    I concluded that since the the direction vectors are the same, therefore the lines are parallel.
    correct?
    (b) Not sure what to do here. Do I use the distance formula for between two points?
    ie.
    [tex]\sqrt{(2-5)^{2}+(-1-2)^{2}+(3--1)^{2}}[/tex]
    I then get [tex]\sqrt{34}[/tex]
    correct? (I have the nagging feeling it ain't!)
     
    Last edited: Feb 27, 2008
  2. jcsd
  3. Feb 26, 2008 #2

    HallsofIvy

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    Yes, the vector equations and the conclusion that the lines are parallel because they have the same direction vector are correct.

    Okay, but what two points? Do you have any reason to believe that the line from (2, -1, 3), on the first line, to (5, 2, -1), on the second line, is perpendicular to the lines? One way to check is to calculate the vector from one of those points to the other:(5-2, 2-(-1), -1-3)= (3, 3, -4) and then find its dot product with the direction vector: (3, 3, -4)[itex]\cdot[/itex](-1, 1, 2)= -3+ 3- 8= -8. Since that is NOT 0, the two vectors are NOT perpendicular and the line between the two points is NOT perpendicular to the two lines.

    Try this instead: the vector (-1, 1, 2), the direction vector, is in the direction of the line and so perpendicular to any plane perpendicular to the line. That means that the plane perpendicular to the first line and containing the point (2, -1, 3) has equation -1(x- 2)+ 1(y-(-1))+ 2(z- 3)= 0 or -x+ y+ 2z= 3. Where does the second line x= 5- s, y= 2+ s, z= -1+ 2s cross that line? (Replace x, y and z in the equation of the plane with those formulas in terms of s and solve for s.) Now you can find the distance between (2, -1, 3) and that point.
     
  4. Feb 27, 2008 #3
    ah-ha! I knew I was missing a step there. It seemed too easy, just using the two points provided but I had a massive brainfreeze as to why this wasn't correct.

    So in the equation you supplied (TQ very much!):
    -x+ y+ 2z= 3,
    we sub
    x= 5- s
    y= 2+ s
    z= -1+ 2s
    and solve for s:

    -(5-s) + (2+s) + 2(-1+2s) = 3
    -5+2-2 + s + s + 4s = 3
    6s=6
    s=1

    point on Q which is perpendicular to P at (2,-1,3) is (5,2,-1) + 1(-1,1,2) = (4,3,1).
    Then use distance formula:
    [tex]\sqrt{(2-4)^{2}+(-1-3)^{2}+(3-1)^{2}}[/tex]
    =[tex]\sqrt{24}[/tex]

    yes/no?
     
    Last edited: Feb 27, 2008
  5. Feb 27, 2008 #4

    HallsofIvy

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    Staff Emeritus
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    Well, I was under the impression that -1- 2 was equal to -3, not 1!
     
  6. Feb 27, 2008 #5
    D'oh!

    okay.got it now. thanks for the help.
     
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