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Determine a direction vector when a line perpendicular and a point is given

  1. Mar 1, 2012 #1
    Find the direction vector of the line perpendicular to the line r=[-1,0,1] + t [2,3,-2] and passing through the point (-7,-9,7)
     
  2. jcsd
  3. Mar 1, 2012 #2

    tiny-tim

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    welcome to pf!

    hi euro94! welcome to pf! :wink:

    show us what you've tried, and where you're stuck, and then we'll know how to help! :smile:
     
  4. Mar 1, 2012 #3
    I tried finding the dot product of the line's direction vector with a vector from (-7,9,7). I tried (-1+2t, 3t, 1-2t) - (-7,-9,7) and I dotted it with (2,-3,2) and made it equal to zero, and then i got stuck :(
     
  5. Mar 1, 2012 #4

    tiny-tim

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    (you mean (2,3,-2)?)

    that should work …

    what did you get?​
     
  6. Mar 1, 2012 #5
    I got (6+2t,3t+9,6-2t) and i dotted it with (2,3,-2) and then i made it equal to zero and i got 12+4t+9t+27-12+4t and i solved for t and i got -27/17, but i'm not sure what to do next..
     
  7. Mar 1, 2012 #6
    i mean i got a value of -27/14 for t, i had a miscalculation, but im not sure what step to take next
     
  8. Mar 2, 2012 #7

    tiny-tim

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    hi euro94! :smile:

    (just got up :zzz:)

    that value of the parameter (t) gives you the foot of the perpendicular …

    so put it into [-1,0,1] + t [2,3,-2] to give you the actual coordinates,

    and then join that to [-7,-9,7] :smile:
     
  9. Mar 2, 2012 #8
    oppsss, i got t=-3 and i plugged it into the equation and i got that my direction vector is [-7,-9,7] and so my overall equation is r=[-7,-9,7] + t[-7,-9,7]
     
  10. Mar 2, 2012 #9
    is that normal?
     
  11. Mar 2, 2012 #10

    tiny-tim

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    you mean it goes through the origin? :smile:

    can you please write it all out in one go, if you want it checked? :wink:
     
  12. Mar 2, 2012 #11
    okay :)
    (-1+2t, 3t, 1-2t) - (-7,-9,7) = (6+2t, 3t+9, -6-2t)
    dot product:
    (6+2t, 3t+9, -6-2t) dot (2,3,-2) =0
    12+4t+9t+27+12+4t =0
    51+17t=0
    -51/17=t
    -3=t
    r=[-1,0,1] + 3 [2,3,-2]
    = [-7,-9,7]
     
  13. Mar 2, 2012 #12
    is that right? :)
     
  14. Mar 2, 2012 #13

    tiny-tim

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    hmm … that's weird! :rolleyes:

    yes, your caluculations are fine :smile:

    looking back at the original question now, obviously [-7,-9,7] does actually lie on the original line …

    so (since we're in three dimensions), asking for "the line perpendicular" makes no sense, since there's an infinite number of them! :confused:
     
  15. Mar 2, 2012 #14
    thank youu :)
     
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