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Find point d on the line l closest to point c

  1. Nov 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Find point d on the line l closest to the point c (1,1,7). Point c is on the end of a vector who's origin (1,1,2) is on line l. There is an imaginary line that connects point c to point d. This imaginary line is perpendicular to the line l. This problem is relating to vectors and just before this problem I found the parametric equation of the line l so I guess Im going to need that to work this problem.


    2. Relevant equations
    parametric equation for l x=2+t u=3+2t z=5+3t
    the angle between the vector with point c and line l is 36.7 degrees
    c=(1,1,7)




    3. The attempt at a solution I think there is a specific equation for this but I am not sure. I thought about it for awhile but am still confused.
     
  2. jcsd
  3. Nov 11, 2011 #2

    ehild

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    It would be nice to see a drawing.

    Do you know how to get the projection of a vector onto the direction of an other vector and how it is related to the dot product of those vectors?

    ehild
     
  4. Nov 11, 2011 #3
    I just posted a better picture
     

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  5. Nov 12, 2011 #4

    ehild

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    Thanks for the picture. You see the right triangle. The vectors enclose the angle of 36.7 °.

    The closest distance between a fixed point and any point of a line is the length of the perpendicular light segment drawn from point C to the line. What is this length, knowing the hypotenuse of the right triangle?

    ehild
     
  6. Nov 12, 2011 #5
    Yes I know right triangles in two dimensions however this is in three dimensions. Thats what is confusing this for me.
     
  7. Nov 12, 2011 #6
    I know that the magnitude of the vector with the point c is 5. So using cos(36.7)=x/5
    I get x=4 this is the length along line l. However this is in three dimensions so I thought that there was something else I had to do.
     
  8. Nov 12, 2011 #7

    ehild

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    You need the length of the side opposite to the angle. It is the distance of point C from the line.

    The problem is in three dimensions, but the line and the vector define a plane.

    ehild
     
  9. Nov 12, 2011 #8
    oh sorry, the length between the vector and line l is 3.
     
  10. Nov 12, 2011 #9

    ehild

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    So the distance of point C from the line is 3. I just re-read the problem, and it asks the position of the point of the line which is closest to C. You gave the distance to the closest point on the line from point (1,1,2), it was 4. Now you need to find the coordinates of the closest point.
    Hint: how far do you reach along the line if you change the parameter t by 1? What is the change of t to the closest point?

    ehild
     
    Last edited: Nov 12, 2011
  11. Nov 12, 2011 #10
    idk if you change it by one I guess you reach along the line one. I just don't know this stuff because we haven't covered it yet in class yet we have homework over it.
     
  12. Nov 12, 2011 #11

    ehild

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    The direction vector is [itex]\vec{t}[/itex]=(1,2,3) (from the parametric equation). Its length is √14=3.74. If you move by one [itex]\vec{t}[/itex], your displacement is 3.74. The distance (OP) along the line is 4. How much does the parameter t change along this distance? You need to find the coordinates of P. Both O and P fulfil the parametric equation of the line. What is t at point O? What is t at point P?

    ehild
     

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  13. Nov 12, 2011 #12

    HallsofIvy

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    Looks to me like you need the cosine of that angle.
     
  14. Nov 12, 2011 #13
    The point I picked on the line l for the parametric equation was not point (1,1,2)... So whatever point you pick on the line for your parametric equation is that your t=0 point?
    If I were to pick point (1,1,2) on the line l for the parametric equation then it becomes x=1+t u=1+2t z=2+3t then t would =1.0695 to move 4. so x=2.0695 u=3.139 z=5.209
    edit: I chose the point (1,1,2) because that is the coordinates for point O.
     
  15. Nov 12, 2011 #14

    ehild

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    Superb!!! Correct!!!:smile:

    ehild
     
  16. Nov 12, 2011 #15
    Thank you for your assistance:biggrin:
     
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