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Find a point on the line closest to another point

  1. Dec 10, 2014 #1
    1. The problem statement, all variables and given/known data
    (2 part problem) a) A plane passes through the point P(3,1,4) and is orthogona to the line (x-1)/2 = (2-y)/-7 = z-3
    b) Find the point on the line closest to point (3,1,4)

    2. Relevant equations
    Symmetric equation --> (x-x1)/t = (y-y1)/t = (z-z1)/t ( anyway i think that's what it is

    3. The attempt at a solution
    I found the equation of the plane using the direction vector as (2,-7,1)
    and used this form 2(x-3)-7(y-1)+(z-4)= 0

    I can't seem to be able to go on to start b. I would think of plugging P in but that seems way too easy.

    Any idea on how I can approach this?
     
  2. jcsd
  3. Dec 10, 2014 #2

    haruspex

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    Where will that closest point be in relation to the plane?
     
  4. Dec 10, 2014 #3

    ehild

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    Check the sign of the y component of the direction vector.

    The distance of P from the normal line is the length of the line drawn perpendicularly to the normal...Where is that line?
     
  5. Dec 10, 2014 #4
    The normal, but... it seems kinda redundant if it lies on the same line , anyway to me at least.
     
  6. Dec 11, 2014 #5

    ehild

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    Of course, it intersects the normal, but what is the position of the line drawn from P and perpendicular to the normal, with respect to the plane? Try to draw a picture.

    If you can not see it, write the distance of any point of the normal line from point P. When is it minimum, and what is that minimum distance?
     
  7. Dec 11, 2014 #6

    Mark44

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    Drawing a picture, as ehild suggested, is an excellent idea. Having an image to look at gives you insights that formulas and equations simply can't provide. This took me a couple of minutes to draw.
    Plane_and_Line.png
     
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