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Tricky minimum distance vector problem

  1. Oct 29, 2011 #1
    1. The problem statement, all variables and given/known data

    The line intersects a plane at an angle alpha=2pi/3. The line is defined by rn, and the plane by r.m=0, with n and m unit vectors. Calculate the shortest distance from the plane to the point on the line with μ=2.

    2. Relevant equations

    ?

    3. The attempt at a solution

    I want to find the value of μ at the point of intersection. Then it will be easy to find the distance, as d=(2-μ at intersection)sin(2pi/3). However, I don't know how to find μ. Any hints?
     
  2. jcsd
  3. Oct 29, 2011 #2
    I'm still stuck. Any ideas would be much appreciated!
     
  4. Oct 29, 2011 #3

    I like Serena

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    Hi Lucy Yeats! :smile:

    To find μ at the point of intersection you need to substitute the equation of the line into the equation of the plane.
    That is (μn).m=0. From this it follows that μ=0.

    That's funny!!
    Yes, the origin is on the line and the origin is also in the plane.

    Btw, the regular way to calculate the distance of a point to a plane, is to project any vector from the plane to the point, onto the unit normal vector.
    You can do this with the dot product.
     
  5. Oct 29, 2011 #4
    If n and m are perpendicular, surely mu could be anything? How do you know that they aren't perpendicular?
     
  6. Oct 29, 2011 #5
    I'm guessing that you have to use the angle 2pi/3 in some way?

    Thanks for helping, btw!
     
  7. Oct 29, 2011 #6

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    Yes.
    What would happen (geometrically speaking) with the point of intersection if n and m are perpendicular?



    Yes.
    But first you need the formula for the distance of a point to a plane:
    distance = (any vector from plane to point) . (normal vector of plane)
     
    Last edited: Oct 29, 2011
  8. Oct 29, 2011 #7

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    Actually, your distance formula d=(2-μ at intersection)sin(2pi/3) also works.

    Just fill in the μ we just found for the intersection.



    Btw, did you know that n.m=sin(2pi/3)?
     
  9. Oct 29, 2011 #8
    Ah, I see, they can't be perpendicular because then the angle wouldn't be 2pi/3. so mu has to be zero.

    Brilliant, thank you very much!
     
  10. Oct 29, 2011 #9

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    You're welcome! :smile:
     
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