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Vectors and 3D Coordinate Geometry

  1. Oct 1, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose that II c R3 is a plane, and that P is a point not on II. Assume that Q is a point in II whose distance to P is minimal; in other words, the distance from P to Q is less than or equal to the distance from P to any other point in II. Show that the vector PQ is orthogonal to II.
    Hint given: Define a differentiable vector function r(t) with r(t) a subset of II, and r(0) = Q. Let p be the vector with components given by the coordinates of P. Let
    f(t) = |r(t)-p|2 = (r(t)-p) . (r(t)-p)
    What can u say about
    df(t)/dt |t=0 ?


    3. The attempt at a solution
    Not quite sure how to do it
    I let r(t) = ro + vt, where v is any vector in II and r(0) = Q
    So r(t) = Q + vt
    Then,
    f(t) = (Q + vt -p) . (Q + vt -p)
    df(t)/dt = (Q + vt - p) . (v) + (Q + vt - p) . (v) (wasn't quite sure how to differentiate dot products but i used the product rule)

    So, df(t)/dt |t=0 = 2(Q-p) . (v)

    I don't know where to go from here. I guess I must show that equation is zero to prove that it's ortogonal but I don't know how to....
    Is this method even right? Any help would be much appreciated :)
     
    Last edited: Oct 1, 2008
  2. jcsd
  3. Oct 1, 2008 #2

    Dick

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    You should make it clear the Q=r(0) is the closest point in the plane to P. That makes r(t).r(t) a minimum at t=0. What's the derivative of a function at a minimum?
     
  4. Oct 1, 2008 #3

    tiny-tim

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    Hi kehler! :smile:

    I think you're missing the point …

    r(t) is a vector in II.

    In other words, r defines a perfectly general curve in II, with a parameter t …

    r(t) is the vector (from the origin) to the point on the curve with parameter t.

    Then you get a relationship between r'(t) and r(t). :smile:

    (and you're right, you can use the product rule on a dot-product!)
     
  5. Oct 1, 2008 #4
    I still don't really get it, tiny-tim :S. So did I do it correctly? Does this mean r(t) is not ro + vt because this is the equation of a line not a curve?
    r'(t) is the slope of r(t) isn't it?

    Thd derivative should be zero but I don't really know how to get it to be zero :S


    Thanks guys for your help :). Sorry I'm a bit slow to get stuff :(
     
  6. Oct 1, 2008 #5

    tiny-tim

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    Yes. It's a line, and therefore it's not in II.

    r(t) is in II. :smile:
    Hold it!

    r(t) is a vector.

    So r'(t) is a vector also (not a number, such as a slope).

    And in fact r'(t) is the vector which is … ? :smile:
     
  7. Oct 1, 2008 #6
    How then do I get an equation of a curve, tiny-tim? :S
    r(t) = At2 + Bt + C?

    Would it be the vector tangential to the point on the curve then?
     
  8. Oct 1, 2008 #7

    tiny-tim

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    Hi kehler! :smile:
    No, r(t) is the equation of the curve!!

    r is an "unknown curve", just as x can be an "unknown number". :wink:

    You don't need to know what r(t) is!!
    Yes. r'(t) is tangential to the (unknown, general) curve at that point.

    Now combine that with the differential of that dot-product. :smile:
     
  9. Oct 1, 2008 #8
    Hmm ok. So df(t)/dt = (r(t) - p) . (r'(t)) + (r(t) - p) . (r'(t)) = 2 (r(t) - p) . (r'(t))
    So, df(t)/dt |t=0 = 2(Q-p) .(r'(0))
    I guess this should be zero..... But I don't know how to show it without already knowing that they are ortogonal :S.
     
  10. Oct 1, 2008 #9

    tiny-tim

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    Are you forgetting how this started? :wink:

    When the distance is a minimum, df(t)/dt = … ? :smile:

    (going to sleep now … :zzz:)
     
  11. Oct 1, 2008 #10
    Hmm ok. I'll think about it some more and come back if I still don't get it. Thanks tiny-tim for your patience! Goodnight :D
     
  12. Oct 1, 2008 #11

    HallsofIvy

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    A very simple way to do this is purely geometric. Suppose PQ is NOT perpendicular to the plane. Draw the perpendicular from P to the plane and call the point at which the perependicular crosses the plane R. Then PQR is a right triangle with right angle at R. PR is a leg of that right triangle and PQ is the hypotenuse. By the Pythagorean theorem, [itex]|PQ|= \sqrt{|PR|^2+ |QR|^2}< |PR|[/itex] contradicting the fact that |PQ| is minimal. (| | here indicates the length of the line.)
     
  13. Oct 2, 2008 #12
    Oh wow, that method's a lot simpler :). Thanks HallsOfIvy.

    Tiny-tim, is my solution correct now:
    Let r(t) be a differentiable vector function r(t) with r(t) a subset of II, and r(0) = Q.
    Let p be the vector with components given by the coordinates of P.
    Let f(t) = |r(t)-p|2 = (r(t)-p) . (r(t)-p)
    So, f'(t) = (r(t)-p).r'(t) + (r(t)-p).r'(t) = 2(r(t)-p).r'(t)
    Since the distance from P to Q is less than or equal to the distance from P to any other point in II, f'(t) = 0 when r(t) = Q. This occurs when t=0 by definition
    So f'(0) = 0
    But f'(0) = 2(r(0)-p).r'(0) = 2(Q-p).r'(0)
    Therefore, 2(Q-p).r'(0) = 2(PQ).r'(0) = 0
    Since r'(0) lies in II, PQ is orthogonal to II
     
  14. Oct 2, 2008 #13

    tiny-tim

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    Hi kehler! :smile:

    Yes, that's fine, except for the last line.

    All you've proved is that PQ is perpendicular to the tangent of that particular curve, r(t).

    So you should add "but this applies for any curve through Q, and so PQ is perpendicular to the tangent of every curve in II through Q, ad so is perpendicular to II."

    btw, HallsofIvy's proof works fine in this case, where II is a plane, but the proof above will work for any surface, to prove that pQ is perpendicular to the tangent plane at Q. :wink:
     
  15. Oct 2, 2008 #14
    Thanks tiny-tim! :D
     
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