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Math Help - Point and Line - parameter (t)

  1. #1
    Member mybrohshi5's Avatar
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    Point and Line - parameter (t)

    For the point P = (1,2,3) and the line x = 8-2t, y = -5+t, z = -4 + 3t

    Find the distance between P and an arbitrary point on the line, in terms of the parameter t.

    i chose t = 0 so the arbitrary point Q = (8,-5,-4)

    then,

    PQ = <7,-7,-7>

    so the distance would be

     d = \sqrt(7^2 + 7^2 + 7^2) = \sqrt(147)

    Does this look correct?

    Then the next part of this question states:

    Find the value of t that minimizes the distance function above

    i know that to find a minimization you take the derivative of the function but im not sure what to take the derivative of in this case.... or if thats what i even need to do.

    Thanks for any help
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    To apply projection of linear algebra, it's more easy than derivative
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    Quote Originally Posted by mybrohshi5 View Post
    For the point P = (1,2,3) and the line x = 8-2t, y = -5+t, z = -4 + 3t

    Find the distance between P and an arbitrary point on the line, in terms of the parameter t....
    You have to have a distance as a function in variable t. You need to find the distance between points
    P(1,2,3),\quad and\quad T(8-2t,-5+t,-4+3t)
    d(P,T)=\sqrt{(8-2t-1)^2+(5+t-2)^2+(-4+3t-3)^2}=\sqrt{14 t^2-64t+107}

    There it is. Now you have to find the value t that minimizes that function. You know how to do that, so chaaarge!
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    Quote Originally Posted by math2009 View Post
    To apply projection of linear algebra, it's more easy than derivative
    Yes. At least its more appropriate.
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    ref attachment
    Attached Thumbnails Attached Thumbnails Point and Line - parameter (t)-math_q128.pdf  
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  6. #6
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    Quote Originally Posted by mybrohshi5 View Post
    For the point P = (1,2,3) and the line x = 8-2t, y = -5+t, z = -4 + 3t

    Find the distance between P and an arbitrary point on the line, in terms of the parameter t.

    i chose t = 0 so the arbitrary point Q = (8,-5,-4)
    No, it is not! An "arbitrary point" means just that. Once you choose a point, it is not arbitrary! Did you see where the problem says "in terms of the parameter t"? What they mean is to find the distance between (1, 2, 3) and the "arbitrary point" (8- 2t, -5+ t, -4+ 3t).

    then,

    PQ = <7,-7,-7>

    so the distance would be

     d = \sqrt(7^2 + 7^2 + 7^2) = \sqrt(147)
    No, it isn't. The distance between two point (x, y, z) and (a, b, c) is given by \sqrt{(x- a)^2+ (y- b)^2+ (z- c)^2}. What you have is \sqrt{(7- 0)^2+ (-7- 0)^2+ (-7- 0)^2} the distance from your "arbitrary" point (7, -7, 7) and the origin. What happened to the point (1, 2, 3)?

    The distance between (1, 2, 3) and an arbitrary point on the line x = 8-2t, y = -5+t, z = -4 + 3t and (1, 2, 3) is given by \sqrt{(8- 2t- 1)^2, (-5+ t- 2)^2+ (-4+ 3t- 3)^2}.
    What is that equal to?


    Does this look correct?
    You have completely misunderstood the question. See above.

    Then the next part of this question states:

    Find the value of t that minimizes the distance function above

    i know that to find a minimization you take the derivative of the function but im not sure what to take the derivative of in this case.... or if thats what i even need to do.

    Thanks for any help
    Yes, you do- but having chosen a specific point rather than taking an "arbitrary point" "in terms of t", you don't have function of t!

    Take the derivative of \sqrt{(8- 2t- 1)^2, (-5+ t- 2)^2+ (-4+ 3t- 3)^2} with respect to t.

    (Hint: the derivative of \sqrt{f(x)}= (f(x))^{1/2} is (1/2)(f(x))^{-1/2} f'(x) and is 0 if and only if f'(x) is 0.)
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  7. #7
    Member mybrohshi5's Avatar
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    I was completely looking at the question in the wrong way i got it all figured out now

    Thanks everyone!
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  8. #8
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    Quote Originally Posted by mybrohshi5 View Post
    I was completely looking at the question in the wrong way i got it all figured out now

    Thanks everyone!
    In a complete arbitrary way!!
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