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Math Help - Proof of w = u + {(p-u).T}T = p - {(p-u).N}N and couple of questions?

  1. #1
    Senior Member x3bnm's Avatar
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    Proof of w = u + {(p-u).T}T = p - {(p-u).N}N and couple of questions?

    I've some questions about the proof of w = u + \{(p-u) \cdot T\}T = p - \{(p-u) \cdot N\}N

    As you can see there are two parts to the proof.



    Let L be a line and let p be a point. There is a unique point w on L for which (w - p) is orthogonal to L. Moreover, w is given by
    the following formulas:

    w = u + \{(p - u) \cdot T\}T = p - \{(p - u) \cdot N\}N ,

    where u is any point on L, T is a unit vector in the direction of L, and N is a unit vector orthogonal to T.


    Proof:

    L can be parameterized as z(t) = u + tT(Vector equation of line). We seek a point z(t) for which z(t) - p is orthogonal to the direction T of the line, i.e. for which
    0 = (p - z(t)) \cdot T = (p - u - tT) \cdot T This relation has a unique solution t = (p - u) \cdot T Substituting for t in z(t) gives the first
    formula(first proof) for w in the statement of the result.

    My question is:

    How did the author find the solution t = (p - u) \cdot T from the equation? What method he used for this?
    How did he conclude that w = u + \{(p-u) \cdot T\}T?


    And back to the proof.

    Substituting v = p - u into:

    v = (v \cdot T) T + (v \cdot N) N

    gives the second part of the proof.

    How to prove w = p - \{(p-u) \cdot N\}N by substituting v = p - u into v = (v \cdot T) T + (v \cdot N) N?
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  2. #2
    Senior Member x3bnm's Avatar
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    Re: Proof of w = u + {(p-u).T}T = p - {(p-u).N}N and couple of questions?

    Don't worry I've found the answers.

    For the first question i had: we know that T is a unit vector. So T \cdot T = 1

    \begin{align*}(p - z(t)) \cdot T =& 0 \\ (p - u - tT) \cdot T =& 0 \\ (p - u) \cdot T - t (T \cdot T) =& 0 \\ t.1 =& (p - u) \cdot T \\ t =& (p - u) \cdot T \end{align*}



    If I plugin t = (p - u) \cdot T into z(t) = u + tT I get:

    z(t) = u + \{(p - u) \cdot T\}T



    Now for the second part of the proof if I plugin v = p - u into  v = (v \cdot T) T + (v \cdot N) N I get:

    \begin{align*} p - u = & \{(p - u) \cdot T\} T + \{(p - u) \cdot N\} N  \\ p - u = & t T + \{(p - u) \cdot N\} N \text{       because I know that } t = & (p - u) \cdot T \\ u + tT = & p - \{(p - u) \cdot N\} N \text{      because } z(t) = u + tT \\ z(t) = & p - \{(p - u) \cdot N\} N \end{align*}

    And that's it. Q.E.D.
    Last edited by x3bnm; August 31st 2012 at 11:33 AM.
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    Senior Member x3bnm's Avatar
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    Re: Proof of w = u + {(p-u).T}T = p - {(p-u).N}N and couple of questions?

    I understand the proof. But what makes one think that z(t) is actually the projection of p which is w and not just any other point on the line L?

    Because z(t) = u + tT why is this not a projection of point p? Why do only z(t) that satisfies this w = z(t) = u + \{(p - u) \cdot T\}T = p - \{(p - u) \cdot N\}N have to be the projection of p?
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    Senior Member x3bnm's Avatar
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    Re: Proof of w = u + {(p-u).T}T = p - {(p-u).N}N and couple of questions?

    Quote Originally Posted by x3bnm View Post
    I understand the proof. But what makes one think that z(t) is actually the projection of p which is w and not just any other point on the line L?

    Because z(t) = u + tT why is this not a projection of point p? Why do only z(t) that satisfies this w = z(t) = u + \{(p - u) \cdot T\}T = p - \{(p - u) \cdot N\}N have to be the projection of p?
    I also found the answer.

    Because: [z(t) - p] \cdot T = 0 only for w =  z(t) = u + \{(p - u) \cdot T\}T = p - \{(p - u) \cdot N \} N
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