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Math Help - Vector: Simple Line Question

  1. #1
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    Vector: Simple Line Question

    I partially understand the method but not fully so can someone please briefly go through the method. Help would be appreciated.

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    Q: The positions vectors with respect to the origin O are:

    B: -4 \bold i + 4 \bold j - \bold k,
    C: 5 \bold i - 2 \bold j + 11 \bold k.

    Find a vector equation for the line BC.

    _______________________________________________


    The Correct Answer:

    \bold r = \begin{pmatrix} -4 \\ 4 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix} .
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  2. #2
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    That happens to be just B + \frac{\mu }{3}\left( {C - B} \right).
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    Quote Originally Posted by Plato View Post
    That happens to be just B + \frac{\mu }{3}\left( {C - B} \right).
    I can understand that the first column vector is the position vector but the second one near \mu, is it the postion vecots made into a vector or both position vectors dotted or crossed.

    Fro example is it:

    \vec {BC} or B . C or B \bold x C or something else?

    EDIT: Are you allowed to just remove a common denominator from the second bracket near \mu and leave the equation without it?
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  4. #4
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    I am not sure that I understand your question. But
    \left\{ \begin{array}{l}<br />
 B =  - 4i + 4j - k \\ <br />
 C = 5i - 2j + 11k \\ <br />
 \end{array} \right.\quad  \Rightarrow \quad C - B = 9i - 6j + 12k = 3\left( {3i - 2j + 4k} \right)<br />
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    Quote Originally Posted by Plato View Post
    I am not sure that I understand your question. But
    \left\{ \begin{array}{l}<br />
 B =  - 4i + 4j - k \\ <br />
 C = 5i - 2j + 11k \\ <br />
 \end{array} \right.\quad  \Rightarrow \quad C - B = 9i - 6j + 12k = 3\left( {3i - 2j + 4k} \right)<br />
    I was wondering where the 3 that was removed from the bracket went as the answer was <br />
\bold r = \begin{pmatrix} -4 \\ 4 \\ -1 \end{pmatrix} + \mu \begin{pmatrix} 3 \\ -2 \\ 4 \end{pmatrix}<br />
.

    I hope my question is clear.
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  6. #6
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    If D is the direction vector of a line the any scalar multiple of D is also a direction vector.
    Given any to points B & C the line determined by them is B+t(C-B).
    Because we can factor out a 3 in C-B is just makes the direction vector simplier.
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