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Thread: c4 vector question

  1. #1
    Junior Member
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    Jun 2008
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    c4 vector question

    The point A with coordinates (0,-5,11) lies on the line l1, which has equation

    r=6i+19j-k+(lamda)
    (i+4-2k)

    The point p lies on l1 and is such that OP is perindicular to l1, where O is the origin.

    Find the position vectore of point P.

    Given that B had coordinates (5,15,1)

    Show that the points A, P and B are collinear and find ratio AP:PB

    Thanks

    Bryn
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  2. #2
    Super Member
    Joined
    Oct 2007
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    London / Cambridge
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    Have you found the values of $\displaystyle \lambda$ that corresponds to each point a , b and p ?

    Once you find them you can consider differences of values of $\displaystyle \lambda$ as distances between points on the line, that should help with finding the ratio.

    Finding the values of $\displaystyle \lambda$ for a should not be a problem

    for p, your given it lies on $\displaystyle l_{1}$ so the position of p is $\displaystyle \left(\begin{array}{c}6 + \lambda \\19 + 4 \lambda \\-1 - 2 \lambda\end{array}\right)$ and you're given that OP it is perpendicular to the the line $\displaystyle l_{1}$ so $\displaystyle \left(\begin{array}{c}6 + \lambda \\19 + 4 \lambda \\-1 - 2 \lambda\end{array}\right) \cdot \left(\begin{array}{c}6 \\19 \\-1\end{array}\right) = 0$, So now you can solve for $\displaystyle \lambda$

    Bobak
    Last edited by bobak; Jun 7th 2008 at 10:23 AM.
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