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Thread: so confused, vector length exam tommorow!

  1. #1
    Member i_zz_y_ill's Avatar
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    so confused, vector length exam tommorow!

    given length BD is 15m and vector equation of line BD is

    (-1) (15) i.e +lamda(15,-20,0)
    (-7) + lamda (-20)
    (11) (0)
    find corordinates of D which is in mark scheme (8,-19,11)
    This question is so easy but im so annoyed i cant understand it. I wouldnt have thought 15 comes from pythagarus of each point and someow thats incorporated, Obviously not as the mark scheme tells me to get lamda as 3 or 3/5 as appropriate(HOW) then sub in! where does lamda come from....????
    Last edited by i_zz_y_ill; May 20th 2008 at 10:45 AM. Reason: vsorry 3 or 3/5f
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  2. #2
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    earboth's Avatar
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    Quote Originally Posted by i_zz_y_ill View Post
    given length BD is 15m and vector equation of line BD is

    (-1) (15) i.e +lamda(15,-20,0)
    (-7) + lamda (-20)
    (11) (0)
    find corordinates of D which is in mark scheme (8,-19,11)
    This question is so easy but im so annoyed i cant understand it. I wouldnt have thought 15 comes from pythagarus of each point and someow thats incorporated, Obviously not as the mark scheme tells me to get lamda as 3 or 3/5 as appropriate(HOW) then sub in! where does lamda come from....????
    The equation of a line is written as:

    $\displaystyle \underbrace{(x,y,z)}_{coordinates\ of \ points \ on \ the \ line}=\underbrace{(a,b,c)}_{startingpoint} + t \cdot \underbrace{(u,v,w)}_{direction\ vector}$. So you get:

    $\displaystyle \begin{array}{l}x=a+t \cdot u \\
    y=b+t \cdot v \\
    z=c+t \cdot w
    \end{array}$
    Instead of t some authors prefer $\displaystyle \lambda, ~ \mu$

    Obviously B(-1, -7, 11) and the direction vector is $\displaystyle \vec u=(15, -20,0)$

    So you can calculate the coordinates of D by:

    $\displaystyle D(-1+\lambda \cdot 15~,~ -7 +\lambda \cdot (-20)~~,~11+\lambda \cdot 0)$

    Now calculate $\displaystyle \overrightarrow{BD}=(15\lambda~,~-20\lambda~,~0)$

    The length of $\displaystyle |\overrightarrow{BD}|=\sqrt{625\lambda^2}=15$

    Solve for $\displaystyle \lambda$ and you'll get: $\displaystyle \lambda=\frac35~\vee~\lambda=-\frac35$

    Plug in this value into the equation of the line to calculate the coordinates of D:

    $\displaystyle D_1(8, -19,11)$ or $\displaystyle D_2(-10, 5, 11)$
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