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Thread: Vectors problem

  1. #1
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    Vectors problem

    Vectors problem-img00707-20100813-1747.jpgAny help is appreciated.
    To begin with i did when x=0
    y=c1
    so i got the (0 c1) part
    I don't know how to proceed.
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  2. #2
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    Quote Originally Posted by kandyfloss View Post
    Click image for larger version. 

Name:	IMG00707-20100813-1747.jpg 
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ID:	18563Any help is appreciated.
    To begin with i did when x=0
    y=c1
    so i got the (0 c1) part
    I don't know how to proceed.
    1. Determine the coordinates of the x- and y-intercept of L1:

    $\displaystyle A(0, c_1)$ and $\displaystyle B\left(-\frac{c_1}{m_1}~,~0\right)$

    2. The line L1 passes through A with the staionary vector $\displaystyle \vec a = (0,c_1)$ and B with the staionary vector $\displaystyle \vec b = \left(-\frac{c_1}{m_1}~,~0\right)$

    3. $\displaystyle L_1:\vec r = (0,c_1)+\mu \underbrace{\left(\left(0,c_1\right) - \left(-\frac{c_1}{m_1}~,~0\right) \right)}_{direction\ of\ L_1}$

    $\displaystyle L_1:\vec r = (0,c_1)+\mu \underbrace{\left( \frac{c_1}{m_1}~,~c_1 \right)}_{direction\ of\ L_1}$

    Factor out $\displaystyle \frac{c_1}{m_1}$ and set $\displaystyle \lambda = \mu \cdot \frac{c_1}{m_1}$ and you'll get the given result.
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  3. #3
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    To start with, to show that

    $\displaystyle \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}\phantom{-}0\\-\frac{c}{b}\end{matrix}\right) + \lambda \left(\begin{matrix}\phantom{-}b\\-a\end{matrix}\right)$ is another way of writing $\displaystyle ax + by + c = 0$, you will need to evaluate $\displaystyle \lambda$...


    By expanding and simplifying we find that

    $\displaystyle \left(\begin{matrix}x \\ y\end{matrix}\right) = \left(\begin{matrix}\phantom{-}\lambda b\\-\frac{c+\lambda ab}{b}\end{matrix}\right)$

    and so

    $\displaystyle x = \lambda b$ and $\displaystyle y = -\frac{c + \lambda ab}{b}$.


    From the first equation we can see $\displaystyle \lambda = \frac{x}{b}$ and substituting into the second equation gives

    $\displaystyle y = -\frac{c + \left(\frac{x}{b}\right)ab}{b}$

    $\displaystyle y = -\frac{c + ax}{b}$

    $\displaystyle -by = c + ax$

    $\displaystyle 0 = ax + by + c$ as required...
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