Attachment 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.

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- Aug 14th 2010, 04:14 AMkandyflossVectors problem
Attachment 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. - Aug 14th 2010, 04:32 AMearboth
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. - Aug 14th 2010, 04:33 AMProve It
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...