# Thread: Vectors problem

1. ## Vectors problem

Any 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.

2. Originally Posted by kandyfloss
Any 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:

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

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

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

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

Factor out $\frac{c_1}{m_1}$ and set $\lambda = \mu \cdot \frac{c_1}{m_1}$ and you'll get the given result.

3. To start with, to show that

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

By expanding and simplifying we find that

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

and so

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

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

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

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

$-by = c + ax$

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