Here's the question you clicked on:
theanonymous27
Find a parametrization r(t) = h x(t), y(t), z(t)i of the straight line passing through the origin in 3-space whose projection on the xy-plane is a line with slope 4, while its projection on the yz-plane is a line with slope −3, i.e., ∆y/∆x= 4,∆z/∆y= −3.
Can you come up with a vector which goes in the correct direction?
First, it doesn't matter what the magnitude of the vector is... so we get to chose one of the components as long as the others are correctly proportional.
Our vector only has to have the right direction.
@theanonymous27 Does this make sense?
I understand, but how can I find that vector? what I dont get is how can I use the slopes here to find what I need
So we can start by letting our \(x\) component be 1:\[ \mathbf{v} = \begin{bmatrix} x_1 \\ y_1 \\ z_1 \end{bmatrix} = \begin{bmatrix} 1 \\ y_1 \\ z_1 \end{bmatrix} \]
Since \(\Delta y/\Delta x= 4\) and \(\Delta x = 1\), what must \(\Delta y\) be?
We use the slopes to figure out the proportions between each component, @theanonymous27
so <Y is 4, and then we get that Z has to be -12
so does the slope represent the component of the vector?
we end up with r(t) = <t, 4, -12t> ? is that right?
So given the direction vector \(\mathbf{v}\) and some point which the line goes though \(\mathbf{b}\) the parametrization for our line is: \[ \mathbf{r}(t) = \mathbf{v}t+\mathbf{b} \]
\[ r(t) = \begin{bmatrix} 1 \\ 4 \\ -12 \end{bmatrix} t + \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} t \\ 4t \\ -12t \end{bmatrix} \]
Thank you!, One last question... Why did you pick x=1 first? what would have happened if you choose y=1 instead
If you pick \(y=1\) then you get \(x=1/4\) and \(z=-3\)