anonymous 3 years ago 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.

1. anonymous

Can you come up with a vector which goes in the correct direction?

2. anonymous

How is that?

3. anonymous

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.

4. anonymous

Our vector only has to have the right direction.

5. anonymous

@theanonymous27 Does this make sense?

6. anonymous

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

7. anonymous

So we can start by letting our $$x$$ component be 1:$\mathbf{v} = \begin{bmatrix} x_1 \\ y_1 \\ z_1 \end{bmatrix} = \begin{bmatrix} 1 \\ y_1 \\ z_1 \end{bmatrix}$

8. anonymous

Since $$\Delta y/\Delta x= 4$$ and $$\Delta x = 1$$, what must $$\Delta y$$ be?

9. anonymous

We use the slopes to figure out the proportions between each component, @theanonymous27

10. anonymous

so <Y is 4, and then we get that Z has to be -12

11. anonymous

Yes

12. anonymous

so does the slope represent the component of the vector?

13. anonymous

we end up with r(t) = <t, 4, -12t> ? is that right?

14. anonymous

So given the direction vector $$\mathbf{v}$$ and some point which the line goes though $$\mathbf{b}$$ the parametrization for our line is: $\mathbf{r}(t) = \mathbf{v}t+\mathbf{b}$

15. anonymous

$r(t) = \begin{bmatrix} 1 \\ 4 \\ -12 \end{bmatrix} t + \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} t \\ 4t \\ -12t \end{bmatrix}$

16. anonymous

Thank you!, One last question... Why did you pick x=1 first? what would have happened if you choose y=1 instead

17. anonymous

If you pick $$y=1$$ then you get $$x=1/4$$ and $$z=-3$$

18. anonymous

It's the same direction.

19. anonymous

Thanks!