## theanonymous27 2 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. wio

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

2. theanonymous27

How is that?

3. wio

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. wio

Our vector only has to have the right direction.

5. wio

@theanonymous27 Does this make sense?

6. theanonymous27

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. wio

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. wio

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

9. wio

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

10. theanonymous27

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

11. wio

Yes

12. theanonymous27

so does the slope represent the component of the vector?

13. theanonymous27

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

14. wio

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. wio

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

16. theanonymous27

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

17. wio

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

18. wio

It's the same direction.

19. theanonymous27

Thanks!