anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
Can you come up with a vector which goes in the correct direction?
anonymous
  • anonymous
How is that?
anonymous
  • 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.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
Our vector only has to have the right direction.
anonymous
  • anonymous
@theanonymous27 Does this make sense?
anonymous
  • 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
anonymous
  • anonymous
So we can start by letting our \(x\) component be 1:\[ \mathbf{v} = \begin{bmatrix} x_1 \\ y_1 \\ z_1 \end{bmatrix} = \begin{bmatrix} 1 \\ y_1 \\ z_1 \end{bmatrix} \]
anonymous
  • anonymous
Since \(\Delta y/\Delta x= 4\) and \(\Delta x = 1\), what must \(\Delta y\) be?
anonymous
  • anonymous
We use the slopes to figure out the proportions between each component, @theanonymous27
anonymous
  • anonymous
so
anonymous
  • anonymous
Yes
anonymous
  • anonymous
so does the slope represent the component of the vector?
anonymous
  • anonymous
we end up with r(t) = ? is that right?
anonymous
  • anonymous
So given the direction vector \(\mathbf{v}\) and some point which the line goes though \(\mathbf{b}\) the parametrization for our line is: \[ \mathbf{r}(t) = \mathbf{v}t+\mathbf{b} \]
anonymous
  • anonymous
\[ r(t) = \begin{bmatrix} 1 \\ 4 \\ -12 \end{bmatrix} t + \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} t \\ 4t \\ -12t \end{bmatrix} \]
anonymous
  • anonymous
Thank you!, One last question... Why did you pick x=1 first? what would have happened if you choose y=1 instead
anonymous
  • anonymous
If you pick \(y=1\) then you get \(x=1/4\) and \(z=-3\)
anonymous
  • anonymous
It's the same direction.
anonymous
  • anonymous
Thanks!

Looking for something else?

Not the answer you are looking for? Search for more explanations.