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theanonymous27
Group Title
Find a parametrization
r(t) = h x(t), y(t), z(t)i
of the straight line passing through the origin
in 3space whose projection on the xyplane
is a line with slope 4, while its projection on
the yzplane is a line with slope −3, i.e.,
∆y/∆x= 4,∆z/∆y= −3.
 one year ago
 one year ago
theanonymous27 Group Title
Find a parametrization r(t) = h x(t), y(t), z(t)i of the straight line passing through the origin in 3space whose projection on the xyplane is a line with slope 4, while its projection on the yzplane is a line with slope −3, i.e., ∆y/∆x= 4,∆z/∆y= −3.
 one year ago
 one year ago

This Question is Closed

wio Group TitleBest ResponseYou've already chosen the best response.1
Can you come up with a vector which goes in the correct direction?
 one year ago

theanonymous27 Group TitleBest ResponseYou've already chosen the best response.0
How is that?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
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.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Our vector only has to have the right direction.
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
@theanonymous27 Does this make sense?
 one year ago

theanonymous27 Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
So we can start by letting our \(x\) component be 1:\[ \mathbf{v} = \begin{bmatrix} x_1 \\ y_1 \\ z_1 \end{bmatrix} = \begin{bmatrix} 1 \\ y_1 \\ z_1 \end{bmatrix} \]
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
Since \(\Delta y/\Delta x= 4\) and \(\Delta x = 1\), what must \(\Delta y\) be?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
We use the slopes to figure out the proportions between each component, @theanonymous27
 one year ago

theanonymous27 Group TitleBest ResponseYou've already chosen the best response.0
so <Y is 4, and then we get that Z has to be 12
 one year ago

theanonymous27 Group TitleBest ResponseYou've already chosen the best response.0
so does the slope represent the component of the vector?
 one year ago

theanonymous27 Group TitleBest ResponseYou've already chosen the best response.0
we end up with r(t) = <t, 4, 12t> ? is that right?
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
So given the direction vector \(\mathbf{v}\) and some point which the line goes though \(\mathbf{b}\) the parametrization for our line is: \[ \mathbf{r}(t) = \mathbf{v}t+\mathbf{b} \]
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
\[ r(t) = \begin{bmatrix} 1 \\ 4 \\ 12 \end{bmatrix} t + \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} t \\ 4t \\ 12t \end{bmatrix} \]
 one year ago

theanonymous27 Group TitleBest ResponseYou've already chosen the best response.0
Thank you!, One last question... Why did you pick x=1 first? what would have happened if you choose y=1 instead
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
If you pick \(y=1\) then you get \(x=1/4\) and \(z=3\)
 one year ago

wio Group TitleBest ResponseYou've already chosen the best response.1
It's the same direction.
 one year ago

theanonymous27 Group TitleBest ResponseYou've already chosen the best response.0
Thanks!
 one year ago
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