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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 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.
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 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.

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Can you come up with a vector which goes in the correct direction?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0First, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Our vector only has to have the right direction.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@theanonymous27 Does this make sense?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.0Since \(\Delta y/\Delta x= 4\) and \(\Delta x = 1\), what must \(\Delta y\) be?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0We use the slopes to figure out the proportions between each component, @theanonymous27

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so <Y is 4, and then we get that Z has to be 12

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so does the slope represent the component of the vector?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0we end up with r(t) = <t, 4, 12t> ? is that right?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0So 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
 3 years ago
Best ResponseYou've already chosen the best response.0\[ r(t) = \begin{bmatrix} 1 \\ 4 \\ 12 \end{bmatrix} t + \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} t \\ 4t \\ 12t \end{bmatrix} \]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Thank you!, One last question... Why did you pick x=1 first? what would have happened if you choose y=1 instead

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0If you pick \(y=1\) then you get \(x=1/4\) and \(z=3\)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0It's the same direction.
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