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theanonymous27

  • one year 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.

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  1. wio
    • one year ago
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    Can you come up with a vector which goes in the correct direction?

  2. theanonymous27
    • one year ago
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    How is that?

  3. wio
    • one year ago
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    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
    • one year ago
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    Our vector only has to have the right direction.

  5. wio
    • one year ago
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    @theanonymous27 Does this make sense?

  6. theanonymous27
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    Since \(\Delta y/\Delta x= 4\) and \(\Delta x = 1\), what must \(\Delta y\) be?

  9. wio
    • one year ago
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    We use the slopes to figure out the proportions between each component, @theanonymous27

  10. theanonymous27
    • one year ago
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    so <Y is 4, and then we get that Z has to be -12

  11. wio
    • one year ago
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    Yes

  12. theanonymous27
    • one year ago
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    so does the slope represent the component of the vector?

  13. theanonymous27
    • one year ago
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    we end up with r(t) = <t, 4, -12t> ? is that right?

  14. wio
    • one year ago
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    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
    • one year ago
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    \[ r(t) = \begin{bmatrix} 1 \\ 4 \\ -12 \end{bmatrix} t + \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} t \\ 4t \\ -12t \end{bmatrix} \]

  16. theanonymous27
    • one year ago
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    Thank you!, One last question... Why did you pick x=1 first? what would have happened if you choose y=1 instead

  17. wio
    • one year ago
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    If you pick \(y=1\) then you get \(x=1/4\) and \(z=-3\)

  18. wio
    • one year ago
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    It's the same direction.

  19. theanonymous27
    • one year ago
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    Thanks!

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