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Zenmo

  • one year ago

(Give me your strength! :) Help me with this to prepare for finals soon.) Find parametric equations of their line of intersection of two planes. 3x - 4y + 5z = 6 x + y - z = 2

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  1. Zenmo
    • one year ago
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    |dw:1435140954274:dw|

  2. Zenmo
    • one year ago
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    |dw:1435141299438:dw|

  3. Zenmo
    • one year ago
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    |dw:1435141995576:dw|

  4. Zenmo
    • one year ago
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    Not sure if I'm doing it correctly.

  5. Zenmo
    • one year ago
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    I know that once the point ( X, Y, Z) is found, then I can add it to the cross product to find the set of parametric equations.

  6. ganeshie8
    • one year ago
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    Looks good! next maybe let z=7, and find x, y values

  7. Zenmo
    • one year ago
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    where did you get z=7?

  8. ganeshie8
    • one year ago
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    you can let z anything

  9. ganeshie8
    • one year ago
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    y = 8z/7 x = 2 - z/7 z = anything

  10. Zenmo
    • one year ago
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    By letting z=7: x=1, y=8. Plugging them into the original equation of x+y-z=2 to find Z: z=7.

  11. Zenmo
    • one year ago
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    (1, 8, 7) + T< -1, 8, 7> x = 1 - t, y = 8 + 8t, z = 7+ 7t

  12. Zenmo
    • one year ago
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    Those are likely wrong

  13. ganeshie8
    • one year ago
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    thats perfect!

  14. Loser66
    • one year ago
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    Represent the line under matrix form, you have \[\left[\begin{matrix}x\\y\\z\end{matrix}\right]=\left[\begin{matrix}2-(1/7)z\\(8/7)z\\z\end{matrix}\right]\] ok?

  15. Zenmo
    • one year ago
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    The solution from the book is: x = -t + 2, y = 8t, z = 7t.

  16. ganeshie8
    • one year ago
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    that is also correct

  17. Zenmo
    • one year ago
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    Is my answer in a different format?

  18. ganeshie8
    • one year ago
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    your book is using a different point on the line, thats all. both equations are correct

  19. ganeshie8
    • one year ago
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    both equations represent the same line

  20. Zenmo
    • one year ago
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    I see. Thanks! Did Loser66 managed to find the book answer format?

  21. Loser66
    • one year ago
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    and let z =7t, hence you have \[\left[\begin{matrix}x\\y\\z\end{matrix}\right]=\left[\begin{matrix}2\\0\\0\end{matrix}\right]+\left[\begin{matrix}(1/7)7t\\(8/7)7t\\7t\end{matrix}\right]\]

  22. Loser66
    • one year ago
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    sorry, the first entry is -1/7t, not 1/7 t pick t out from the far right matrix, you have required form of parametric equation. \[\left[\begin{matrix}x\\y\\z\end{matrix}\right]=\left[\begin{matrix}2\\0\\0\end{matrix}\right]+t\left[\begin{matrix}-1\\8\\7\end{matrix}\right]\]

  23. Loser66
    • one year ago
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    that gives you x =2-t y= 8t z= 7t

  24. Zenmo
    • one year ago
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    Could you show the work on how you gotten ( x, y, z) = (2, 0, 0). I may need a quick refresher on substitution/elimination.

  25. Zenmo
    • one year ago
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    Other than that. I'm all set for this problem! :)

  26. Loser66
    • one year ago
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    x = 2-(1/7)t ok? so you get 2 for x,

  27. Loser66
    • one year ago
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    y = (8/7)t =0 + (8/7)t, hence you get 0 for y same as z

  28. Loser66
    • one year ago
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    |dw:1435150654823:dw|

  29. Loser66
    • one year ago
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    |dw:1435150706384:dw|