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Calcmathlete

  • 3 years ago

Find the coordinates of the point where the vector line \(\vec{r} = (6 + \frac{7}{9}d)\vec{i} + (3 + \frac{4}{9}d)\vec{j} + (2 - \frac{4}{9}d)\vec{k}\) intersects the plane 3x - 7y + z = 5. I'd like an explanation or better yet, a guide through it rather than just a plain answer.

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  1. LockeMcDonnell
    • 3 years ago
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    holy good luck with that!

  2. Calcmathlete
    • 3 years ago
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    -_-, thanks lol

  3. hieuvo
    • 3 years ago
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    you know that: x=6+7/9d, y=3+4/9d, and z=2-4/9d ?

  4. Calcmathlete
    • 3 years ago
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    Yes. I actually think I might have an answer since I asked this yesterday. Could you check it?

  5. ScottB05
    • 3 years ago
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    I sence De Moivre's Theorum comming on here =)

  6. Calcmathlete
    • 3 years ago
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    d = -54/11, so (24/11, 9/11, 46/11) Is this right?

  7. ScottB05
    • 3 years ago
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    In your question you've mentioned i..... Is that i as in i, or i as in the Square root of -1???

  8. Calcmathlete
    • 3 years ago
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    i as in unit vector \(\vec{i}\)

  9. Calcmathlete
    • 3 years ago
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    How so?

  10. ScottB05
    • 3 years ago
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    Of course. Thank you :)

  11. hieuvo
    • 3 years ago
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    that 7/9 or what, cant see the problem clearly

  12. Calcmathlete
    • 3 years ago
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    7/9

  13. hieuvo
    • 3 years ago
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    maybe I'm wrong in sth

  14. hieuvo
    • 3 years ago
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    d= -54/11

  15. Calcmathlete
    • 3 years ago
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    Here's my work if it helps... \[\vec{r} = (6 + \frac{7}{9}d)\vec{i} + (3 + \frac{4}{9}d)\vec{j} + (2 - \frac{4}{9}d)\vec{k}\]\[3(6 + \frac79d) - 7(3 + \frac49d) + (2 - \frac49d) = 5\]\[18 + \frac{21}{9}d - 21 - \frac{28}{9}d + 2 - \frac49d = 5\]\[-\frac{11}9d = 6\]\[d = 6 \times -\frac9{11}\]\[d = -\frac{54}{11}\]

  16. hieuvo
    • 3 years ago
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    lol, that's right, sr

  17. Calcmathlete
    • 3 years ago
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    lol ok, thanks :)

  18. hieuvo
    • 3 years ago
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    So just plug in to find coordinate. Is this cal 3?

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