anonymous
  • anonymous
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
holy good luck with that!
anonymous
  • anonymous
-_-, thanks lol
anonymous
  • anonymous
you know that: x=6+7/9d, y=3+4/9d, and z=2-4/9d ?

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anonymous
  • anonymous
Yes. I actually think I might have an answer since I asked this yesterday. Could you check it?
anonymous
  • anonymous
I sence De Moivre's Theorum comming on here =)
anonymous
  • anonymous
d = -54/11, so (24/11, 9/11, 46/11) Is this right?
anonymous
  • anonymous
In your question you've mentioned i..... Is that i as in i, or i as in the Square root of -1???
anonymous
  • anonymous
i as in unit vector \(\vec{i}\)
anonymous
  • anonymous
How so?
anonymous
  • anonymous
Of course. Thank you :)
anonymous
  • anonymous
that 7/9 or what, cant see the problem clearly
anonymous
  • anonymous
7/9
anonymous
  • anonymous
maybe I'm wrong in sth
anonymous
  • anonymous
d= -54/11
anonymous
  • anonymous
Here's my work if it helps... \[\vec{r} = (6 + \frac{7}{9}d)\vec{i} + (3 + \frac{4}{9}d)\vec{j} + (2 - \frac{4}{9}d)\vec{k}\]\[3(6 + \frac79d) - 7(3 + \frac49d) + (2 - \frac49d) = 5\]\[18 + \frac{21}{9}d - 21 - \frac{28}{9}d + 2 - \frac49d = 5\]\[-\frac{11}9d = 6\]\[d = 6 \times -\frac9{11}\]\[d = -\frac{54}{11}\]
anonymous
  • anonymous
lol, that's right, sr
anonymous
  • anonymous
lol ok, thanks :)
anonymous
  • anonymous
So just plug in to find coordinate. Is this cal 3?

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