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chrisplusian

  • one year ago

Show that there are no vectors "u" and "v" in R^3 such that ||u||=1, ||v||=2, and <v,u>=3.

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  1. Loser66
    • one year ago
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    Suppose yes, then \(cos (u, v) =\dfrac{<u, v>}{||u||||v||}= \dfrac{3}{2}\) does it make sense? You give out the conclusion.

  2. chrisplusian
    • one year ago
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    I honestly don't follow that way.

  3. Loser66
    • one year ago
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    why not? \(-1\leq cos (\theta) \leq 1\) That is the restriction of a cos function, and the result is out of the range of cos. Hence the "supposing" does not hold .

  4. Loser66
    • one year ago
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    I use contradiction method and it is the easiest way to use all of the given information. But if you don't get it. I am sorry.

  5. chrisplusian
    • one year ago
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    Thank you for the attempt, and I don't doubt what you have done. I just think there has to be a different way to show it. Can you think of a different way to do it?

  6. Loser66
    • one year ago
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    @oldrin.bataku

  7. anonymous
    • one year ago
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    $$\langle v,u\rangle=\|u\|\|v\|\cos\theta=2\cos\theta$$however \(\cos\theta\) has a maximum value of \(1\) so \(2\cos\theta\le 2\), yet \(3\not\le 2\) so there can be no such vectors

  8. anonymous
    • one year ago
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    geometrically, we have that \(\langle v,u\rangle\le\|v\|\|w\|\), and \(3\not\le2\)

  9. chrisplusian
    • one year ago
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    Is there a way to expand the dot product to show this?

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