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chrisplusian

  • one year ago

Linear Algebra.... Need help with #6 part III.... Please see attachment

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  1. chrisplusian
    • one year ago
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  2. anonymous
    • one year ago
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    the inner product is bilinear, so $$\langle a+b,c+d\rangle=\langle a+b,c\rangle+\langle a+b,d\rangle=\langle a,c\rangle +\langle b,c\rangle+\langle a,d\rangle+\langle b,d\rangle$$

  3. chrisplusian
    • one year ago
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    I am about to upload what I have done... don;t know where to go from there

  4. chrisplusian
    • one year ago
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  5. anonymous
    • one year ago
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    so here we have $$\langle 2u+3v-w,u-v+w\rangle\\\quad =\langle 2u,u\rangle+\langle2u,-v\rangle+\langle 2u,w\rangle\\\qquad +\langle 3v,u\rangle+\langle3v,-v\rangle+\langle 3v,w\rangle\\\qquad+\langle -w,u\rangle+\langle-w,-v\rangle+\langle -w,w\rangle$$

  6. anonymous
    • one year ago
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    now we're told that \(u,v,w\) are unit vectors, so \(\|u\|=\|v\|=\|w\|=1\) and we're also told that they're each separated from one another by an angle of \(\pi/4\) in space

  7. chrisplusian
    • one year ago
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    I think your looking at the wrong problem.

  8. chrisplusian
    • one year ago
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    I have that one finished

  9. chrisplusian
    • one year ago
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    I am working on part III

  10. anonymous
    • one year ago
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    remember we have that \(\langle x,x\rangle=\|x\|^2\) and more generally \(\langle x,y\rangle =\|x\|\|y\|\cos\theta\) where \(\theta\) is the angle between \(x,y\), so now evaluate all the inner products

  11. anonymous
    • one year ago
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    oh, oops -- I misread the title and thought it said part II

  12. chrisplusian
    • one year ago
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    Lol No big deal, could you look at my work?

  13. chrisplusian
    • one year ago
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    I also eliminated all the inner products that were not of the same vector because they are orthogonal

  14. chrisplusian
    • one year ago
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    \[\sqrt{<v,v>+<u,u>+<w,w>}\]

  15. chrisplusian
    • one year ago
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    That is what I am left with but I don't know what to do with that because \[\sqrt{4}+\sqrt{2}\neq \sqrt{4+2}\]

  16. anonymous
    • one year ago
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    for part III, consider we have $$\langle x,x\rangle=\|x\|^2$$so it follows that $$\|u+v+w\|^2=\langle u+v+w,u+v+w\rangle\\\quad=\langle u,u\rangle+\langle v,v\rangle+\langle w,w\rangle+2\langle u,v\rangle+2\langle u,w\rangle+2\langle v,w\rangle$$now since we said \(u,v,w\) are all mutually orthogonal it means those cross terms \(\langle u,v\rangle=\langle u,w\rangle=\langle v,w\rangle=0\) so: $$\|u+v+w\|^2=\|u\|^2+\|v\|^2+\|w\|^2\\\implies \|u+v+w\|\implies\sqrt{\|u\|^2+\|v\|^2+\|w\|^2}$$ indeed

  17. chrisplusian
    • one year ago
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    So I can't just say that its the magnitude of one plus the magnitude of the other etc.

  18. anonymous
    • one year ago
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    now \(\|u\|=\|v\|=2\) and \(\|w\|=3\) gives us $$\|u+v+w\|=\sqrt{2^2+2^2+3^2}=\sqrt{17}$$

  19. chrisplusian
    • one year ago
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    I am confused

  20. anonymous
    • one year ago
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    indeed, you can't (in general) say that the magnitude of the sum is the sum of the magnitudes

  21. chrisplusian
    • one year ago
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    I am trying to explain my confusion but I am not as familiar with the equation tool as you are so give me one second and I will try to show you where I am confused

  22. chrisplusian
    • one year ago
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    Real quick how do you create the brackets that go around the inner product?

  23. anonymous
    • one year ago
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    \langle and \rangle

  24. anonymous
    • one year ago
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    $$\text{\langle}\implies\langle\\\text{\rangle}\implies\rangle$$

  25. chrisplusian
    • one year ago
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    \[\sqrt{<v,v>+<u,u>+<w,w>}\neq \sqrt{<v,v>}+\sqrt{<w,w>}+\sqrt{<u,u>}\]

  26. chrisplusian
    • one year ago
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    But in your explanation you made \[<v,v>=\left| v \right|\]

  27. anonymous
    • one year ago
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    indeed, that is not true

  28. chrisplusian
    • one year ago
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    I meant for that absolute v to have double bars.....

  29. anonymous
    • one year ago
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    yep, by definition \(\|v\|^2=\langle v,v\rangle\) so \(\langle u,u\rangle=2^2=4\), \(\langle v,v\rangle =2^2=4\) and \(\langle w,w\rangle=3^2=9\)

  30. chrisplusian
    • one year ago
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    but the definition in my book shows that \[\left| \right|v \left| \right|=\sqrt{<v,v>}\]

  31. chrisplusian
    • one year ago
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    Oh I see it now, you took the square of both sides to arrive at what you had. I didn't make that connection. THank you

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