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anonymous
 one year ago
Linear Algebra....
Need help with #6 part III.... Please see attachment
anonymous
 one year ago
Linear Algebra.... Need help with #6 part III.... Please see attachment

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the inner product is bilinear, so $$\langle a+b,c+d\rangle=\langle a+b,c\rangle+\langle a+b,d\rangle=\langle a,c\rangle +\langle b,c\rangle+\langle a,d\rangle+\langle b,d\rangle$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am about to upload what I have done... don;t know where to go from there

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so here we have $$\langle 2u+3vw,uv+w\rangle\\\quad =\langle 2u,u\rangle+\langle2u,v\rangle+\langle 2u,w\rangle\\\qquad +\langle 3v,u\rangle+\langle3v,v\rangle+\langle 3v,w\rangle\\\qquad+\langle w,u\rangle+\langlew,v\rangle+\langle w,w\rangle$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think your looking at the wrong problem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have that one finished

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am working on part III

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0remember 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oh, oops  I misread the title and thought it said part II

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Lol No big deal, could you look at my work?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I also eliminated all the inner products that were not of the same vector because they are orthogonal

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt{<v,v>+<u,u>+<w,w>}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0That is what I am left with but I don't know what to do with that because \[\sqrt{4}+\sqrt{2}\neq \sqrt{4+2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0for 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So I can't just say that its the magnitude of one plus the magnitude of the other etc.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now \(\u\=\v\=2\) and \(\w\=3\) gives us $$\u+v+w\=\sqrt{2^2+2^2+3^2}=\sqrt{17}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0indeed, you can't (in general) say that the magnitude of the sum is the sum of the magnitudes

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Real quick how do you create the brackets that go around the inner product?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\text{\langle}\implies\langle\\\text{\rangle}\implies\rangle$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sqrt{<v,v>+<u,u>+<w,w>}\neq \sqrt{<v,v>}+\sqrt{<w,w>}+\sqrt{<u,u>}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0But in your explanation you made \[<v,v>=\left v \right\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0indeed, that is not true

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I meant for that absolute v to have double bars.....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep, by definition \(\v\^2=\langle v,v\rangle\) so \(\langle u,u\rangle=2^2=4\), \(\langle v,v\rangle =2^2=4\) and \(\langle w,w\rangle=3^2=9\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but the definition in my book shows that \[\left \rightv \left \right=\sqrt{<v,v>}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh 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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