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Find II 2e-3f II assuming that e & f are unit vectors such that II e +f II=sqrt(3/2).I tried using the law of cosines but not getting it...

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Draw e+f, and solve for the angle.
Do you need another hint?
not yet, doing math & watching my 3year old and 1 year old sister is interesting...makes solving problems kind of slow...i'll post if i need another hint thanks!

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Ok, sometimes people get stuck but don't speak up until you ask :(
why am I finding the theta between e & e+f....theta=e dot e+f / sqrt(3/2)...
You need theta between e and f.|dw:1335665267401:dw|
but then theta=e dot f what is that...
Where is e+f on the picture? You have a triangle and you know the length of all 3 sides.
got it got it got it....
ok, that saves me some drawing :)
now i can move on lol i hope that you aren't much younger than me, i'm very competitive but i gotta hand it to you! idk why I got stumped i tutor calc 3!!! >:/
I'm probably not :)
\[ (e+f) . (e+f)=e.e + e.f + f.e + f.f=2 +2 e.f= \frac 32\\ e.f =-\frac 1 4\\ (2e -3f).(2e-3f)=4e.e - 6 e.f -6 f.e +9f.f =\\ 13 - 12 e.f= 13 + 3=16\\ || 2e - 3f|| =\sqrt{(2e -3f).(2e-3f)} =\sqrt {16}=4 \] where u.v is the dot product of u and v
Wow, that's actually nicer than calculating the angle between e and f and writing them in a new coordinate system.
Actually for any m and n || m e - n f|| can be calculated and depends only on m and n and e.f
awesome eliassaab!!!

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