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minyayao

  • 3 years ago

Find II 2e-3f II assuming that e & f are unit vectors such that II e +f II=sqrt(3/2)

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  1. wio
    • 3 years ago
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    "e & f are unit vectors" meaning \(||\vec{e}||=1,||\vec{f}||=1\)

  2. wio
    • 3 years ago
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    We got to sort of consider the facts here.

  3. minyayao
    • 3 years ago
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    so e dot e=1 and f dot f also equals 1?

  4. wio
    • 3 years ago
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    yeah

  5. minyayao
    • 3 years ago
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    Okay thanks I think I know how to do it! :)

  6. wio
    • 3 years ago
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    Please tell.

  7. wio
    • 3 years ago
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    I can see how \[ (\vec{e} +\vec{f}) \cdot (\vec{e} +\vec{f}) = 3/2 \]

  8. minyayao
    • 3 years ago
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    \[||e+f||^{2}=(e+f)(e+f)=ee+2ef+ff=1+2ef+1=2+2ef=3/2\] so ef=-1/4 similarly \[||2e-3f||^{2}=(2e-3f)(2e-3f)=4ee-12ef+9ff=13-12ef=13-12(-1/4)=16\] therefore ||2e-3f||=sqrt 16=4

  9. wio
    • 3 years ago
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    Wonderful thinking.

  10. oldrin.bataku
    • 3 years ago
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    $$e_x^2+e_y^2=1\\f_x^2+f_y^2=1\\(e_x+f_x)^2+(e_y+f_y)^2=\frac32\\e_x^2+2e_xf_x+f_x^2+e_y^2+2e_yf_y+f_y^2=\frac32\\2+2e_xf_x+2e_yf_y=\frac32\\e_xf_x+e_yf_y=-\frac14\\(2e_x-3f_x)^2+(2e_y-3f_y)^2=4e_x^2-12e_xf_x+9f_x^2+4e_y^2-12e_yf_y+9f_y^2\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4+9-12[e_xf_x+e_yf_y]\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =13+3=16\\\Vert 2e-3f\Vert=\sqrt{16}=4$$

  11. minyayao
    • 3 years ago
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    Thank you!

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