Question

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

Answer 1

"e & f are unit vectors"
meaning \(||\vec{e}||=1,||\vec{f}||=1\)

Answer 2

We got to sort of consider the facts here.

Answer 3

so e dot e=1 and f dot f also equals 1?

Answer 4

yeah

Answer 5

Okay thanks I think I know how to do it! :)

Answer 6

Please tell.

Answer 7

I can see how \[
(\vec{e} +\vec{f}) \cdot (\vec{e} +\vec{f}) = 3/2
\]

Answer 8

\[||e+f||^{2}=(e+f)(e+f)=ee+2ef+ff=1+2ef+1=2+2ef=3/2\]
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

Answer 9

Wonderful thinking.

Answer 10

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

Answer 11

Thank you!