## A community for students. Sign up today

Here's the question you clicked on:

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

• This Question is Closed
1. wio

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

2. wio

We got to sort of consider the facts here.

3. minyayao

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

4. wio

yeah

5. minyayao

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

6. wio

Please tell.

7. wio

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

8. minyayao

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

Wonderful thinking.

10. oldrin.bataku

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

11. minyayao

Thank you!

#### Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy