Question

u=<2,-1> what are the vector v1 and v2 with norma 5 and 60 degrees with vector ? got stuck on system resolution.

Answer 1

loser66 helped me till here

let v =<x,y> you have
cos60=(<u,v>)/|u||v|

and cos 60 = 1/2 , |u| = sqrt(5) ; |v| = 5 so, just count the numerator which is <(2,-1), (x,y)>

Answer 2

I tried to solve the system but...

Answer 3

what does "with norma 5 and 60 degrees with vector" mean?

Answer 4

sorry, the question was:
u=<2,-1> what are the vector v1 and v2 with norma 5 and 60 degrees with vector u?\]

Answer 5

x^2+y^2=5;
(2x-y)/3*(x^2+y^2)^(1/2);

Answer 6

I got lost on this system

Answer 7

so norm of a vector is its length

Answer 8

\[x ^{2}+y ^{2}=5;\]

Answer 9

yes

Answer 10

<2,-1> is already at -30 degrees from the +x axis

Answer 11

\[\frac{ 2x-y }{ 3*\sqrt{x^2 + y^2}}\]

Answer 12

sorry?

Answer 13

30 degrees form the u vector

Answer 14

plus and minus 30 degrees form de u vector

Answer 15

v1 = 5<2/sqrt(3),1/sqrt(3)>

v2 = <0,-5>

Answer 16

from the u vector

Answer 17

like this right?

Answer 18

wait

Answer 19

im confusing my 2,1 triangle at the moment ....

Answer 20

exactly

Answer 21

the last pic

Answer 22

the polar representations would be\[r=(5,tan^{-1}(\frac{-1}{2})\pm\frac{\pi}{3})\]

Answer 23

how to solve it algebrically?

Answer 24

a unit for v1 or v2 is of the form\[cos(\alpha)=\frac{u\cdot v_i}{\sqrt{5}}\]
\[\frac{\pi}{3}=\frac{2x_1-1x_2}{\sqrt{5}}\]
hmmm

Answer 25

wait a min. please..

Answer 26

im thinking we could try a longer method ...