Question

Medddallll and Fannnns!!!

Consider the vector v=(5,8)

What is the angle between v=(5,8) and the positive x-axis

Write v in the form (|v|cos theta|v|sin theta)

Answer 1

\[cos\theta=\frac{\vec u\cdot \vec v}{|\vec u||\vec v|}\]

Answer 2

or, tan theta = 8/5

Answer 3

Ok but how would we express the numbers in the equation

Answer 4

the top is a dot product, the bottom is the product of their lengths

Answer 5

Ahhh, ok..

Answer 6

the simplest vector in the posx direction to use is (1,0) its length is 1 and it dots out to the x component of v

Answer 7

(a,b)
(1,0)
-----
a+0 = a

Answer 8

So thats for the vector position.. So what is the equation for the lenths

Answer 9

since the length of (1,0) is 1 ... and 1|v| = |v|

you should be able to determine the length of v by now

Answer 10

\[cos\theta=\frac{v_x}{|v|}\]

Answer 11

sin(theta) = sqrt(1-cos^2)

Answer 12

ok so the length i got was 13... But instead you would want it to be sqrt 13 right

Answer 13

8^2 + 5^2 = 64+25 = 89, id go with sqrt(89)

Answer 14

Ok but how would we express that for the angle between v=(5,8) and the positive x axis?

Answer 15

\[cos\theta=\frac{5}{\sqrt{89}}\]
\[\theta=cos^{-1}\left(\frac{5}{\sqrt{89}}\right)\]

Answer 16

OR\[tan\theta=\frac{8}{5}\]
\[\theta=tan^{-1}\left(\frac{8}{5}\right)\]

Answer 17

Alright, Thank you!

Answer 18

your welcome

Answer 19

So then I can plug it in to make it appear on the graph by the vector