Question

Last one?

Answer 1

well same reasoning as last time, find the dot product of

(-7,8) and (1,0)

Answer 2

-7*1=-7

8*0=0

Answer 3

good, then add them together and find the dot product

Answer 4

-7+0=-7

Answer 5

good

then you need to find the magnitudes of (-7,8) and (1,0)

Answer 6

sqrt(x^2+y^2)
sqrt((-7)^2+8^2)
=sqrt113


sqrt(x^2+y^2)
sqrt(1^2+0^2)
=1

Answer 7

good, then you just need to plug in the dot product and magnitude into

v1 dot v2 = |v1| |v2| cos(theta)

and solve for cos(theta), not theta

Answer 8

sqrt 113 * 1 = |sqrt113| |1| cos(theta)


=6.28

Answer 9

the dot product is -7 not sqrt(113)*1

-7 = sqrt(113) * 1 * cos(theta)

cos(theta) = ?

Answer 10

6.28

Answer 11

-7 = sqrt(113) * 1 * cos(theta)

divide both sides by sqrt(113)

Answer 12

you are solving for cos(theta) not theta

Answer 13

-7/sqrt(113) = ?

Answer 14

-0.66

Answer 15

good
then take the arccos of that to get the angle for part II

Answer 16

arccos(-0.66)=131.1 degrees

Answer 17

awesome

then for part B) find the magnitude of vector v (-7,8)

Answer 18

sqrt(x^2+y^2)

sqrt((-7)^2+8^2)
=sqrt113

Answer 19

good (they want it in radical form so leave it like that) sqrt(113)

then for part II write out the vector

(|v|cos(theta), |v|sin(theta))

just replace |v| with the magnitude and theta with the angle from part A II and don't do anything else

Answer 20

(|sqrt113|cos(131.1), |v|sin(131.1))

Answer 21

almost

you gotta get that other |v| too

((sqrt113)cos(131.1°), (sqrt113)sin(131.1°))

Answer 22

((sqrt113)cos(131.1°), (sqrt113)sin(131.1°))

Answer 23

Got it

Answer 24

do I solve for anything or is it as it is

Answer 25

that's it

anyway for part C we just use the angle formula again

Answer 26

hm. I could have sworn we already calculated the dot product and the magnitude of this one

Answer 27

we already did this lol

Answer 28

oh alright cool what's left

Answer 29

Thats it! Thank you so much for helping me. I really do appreciate it !!!