Question

Find the angle between the given vectors to the nearest tenth of a degree.

u = <8, 7>, v = <9, 7>

-8.3°

1.7°

3.3°

13.3°

Answer 1

@jim_thompson5910

Answer 2

first you need to compute the dot product

Answer 3

what do you get when you do that

Answer 4

kk onesec

Answer 5

{72, 49}

Answer 6

add those two up

Answer 7

121

Answer 8

@jim_thompson5910

Answer 9

now you need to find the modulus of each vector

Answer 10

how do u do that

Answer 11

mod of u = sqrt(8^2 + 7^2)

Answer 12

mod of v = sqrt(9^2 + 7^2)

Answer 13

sqrt(8^2 + 7^2)=113^(1/2)
sqrt(9^2 + 7^2)=130^(1/2)

Answer 14

ok now

cos x = dot product / ( mod u * mod v)

where x is the required angle

Answer 15

what do you get for modu * mod v ?

Answer 16

121 / ( 113^(1/2) *130^(1/2)) ???

Answer 17

modu * mod v =
Sqrt[14690]

Answer 18

no
use your calculator to work out the 2 square roots seperately then multiply them

Answer 19

121.202

Answer 20

sqrt 113 = 10.63015 right?

Answer 21

yes

Answer 22

modu * mod v =121

Answer 23

ok so we have cos x = 121 / 121.202

Answer 24

cos(x) = 0.998333

Answer 25

right now use your calculator to find x

Answer 26

its the cos-1 key or ( on some its arcos)

Answer 27

3.3

Answer 28

yep

Answer 29

the important formula to remember is

u.v = |u||v| cos x

where u and v are the vectors and |u| and |v| are the modulus of u and v

Answer 30

@jim_thompson5910

is it modulus or magnitude the correct term?

Answer 31

they are the same thing pretty much

Answer 32

the modulus of a vector = the magnitude of a vector

Answer 33

oh ok