Question

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

u = <6, -1>, v = <7, -4>

Answer 1

3 things to find

u dot v

the length of u
the length of v

Answer 2

\[cos(\alpha)=\frac{u\cdot v}{|u||v|}\]
\[\alpha=\cos^{-1}\left(\frac{u\cdot v}{|u||v|}\right)\]

Answer 3

u dot v is 46

Answer 4

i dont know how to do the rest

Answer 5

the length of a vector is the sqrt of its own dot product

|u| = sqrt(u.u)
|v| = sqrt(v.v)

Answer 6

essentially the bottom part is:

sqrt(u.u, times v.v)

Answer 7

so |u| would be sqrt 7?

Answer 8

sqrt(36+1) = sqrt(37) but yes

Answer 9

6, -1
6, -1
-----
36+1 = 37

|u| = sqrt(37)

Answer 10

ohh so |v| would be sqrt(65)

Answer 11

correct

Answer 12

\[\cos(\alpha)=\frac{46}{\sqrt{37*65}}\]
then inverse cosine each side

Answer 13

arccos46/sqrt 37*65 ?

Answer 14

yes :)

which should come out to about 0.353997... radians
or 20.28 degrees


radians times, 180/pi if you need to convert

Answer 15

thanks so much!!!!!!!!

Answer 16

your welcome, and good luck :)