Question

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

u = <-5, -4>, v = <-4, -3>

Answer 1

use the dot product ....\[cos~\alpha=\frac{\vec u \cdot \vec v}{|\vec u||\vec v|}\]

Answer 2

inverse cosine to account for the angle ....

Answer 3

so do I do something like subtract the x(u) from the x(v) to get the u ontop?

Answer 4

i just stack u and v, multiply down, and add across

u = <-5, -4>
v = <-4, -3>
---------
20 + 12 = u.v

Answer 5

now divide by the lengths ...

Answer 6

ohh ok

Answer 7

the lengths of what? I'm sorry

Answer 8

the lengths of the vectors of course ....

Answer 9

how long is u? how long is v ?

Answer 10

by the pythag thrmt:\[|\vec a|=\sqrt{(\vec a \cdot \vec a)}\]

Answer 11

what would the a stand for in this instance

Answer 12

some generic vector, any vector. It is a property that is shared by all vectors and so the vectors in this problem use it as well.

|v| v = <-4, -3>
v = <-4, -3>
--------------
sqrt( 16 + 9 )

Answer 13

okay now we just do it for the other, then multiply it

Answer 14

yep

Answer 15

5 * sqrt(41)

Answer 16

?

Answer 17

25
16
----
41

yep

Answer 18

so, we have:\[cos~\alpha=\frac{20+12}{5\sqrt{41}}\]
solve for alpha

Answer 19

= .99951

Answer 20

?

Answer 21

cos a = .99951

a = cos^-1 (.99951), but this gets us radians

multiply by pi/180 to convert to degrees

Answer 22

http://www.wolframalpha.com/input/?i=arccos%2832%2F%285sqrt%2841%29%29%29

Answer 23

i spose it depends on if your calculator is in degree mode or radian mode ... degree mode gives out 1.7937...

Answer 24

okay, thanks!

Answer 25

youre welcome