Question

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

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

Answer 1

use dot product

u.v = ||u|| ||v|| cos\(\theta\)

Answer 2

<-5, -4> . <-4, -3> = ||<-5, -4> || ||<-4, -3> || cos \(\theta\)

solve \(\theta\)

Answer 3

I keep getting 5 squareroot 41

Answer 4

yes i also got 5 squareroot41 :)

<-5, -4> . <-4, -3> = ||<-5, -4> || ||<-4, -3> || cos \(\theta\)

20+12 = \(\sqrt{41} \times 5\) cos \(\theta\)

32 = \(5\sqrt{41} \) cos \(\theta\)

Answer 5

you still need to solve \(\theta\)

Answer 6

\(\large \frac{32}{5\sqrt{41}} = \cos \theta\)

Answer 7

use ur calculator and find \(\theta\)

Answer 8

\(0.999512 = \cos \theta\)

\(\cos \theta = 0.999512 \)

can u find \(\theta\) ?

Answer 9

Wait, 0.9 is the answer, if I am not mistaken.

Answer 10

nopes

Answer 11

when u have,

\(\cos \theta = 0.999512\)

you must take cos inverse both sides, to isolate \(\theta\)

Answer 12

\(\cos \theta = 0.999512\)

\(\cos^{-1}\cos \theta = \cos^{-1}0.999512\)

\( \theta = \cos^{-1}0.999512\)

Answer 13

do you knw how to take \(\cos^{-1} 0.999512\) in your calcualtor ? :)

Answer 14

you may use wolframalpha :-

http://www.wolframalpha.com/input/?i=arccos%280.99951207%29

Answer 15

1.79 = 1.8.

Answer 16

perfect !

Answer 17

Once again, Thank you, very much.

Answer 18

you're wlcme :)