Question

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

Answer 1

U=<-5,8> V=<-4,8>

Answer 2

Are these coordinates cartesian? If so you should consider using the tan(theta)=opposite/adjacent, and solve for the angle for the two vectors, and then subtract the two.

Answer 3

I don't know what Cartesian means. How do I find the opposite and adjacent

Answer 4

dot them.

Answer 5

you know what the dot prod is, right?

Answer 6

<20,64>

Answer 7

close, but nah

<ax, ay>•<bx,by> = ax*bx + ay*by

two vectors are combined by the dot or ***scalar product*** to give you a scalar, ie a number.

Answer 8

so A • B = <ax, ay>•<bx,by> = ax*bx + ay*by
but A•B also = |A||B| cos theta

hope you see a solution now....

Answer 9

Umm -40 +(-32)= -72

Answer 10

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

Answer 11

So of cos^-1 of(-72)

Answer 12

Oh its supposed to be 84 instead right ?

Answer 13

What are you doing?

Answer 14

I did u -5 ×-4 =20
v 8×8 =64
20 + 64

Answer 15

@allydiaz
do you know what |A| or |B| means?

Answer 16

Absolute value?

Answer 17

yes, that's whats missing.

using the u v thingy posted by @wio, you need also calc |u| and |v| before you can get the angle

u•v = |u| |v| cos theta
you now have u•v, now get |u| and |v|

Answer 18

|u|=20
|v| =64?

Answer 19

@IrishBoy123

Answer 20

for u it's\[ \sqrt { -5^2 + 8^ 2 }\]

yes?

Answer 21

Clearly that's (-5)^2