Question

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

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

Answer 1

@mathmate

Answer 2

@UnkleRhaukus

Answer 3

HI!!

Answer 4

first you need the dot product,
then you have to divide by the product of the absolute values
then take the arc cosine

Answer 5

I'm not sure how to do this, I'm new to it

Answer 6

\[\large <x_1,y_1>\cdot<x_2,y_2>=x_1x_2+y_1y_2\]

Answer 7

so in your case \[(-5)(-4)+(-4)(-3)\]

Answer 8

Okay I got 32

Answer 9

then you need the absolute values \[||<x,y>||=\sqrt{x^2+y^2}\]

Answer 10

you need them for each one

Answer 11

How do I set it up?

Answer 12

Like the square root of -5^2+-4?

Answer 13

\[\sqrt{5^2+4^2}\] for the first one

Answer 14

-4^2*

Answer 15

for the second \[\sqrt{4^2+3^2}\]

Answer 16

you should get \(\sqrt{41}\) and \(5\) respectively

Answer 17

Okay thank you

Answer 18

then \[\cos(\theta)=\frac{32}{5\sqrt{41}}\]

Answer 19

making \[\theta=\cos^{-1}(\frac{32}{5\sqrt{41}})\]

Answer 20

Is it 1.78

Answer 21

Choices:
-9.1°
1.8°
0.9°
11.8°

Answer 22

Is it 1.8?

Answer 23

that is what i get, yes
http://www.wolframalpha.com/input/?i=arccos(32%2F(5sqrt(41)))