Question

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

u = <2, -4>, v = <3, -8>

Answer 1

\[a\cdot b=|a|*|b|*cos(\theta), where\space \theta\space is\space the\space angle\space between\space the\space vectors\]

Answer 2

I tried everything, but I'm not getting the answer I need...

Answer 3

\[ \cos \theta = \frac {a \cdot b} {|a| |b| } \]
take the arccos to find \[ \theta\]

Answer 4

I know that's how you solve it, however my issue is that I am not getting the right answer. Obviously, I'm doing something wrong.

Answer 5

http://www.wikihow.com/Find-the-Angle-Between-Two-Vectors

Answer 6

\[a \cdot b = 38\\
|a|*|b| = \sqrt{4+16}*\sqrt{73}=2\sqrt{5}\sqrt{73
}\\\theta=\cos^{-1}(\frac{19}{\sqrt{5}\sqrt{73}})\approx0.10487\]

Answer 7

@Loveiskey18 do you understand?

Answer 8

\[
\Large \cos \theta = \frac {a \cdot b} {|a| |b| }\\

\Large \cos \theta = \frac {(2*3)+(-4*-8)} {\sqrt{2^2+3^2}*\sqrt{4^4+8^2} }\\
\]

Answer 9

your vectors are wrong on the bottom @cinar
shuold be
\[\sqrt{2^2+4^2}\\and\\\sqrt{3^2+8^2}\]

Answer 10

@zzr0ck3r Sorry I took so long, I was attempting in understanding this but I am not. I just keep getting stuck at the part where you input the vectors.

Answer 11

you are right

Answer 12

where I input the vectors?
do you know what the dot product is, and the TWO different ways it is defined?

Answer 13

you need to know this before you can answer this.

Answer 14

@Loveiskey18 this will go allot faster if you answer:)

Answer 15

I'm sorry, I was attempting it on my own...again

Answer 16

ok just stay with me and ill walk you through it?

Answer 17

My ending results: arccos (38/√(13)√(-80))

Answer 18

nope

Answer 19

can you tell me what \[<2,-4>\cdot<3,-8>=?\]

Answer 20

arccos (38/√(-1040)

Answer 21

square root of negative number is the first thing that I can see is wrong

Answer 22

please just stick with me and stop posting answers:)

Answer 23

Laughing out loud, okay.

Answer 24

\[<2,-4>\cdot<3,-8>=?\]

Answer 25

I think you did this part already, its 38 right?

Answer 26

Yes.

Answer 27

do you understand \[\theta=cos^{-1}(\frac{a\cdot b}{|a||b|})\]

Answer 28

Yes.

Answer 29

so we need |a|*|b| correct?

Answer 30

Correct.

Answer 31

\[a=<2,-4>
\\|a| =\sqrt{2^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}=\sqrt{4*5}=\sqrt{4}\sqrt{5}=2\sqrt{5}\\correct?\]

Answer 32

Yes.

Answer 33

\[b=<3,-8>\\|b|=\sqrt{3^2+(-8)^2}=\sqrt{9+64}=\sqrt{73}\]correct?

Answer 34

Yes.

Answer 35

if there is anything you dont understand just tell me, I am in no rush

Answer 36

I acknowledge everything completely.

Answer 37

ok then we have\[\theta=\cos^{-1}(\frac{a\cdot b}{|a|*|b|})=cos^{-1}(\frac{38}{2\sqrt{5}\sqrt{73}})=\cos^{-1}(\frac{19}{\sqrt{5*73}})=\cos^{-1}(\frac{19}{\sqrt{365}})\]

Answer 38

\[\approx0.10487\]

Answer 39

To be honest I do understand all of this, and actually I already obtained that answer before I asked for help, however I wasn't sure. The problem is... that is not the answer I am looking for, well it is. I just do not know how to convert it into degrees or the appropriate answer:

My choices:
3.0°

6.0°

-7.0°

16.0°
Nonetheless, I am very much grateful for you putting in the time and energy to help me.

Answer 40

we got radians as the answer\[radian\space answer*\frac{180}{\pi}=degree\space answer\]

Answer 41

\[.10487*\frac{180}{\pi}\approx6.0081\approx6\]

Answer 42

next time you are confused about radians to degree
imagine if you have pi, then you know that is 180 degrees
well
what can we multiply pi by to get 180?
pi(180/pi) = 180

Answer 43

and if you have 180 degrees you know that is pi
well what can we multiply 180 by to get pi?
180*(pi/180)=pi

Answer 44

ok so, after all that....are we good to go now?

Answer 45

Thank You so much :D Do you mind verifying a couple questions for me, please. Just tell me if they are right or wrong.