Question

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

u = <6, -1>, v = <7, -4>

PLEASE HELP MEEE PLEASEE!!! :(

Answer 1

@saifoo.khan

Answer 2

keep in mind that \(\huge {cos(\theta)=\frac{u \cdot v}{||u||\times ||v||}}\)

Answer 3

once you get that value, just arcCosine it to get the angle between them

Answer 4

I don't really know how to use even that??

Answer 5

How do I know what u and v are?

Answer 6

u = <6, -1>, v = <7, -4> <-----

Answer 7

so I plug in both values? and solve twice?

Answer 8

well, the numerator is the scalar value from the dot-product
the denominator is the multiplication of their magnitudes

Answer 9

<a, b> <c, d> dot product, as you should know is a*c+b*d

Answer 10

and the magnitudes of <a, b> is just \(\bf \sqrt{a^2+b^2}\)

Answer 11

ohh okay so how I set it up would be costheta= 6⋅7||6||×||7||?

Answer 12

hmmm, let's first get the so-called "dot product"
\(\bf <6, -1>\cdot <7, -4> \implies 6\times 7 + -1 \times -4\)

Answer 13

now let's get the magnitudes
\(\bf ||u|| = \sqrt{(6)^2+(-1)^2}\\
||v|| = \sqrt{(7)^2+(-4)^2}\)

Answer 14

so once you get those values, you use that to find the
\(\bf cos(\theta)=\cfrac{u \cdot v}{||u||\times ||v||}\)

Answer 15

let me grab my calcc

Answer 16

so v=65

Answer 17

and u=37

Answer 18

well, square root of that, yes

Answer 19

okay! after I have those two do I plug them in to the costheta formula?

Answer 20

yes lemme check something

Answer 21

okay :)

Answer 22

well, I'm getting a number that doesn't work :/

Answer 23

are the signs correct for => u = <6, -1>, v = <7, -4> ?

Answer 24

yeah they are!! there are four answer choices too!

20.3°

10.2°

0.2°

30.3°

Answer 25

if you got one of those it is probably correct

Answer 26

ohh wait

Answer 27

yes, I get a good number

Answer 28

just had a wrong digit

Answer 29

so just plug them into that?

Answer 30

so the dot product is 46, thus
$$\bf
cos(\theta)=\cfrac{46}{\sqrt{37} \times \sqrt{65}}\\
$$

Answer 31

then arcCosine them both sides
$$
cos^{-1}(cos(\theta))=cos^{-1}\left(\cfrac{46}{\sqrt{37} \times \sqrt{65}}\right)\\
\theta = 20.28^o
$$

Answer 32

I GOT THATTTT!!!! :DDDD

Answer 33

WOOOOOOO THANK YOUUUUUUU :) <3

Answer 34

yw