Question

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

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

20.3°

10.2°

0.2°

30.3°

Answer 1

use u dot v = |u| * |v| * cos(A)
A is the angle you want

Answer 2

42+4*cos(A)
46*cos(A)
what is u dot v?

Answer 3

u dot v means multiply corresponding numbers and add

(x, y) dot (a,b) is x*a+ y*b

Answer 4

| (x,y) |^2 = x^2 + y^2

and | (x,y) | = sqrt(x^2 + y^2)

Answer 5

so your first step is find the "length" of each vector
then the dot product of the vectors.
\[ \cos(A) = \frac{ u \cdot v}{|u|\ |v|} \]

Answer 6

what is
| u | ?
u = <6, -1>

Answer 7

notice that the length squared can be written as
u dot u

Answer 8

so cos(a)=46/|46|??

Answer 9

A perhaps more intuitive way to go:

Find the angle each vector makes with basic trig properties. For example, the vector z = <4,3> has magnitude \[\sqrt{(4-0)^2+(3-0)^2} = 5\] and its angle will be \[\theta = \arctan(3/4) \approx 36.87^\circ\]The difference of the angles will be the angle between them.

Answer 10

I would do it step by step

first what is | u | ?
u = <6, -1>

Answer 11

in other words, what is sqrt(6*6 + -1* -1) ?

Answer 12

sqrt 37

Answer 13

yes, now what is the length of v ?
|v | = ?

Answer 14

sqrt 85?

Answer 15

use the "sum of squares" then take the square root
v = <7, -4>
sqrt(7*7 + -4*-4) = sqrt(49+16)= sqrt(65)

not sure how you got 85.

Answer 16

oops i thought it was <7,-6> gotcha

Answer 17

notice that v dot v gives you
<7, -4> dot <7 , -4>= 7*7 + -4*-4
that is |v|^2
then take the sq root to get the length

Answer 18

next, you need
u dot v = ?

Answer 19

u dot v means multiply corresponding numbers and add

(x, y) dot (a,b) is x*a+ y*b

Answer 20

oh ok one second sorry!

Answer 21

46?

Answer 22

yes,

Now we use the fact that the dot product has 2 definitions.
There is the multiply corresponding numbers and add version
and this one:
\[ u \cdot v = |u| \ |v| \ \cos(A) \]

Answer 23

If we solve for cos(A), we get
\[ \cos(A) = \frac{ u \cdot v}{|u|\ |v|} \]

Answer 24

so 46/sqrt65sqrt37?

Answer 25

what is left is to replace u dot v in the top with 46
and |u| with sqrt(37) and |v| with sq rt(65)

Answer 26

time for a calculator.

Answer 27

you should get
cos(A) = 0.93799...

Answer 28

i got .938

Answer 29

now take the inverse cosine of both sides (in degree mode)
A= acos(0,93799)

Answer 30

oh yeah i got that

Answer 31

okay

Answer 32

and round the answer to the nearest tenth of a degree.

Answer 33

20.3!

Answer 34

success!

Answer 35

Thank you for being so patient with me and helping me understand! :)

Answer 36

whenever you see questions about the angle between vectors, think "dot product"

Answer 37

okay! :D

Answer 38

As a check:
\[|\arctan(\frac{-1}{6}) - \arctan(\frac{-4}{7})| = |-29.74^\circ-(-9.462^\circ)| \approx 20.3^\circ\]