Question

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

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

Answer 1

here's a page with the formula and an example

http://www.vitutor.com/geometry/vec/angle_vectors.html

Answer 2

Use \[
\cos \theta = \frac{\mathbf{u}\cdot \mathbf v}{\|\mathbf u\|\|\mathbf v\|}
\]

Answer 3

@jim_thomson5190 im lost I don't see how they got the answer

Answer 4

where are you stuck

Answer 5

I work the problem out on the ex and I dont see how they got 45

Answer 6

do you see how they got

cos(alpha) = sqrt(2)/2

??

Answer 7

yes

Answer 8

well you want the value of alpha, not cos(alpha)

Answer 9

so you must apply the arccosine to both sides

Answer 10

that's how they got 45 degrees

arccos(sqrt(2)/2) = 45 degrees

Answer 11

ok so I do the same for my problem but I think Im getting the wrong answer. arccos((32√(41))/(205)) = 0.0312

Answer 12

what did you get for the dot product

Answer 13

32

Answer 14

@jim_thompson5190?

Answer 15

the length of vector u is sqrt(41)

the length of vector v is 5

so this means


cos(theta) = 32/(5*sqrt(41))

theta = arccos(32/(5*sqrt(41)))

theta = 1.78991060824608 degrees

Answer 16

That is radians, not degrees.

Answer 17

no 1.78991060824608 is in degrees

Answer 18

thank u

Answer 19

That is weird.

Answer 20

the two vectors are actually that close together (their lengths are)