Question

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

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

Answer 1

Dot-product...\[\bf \theta=\cos^{-1}\left( \frac{ \vec{u} \frac{}{} \vec{v} }{ |\vec{u}||\vec{v}| } \right)\]

Answer 2

now find the dot product of the two vectors and plug it in to the numerator. then find the magnitudes of each vector and multiply them then plug that in to the denominator. then finally evaluate the inverse cosine to get the angle between the vectors..

Answer 3

what are the signs on the bottem?

Answer 4

\(\bf \cfrac{\text{dot product}}{\text{product of magnitudes}}\)

Answer 5

like can you just explain it all. not the answer. but finding the dot product of the two vectors on top...is that u^2 and v^2?
and i don't know how to find the product of magnitudes

Answer 6

which begs the question why you have this exercise :(

Answer 7

http://www.mathsisfun.com/algebra/vectors-dot-product.html

Answer 8

i just haven't been taught it well and forget it

Answer 9

ok i get the dot product, just not the magnitudes

Answer 10

the top is 32

Answer 11

magnitude of a vector <a, b> = \(\bf \sqrt{a^2+b^2}\)

Answer 12

5√41

Answer 13

32/5√41

right?

Answer 14

yes, that'd be the cosine for the angle, then you'd just need to get the \(\bf cos^{-1}\) to get the angle