Question

Find the angle between the given vectors to the nearest tenth of a degree.
u = <6, -1>, v = <7, -4>

Answer 1

cosine theta= <u,b>/(sqrt(<u,u>)sqrt(<v,v>))

Answer 2

where <u,b> is an inner product space (in this particular case, <u,u>= u dot u, and <u,v>= u dot v, etc..

Answer 3

That's basically a definition of the dot product algebraically rearranged to give you the angle.

Answer 4

so...
cos theta = ( (6*7)(-1*-4) )/(sqrt (6*7)) (srt (-1*-4))

Answer 5

= (42 * 4)/ (sqrt 42) ( sqrt 4)
= 168 / (sqrt 42) * 2

Answer 6

= 168 / (6.48)(2)
= 168 / 12.96
= 12.96

theta = cos-1 (12.96)

Answer 7

dot product of u, v is 6*7+(-1)(-4). in simple words, x's multiplied together, y;s multiplied together, then you add them up.

Answer 8

Not (x1+x2)*(y1+y2) <-- i think this is the formula you used. :(

Answer 9

= (42 + 4)/ (sqrt (6+7)) ( sqrt (-1+-4))
= 46 / ((sqrt (13))(sqrt(-5))
but that doesn't work since you cn't sqrt -5

Answer 10

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

The point is to work with UNIT vectors. You can do it this way:

\(U = \dfrac{u}{||u||} = \dfrac{<6,-1>}{\sqrt{6^2}+(-1)^2} = \dfrac{<6,-1>}{\sqrt{6^2+(-1)^2}} = \left<\dfrac{6}{\sqrt{37}},\dfrac{-1}{\sqrt{37}}\right>\)

Do the same for v, creating V.

Or, you can do it all at once as in katerine.ok's first post which was not followed very well. You have \(\sqrt{6 + 7} = \sqrt{13}\) where you should have \(\sqrt{37}\).

Answer 11

how did you get sqrt(-5).... you are dotting u by u and v by v... iits always going to yield 0 or greater number...(unless complex number inner product space, which this question is not)

Answer 12

* Not sure what happened to the definition of U. Please ignore the poorly formatted middle expression. I don't feel like retyping the whole thing.