If vector a=-8(vector b) and vector c=7(vector b) and what is angle between a and c ??

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If vector a=-8(vector b) and vector c=7(vector b) and what is angle between a and c ??

Mathematics
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cos(t) = a*c/|a||c|
b*b = |b|^2
<-8xb, -8yb> < 7xb, 7yb> -------------- -xb + -yb = -1

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Other answers:

thats messed up lol
yeah.. i didnt really understand that..
what do u mean in ur 3rd reply..??
a*c = -8 * 7
-56x^2b + -56 y^b is the only values we can get from it right?
-56|b|^2 is the top value
the bottom is -8|b| * 7|b| = -56|b|^2 cos(t) = 1
t = 90 begrees right?
so hard to type in the dark lol
or it might just be 180 degrees; since a and c are scalars of b and they are facing inopposite directions.
I worked it similarly. I don't know if we got same place. \[-8b.-7b =\left| a \right|\left| c \right|\cos \theta\] \[-8b.-7b =8b7b \cos \theta\] \[-1=\cos \theta\] \[\cos^{-1} -1=\theta\] \[\theta = 3.14 (radians)\]
I didn't dot the left hand side. May be that is a mistake that needs to be cleaned up. I should go to sleep now.

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