If #A = <8 ,9 ,4 >#, #B = <6 ,7 ,4 ># and #C=A-B#, what is the angle between A and C?

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Answer 1 · 2016-08-05T12:03:24

#=18.67^o#.
If the measurement is in the opposite sense, the angle is #180^o-18.67^o=161.33^o#

#C=<8, 9, 4> - <6, 7, 4> = <2, 2, 0>#.

Use the the formula

angle# = theta = arc cos (a^2+c^2-b^2)/(2ac) or pi-theta#

#a = |<8, 9, 4>=sqrt(8^2+9^2+4^2)=sqrt 161#. Likewise,

#b=sqrt.(6^2+7^2+4^2)=sqrt(101)#

#c=sqrt(2^2+2^2+0)=sqrt 8#

Now #theta

#= arc cos ((161+8-101)/(2 X sqrt 161 X sqrt 8) )#

#=18.67^o#

If the measurement is in the opposite sense, the angle is #180^o-

theta=161.33^o#

For checking the answer, you can use what I have used.

#theta= arc sin( |AXC|/(|A||C|).)#