If #A = <2 ,3 ,-4 >#, #B = <5 ,1 ,4 ># and #C=A-B#, what is the angle between A and C?

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Answer 1 · 2016-10-20T03:45:43

#theta ~~ 0.8# radians

To obtain the vector C, subtract the components of B from the respective components of a:

#C = < 2 - 5, 3 - 1, -4 - 4>#

#C = < -3, 2, -8>#

There are two ways to compute #A*C#, the first is to add the products of the respective components of vector A and C:

#A*C = (2)(-3) + (3)(2) + (-4)(-8)#

#A*C = 32#

The second way is:

#A*C = |A||C|cos(theta)#

where #theta# is the angle between the two vectors.

We know the value of #A*C# but we need to compute |A| and |C|

#|A| = sqrt(2^2 + 3^2 + (-4)^2)#

#|A| = sqrt(29)#

#|C| = sqrt((-3)^2 + 2^2 + (-8)^2)#

#|A| = sqrt(77)#

#theta = cos^-1((A*C)/(|A||C|))#

#theta = cos^-1((32)/(sqrt(29)sqrt(77)))#

#theta ~~ 0.8# radians