If #A = <4 ,9 ,1 >#, #B = <5 ,8 ,-3 ># and #C=A-B#, what is the angle between A and C?

1 Answer
Sep 18, 2016

#70.9^(circ)#

Explanation:

We have: #A = <4, 9, 1># and #B = <5, 8, - 3>#

First, let's determine #C#:

#=> C = A - B#

#=> C = <4, 9, 1> - <5, 8, - 3>#

#=> C = <(4 - 5), (9 - 8), (1 - (- 3))>#

#=> C = <- 1, 1, 4>#

Then, let's determine the angle between vectors #A# and #C#.

The dot product between two vectors is given as #A cdot B = abs(A) abs(B) cos(theta)#.

We can rearrange this to get:

#=> cos(theta) = (A cdot B) / (abs(A) abs(B))#

#=> cos(theta) = (A cdot C) / (abs(A) abs(C))#

#=> cos(theta) = (<4, 9, 1> cdot <- 1, 1, 4>) / (abs(<4, 9, 1>) abs(<- 1, 1, 4>))#

#=> cos(theta) = ((4 cdot (- 1)) + (9 cdot 1) + (1 cdot 4)) / (sqrt(4^(2) + 9^(2) + 1^(2)) cdot sqrt((- 1)^(2) + 1^(2) + 4^(2)))#

#=> cos(theta) = (- 4 + 9 + 4) / (sqrt(42) cdot sqrt(18))#

#=> cos(theta) = (9) / (sqrt(756))#

#=> cos(theta) = (9) / (sqrt(4 cdot 3 cdot 7 cdot 9))#

#=> cos(theta) = (9) / (6 sqrt(21))#

#=> cos(theta) = (3) / (2 sqrt(21))#

#=> theta = arccos((3) / (2 sqrt(21)))#

#=> theta approx 70.9^(circ)#