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)