# If #A = <3 ,-1 ,5 >#, #B = <5 ,2 ,-9 ># and #C=A-B#, what is the angle between A and C?

##### 1 Answer

#### Explanation:

The angle

# vec A * vec B = ||A|| \ ||B|| \ cos theta #

By convention when we refer to the angle between vectors we choose the acute angle.

So for this problem,

#vec A=<<3,-1,5>>#

#vec B=<<5,2,-9>>#

And so;

# vec C = vec A - vec B #

# \ \ \ \ = <<3,-1,5>> - <<5,2,-9>> #

# \ \ \ \ = <<-2,-3,14>> #

The moduli are given by;

# || vec A || = || <<3,-1,5>> || #

# \ \ \ \ \ \ \ \ = sqrt( (3)^2+(-1)^2+(5)^2 ) #

# \ \ \ \ \ \ \ \ = sqrt( 9 + 1 + 25 ) #

# \ \ \ \ \ \ \ \ = sqrt( 35 ) #

# || vec C || = || <<-2,-3,14>> || #

# \ \ \ \ \ \ \ \ = sqrt( (-2)^2+(-3)^2+(14)^2 ) #

# \ \ \ \ \ \ \ \ = sqrt( 4 + 9 + 196 ) #

# \ \ \ \ \ \ \ \ = sqrt( 209 ) #

And the scaler product is:

# vec A * vec C = || <<3,-1,5>> || * || <<-2,-3,14>> || #

# \ \ \ \ \ \ \ \ \ \ = (3)(-2) + (-1)(-3) + (5)(14)#

# \ \ \ \ \ \ \ \ \ \ = -6 + 3 + 70#

# \ \ \ \ \ \ \ \ \ \ = 67#

And so using

# 67 = sqrt(35) * sqrt(209) * cos theta #

# :. cos theta = (67)/(sqrt(7315))#

# :. cos theta = 0.783371 ... #

# :. theta = 38.43 °# (2dp)

So the acute angle between the vectors is