# If #A = <2 ,1 ,6 >#, #B = <5 ,1 ,3 ># and #C=A-B#, what is the angle between A and C?

##### 1 Answer

Jan 14, 2016

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#### Explanation:

#vecC = vecA - vecB =( 2 , 1 ,6 ) - (5 , 1 ,3 ) =( - 3 , 0 , 3 ) # To calculate the angle

# (theta) between vecA and vecC # use the following formula :

#costheta =( vecA . vecC)/ (|vecA| |vecC|)# The 'dot product'

# vecA . vecC =(2 , 1 , 6 ) . ( - 3 , 0 , 3 )# = -6 + 0 + 18 = 12

and

# |vecA| = sqrt(2^2 +1^2 + 6^2) = sqrt(4 + 1 + 36 ) = sqrt41#

#|vecC| = sqrt( (- 3 )^2 +0^2 + 3^2) = sqrt(9 + 0 + 9) = sqrt18 # substituting these values into the formula :

# costheta = 12/(sqrt41 xx sqrt18) = 0.4417...#

#rArr theta = 63.8^@ #