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

1 Answer
LewisCroney
May 4, 2016

If #C=A-B#
Then #C = <2,1,6> - <5,2,3>#
So #C = <-3,-1,3>#

The dot/scalar product states that for two vectors #vec x#and #vec y# (where #vec x = < x_1 , x_2 , x_3 ># and #vec y = < y_1 , y_2 , y_3 ># )
#vec x * vec y# is equal to two things:
1. = #| vec x| |vec y| cos(Theta)#, where #Theta# denotes the angle between the vectors
2. = #x_1y_1+x_2y_2+x_3y_3#

Therefore it is possible to equate these two and solve for #Theta#

It is important to note, if you didn't know so already, that #|vec x| = sqrt(x_1^2 + x_2^2+ x_3^2)#

So 1. #sqrt(2^2+1^2+6^2)sqrt((-3)^2+(-1)^2+3^2)cos(Theta)#
This is equivalent to #sqrt(41)sqrt(19)cos(Theta)#
Which is: #sqrt(779)cos(Theta)#

  1. #2xx(-3)+1xx(-1)+6xx3#
    This is just #11#

So therefore: #sqrt(779)cos(Theta)=11#
#Cos(Theta)=11/sqrt(779)#
#Theta = 1.17# radians

Hope this helped.

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