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

1 Answer
May 4, 2016

If C=A-BC=AB
Then C = <2,1,6> - <5,2,3>C=<2,1,6><5,2,3>
So C = <-3,-1,3>C=<3,1,3>

The dot/scalar product states that for two vectors vec xxand vec yy (where vec x = < x_1 , x_2 , x_3 >x=<x1,x2,x3> and vec y = < y_1 , y_2 , y_3 >y=<y1,y2,y3> )
vec x * vec yxy 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.