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

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 $\vec{y} = < {y}_{1} , {y}_{2} , {y}_{3} >$ )
$\vec{x} \cdot \vec{y}$ is equal to two things:
1. = $| \vec{x} | | \vec{y} | \cos \left(\Theta\right)$, where $\Theta$ denotes the angle between the vectors
2. = ${x}_{1} {y}_{1} + {x}_{2} {y}_{2} + {x}_{3} {y}_{3}$

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. $\sqrt{{2}^{2} + {1}^{2} + {6}^{2}} \sqrt{{\left(- 3\right)}^{2} + {\left(- 1\right)}^{2} + {3}^{2}} \cos \left(\Theta\right)$
This is equivalent to $\sqrt{41} \sqrt{19} \cos \left(\Theta\right)$
Which is: $\sqrt{779} \cos \left(\Theta\right)$

1. $2 \times \left(- 3\right) + 1 \times \left(- 1\right) + 6 \times 3$
This is just $11$

So therefore: $\sqrt{779} \cos \left(\Theta\right) = 11$
$C o s \left(\Theta\right) = \frac{11}{\sqrt{779}}$
$\Theta = 1.17$ radians

Hope this helped.