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

Jan 17, 2017

$A = < 3 , - 1 , 3 > , B = < 2 , - 2 , 5 > \mathmr{and} C = A - B \implies {\theta}_{A , C} \approx {121.091}^{\circ}$

Explanation:

Algebraically we can get $C = A - B$ by subtracting the components of $B$ from the components of $A$

$\vec{C} = \vec{A} - \vec{B} = < 3 - 2 , - 1 - \left(- 2\right) , 3 - 5 >$

$= < 1 , - 1 + 2 , - 2 > = < 1 , 1 , - 2 >$

Also, we can find the dot product of the vector A anc C

$\vec{A} \cdot \vec{C} = 3 \left(1\right) + \left(- 1\right) \left(1\right) + 3 \left(- 2\right) = 3 \pm 1 \pm 6 = - 4$

Another definition of the dot product is

$\vec{A} \cdot \vec{C} = | \vec{A} | | \vec{C} | \cos \left({\theta}_{A , C}\right)$

So, we need to find $| A |$ and $| C |$

$| \vec{A} | = \sqrt{{3}^{2} + {\left(- 1\right)}^{2} + {3}^{2}} = \sqrt{9 + 1 + 9} = \sqrt{10}$

$| \vec{C} | = \sqrt{{1}^{2} + {1}^{2} + {\left(- 2\right)}^{2}} = \sqrt{1 + 1 + 4} = \sqrt{6}$

$\implies \vec{A} \cdot \vec{C} = | \vec{A} | | \vec{C} | \cos \left({\theta}_{A , C}\right) = \left(\sqrt{10}\right) \left(\sqrt{6}\right) \cos \left({\theta}_{A , C}\right)$

Then we plug in

<=> (vecA*vecC)/((sqrt(10))(sqrt(6))= (-4)/(sqrt(60)$= \frac{- 4}{2 \sqrt{15}}$=(-2)/(sqrt(15)$= \cos \left({\theta}_{A , C}\right)$

$\iff {\cos}^{- 1} \left(- \frac{2}{\sqrt{15}}\right) = \underline{{\theta}_{A , C} \approx {121.091}^{\circ}}$

And we have the angle