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

Apr 8, 2016
• The angle between $\vec{A}$ and $\vec{C}$ =90

#### Explanation:

• $C = A - B = < - 4 , 2 , - 3 >$
• If you consider that the vectors are in rectangular form (the vectors' tail is on the origin), the vector difference between two vectors $A - B$ goes from the head of $\vec{A}$ to the to the head of $\vec{B}$ forming a triangle $A B C$.
• By finding the magnitude of the vectors, you can find the sine of the angle between to $\vec{A}$ and $\vec{C}$ $\left(\angle A C\right)$.
• $| A | = \sqrt{{2}^{2} + {3}^{2} + {\left(- 1\right)}^{2}} = 3.74$
$| B | = \sqrt{{6}^{2} + {1}^{2} + {2}^{2}} = 6.40$
$| C | = \sqrt{{\left(- 4\right)}^{2} + {2}^{2} + {\left(- 3\right)}^{2}} = 5.39$

• Using the sine rule:
$| A \frac{|}{\sin \angle B C} = | C \frac{|}{\sin \angle A B} = | B \frac{|}{\sin \angle A C}$
$\frac{3.74}{\sin \angle B C} = \frac{5.39}{\sin \angle A B} = \frac{6.40}{\sin \angle A C}$

• Doing the algebra:
$\angle A B = 180 - \left(\angle A C + \angle B C\right)$
$\sin \angle A B = \sin \left(180 \left(\angle A C + \angle B C\right)\right) = \sin \left(\angle A C + \angle B C\right)$

$6.40 \sin \angle A B = 5.39 \sin \angle A C$
$6.40 \sin \left(\angle A C + \angle B C\right) = 5.39 \sin \angle A C$

$\sin \left(\angle A C + \angle B C\right) = 8.41 \sin \angle A C$

$\cancel{\sin \angle A C} \cdot \cos \angle B C + \cancel{\cos \angle A C \cdot \sin \angle B C} = 8.41 \cancel{\sin \angle A C}$

$\cos \angle B C = 8.41 \implies \angle B C = 32.75$

$\cos \angle A C \cdot \sin \angle B C = 0$ ; $\sin \angle B C \ne 0$

$\therefore \cos \angle A C = 0 \implies \angle A C = 90$

• The angle between $\vec{A}$ and $\vec{C}$ =90