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

Nov 9, 2016

The angle is =48º

#### Explanation:

Let's calculate vecC=vecA-vecB=〈6,1,5〉-〈3,6,-2〉
=〈3,-5,7〉
To calculate the angle $\theta$, we need to calculate the dot product.
vecA.vecC=∥vecA∥*∥vecC∥costheta
Let's calculate the dot product $\vec{A} . \vec{C}$
=〈6,1,5〉.〈3,-5,7〉=18-5+35=48
The modulus of vecA=∥vecA∥=∥〈6,1,5〉∥
$= \sqrt{36 + 1 + 25} = \sqrt{62}$
The modulus of vecC=∥vecC∥=∥〈3,-5,7〉∥
$= \sqrt{9 + 25 + 49} = \sqrt{83}$

$\therefore \cos \theta = \frac{48}{\sqrt{62} \cdot \sqrt{83}} = 0.67$

theta=48º