# If A = <1 ,6 ,9 >, B = <-9 ,4 ,-8 > and C=A-B, what is the angle between A and C?

Nov 15, 2016

The angle is $= 35.6$º

#### Explanation:

Let's start by calculating $\vec{C}$

vecC=vecA-vecB=〈1,6,9〉-〈-9,4,-8〉=〈10,2,17〉

The angle $\theta$ is given by the dot product definition

vecA.vecC=∥vecA∥*∥vecC∥*costheta

vecA.vecC=〈10,2,17〉.〈1,6,9〉=10+12+153=175

The modulus of vecA=∥vecA∥=∥〈1,6,9〉∥=sqrt(1+36+81)=sqrt118

The modulus of vecC=∥vecC∥=∥〈10,2,17〉∥=sqrt(100+4+289)=sqrt393

Therefore, $\cos \theta = \frac{175}{\sqrt{118} \cdot \sqrt{393}} = 0.81$

$\theta = 35.6$º