# If A = <8 ,9 ,4 >, B = <6 ,7 ,-1 > and C=A-B, what is the angle between A and C?

Oct 31, 2016

The angle is 42.2º

#### Explanation:

Let's start by calculating vecC=〈8,9,4〉-〈6,7,-1〉=〈2,2,5〉
The the angle is calculated from the definition of the dot product
vecA.vecC=∣vecA∣∣vecC∣cos theta
So, costheta=(vecA.vecC)/(∣vecA∣∣vecC∣)

vecA.vecC=〈8,9,4〉.〈2,2,5〉=16+18+20=54

∣vecA∣=sqrt(8^2+9^2+4^2)=sqrt161

∣vecC∣=sqrt(4+4+25)=sqrt33

$\cos \theta = \frac{54}{\sqrt{161} \cdot \sqrt{33}} = 0.741$

theta=42.2º