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

Mar 12, 2017

The angle is $= 54.2$º

Explanation:

Let's start by calculating

$\vec{C} = \vec{A} - \vec{B}$

vecC=〈1,6,9〉-〈-9,-6,7〉=〈10,12,2〉

The angle between $\vec{A}$ and $\vec{C}$ is given by the dot product definition.

vecA.vecC=∥vecA∥*∥vecC∥costheta

Where $\theta$ is the angle between $\vec{A}$ and $\vec{C}$

The dot product is

vecA.vecC=〈1,6,9〉.〈10,12,2〉=10+72+18=100

The modulus of $\vec{A}$= ∥〈1,6,9〉∥=sqrt(1+36+81)=sqrt118

The modulus of $\vec{C}$= ∥〈10,12,2〉∥=sqrt(100+144+4)=sqrt248

So,

costheta=(vecA.vecC)/(∥vecA∥*∥vecC∥)=100/(sqrt118*sqrt248)=0.58

$\theta = 54.2$º