# If A = <-7 ,1 ,-3 >, B = <3 ,6 ,1 > and C=A-B, what is the angle between A and C?

Oct 20, 2016

The angle is $0.566$ rad

#### Explanation:

vecC=vecA-vecB=〈-10,-5,-4〉
The dot product is vecA.vecC=∥vecA∥*∥vecC∥*cos(vecA,vecC)

so cos(vecA,vecC)=(vecA,vecC)/(∥vecA∥*∥vecC∥)

∥vecA∥=sqrt(49+1+9)=sqrt59
∥vecC∥=sqrt(100+25+16)=sqrt141
and $\left(\vec{A} , \vec{C}\right) = \left(\left(- 7 \cdot - 10\right) + \left(1 \cdot - 5\right) + \left(- 3 \cdot - 4\right)\right) = 70 - 5 + 12 = 77$
$\cos \left(\vec{A} , \vec{C}\right) = \frac{77}{\sqrt{59} \cdot \sqrt{141}}$
$\left(\vec{A} , \vec{C}\right) = \arccos \left(\frac{77}{\sqrt{59} \cdot \sqrt{141}}\right)$