Find the angle between the two vectors `vecA =hati-2hatj+3hatk and vecB=2hati+hatj+3hatk`.

Explanation

`vecA=hati-2hatj+3hatk ,vecB=2hati+hatj+4hatk`
`vecA*vecB=1xx2+(-2)xx1+3xx4=12`
`|vecA|=A=sqrt(1^2+(-2)^2+3^2)=sqrt(14)`
`|vecB|=B=sqrt(2^2+1^2+4^2)=sqrt(21)`
Now, `vecA*vecB=AB cos theta [theta` is the angle between the vector `vecA and vecB`.]
or, `cos theta =(vecA*vecB)/(AB)=(12)/(7(sqrt(6)))=(2sqrt(6))/(7) or, theta =cos^(-1) ((2(sqrt(6)))/(7))`