Given two vectors vecA=2hati+4hatj+5hatkandvecB=hati+3hatj+6veck. Find the product vecA.vecB and the magnitudes of vecAandvecB. What is the angle between them?

Given two vectors vecA=2hati+4hatj+5hatkandvecB=hati+3hatj+6veck. Find

1.	Given two vectors vecA=2hati+4hatj+5hatkandvecB=hati+3hatj+6veck. Find the product vecA.vecB and the magnitudes of vecAandvecB. What is the angle between them?
Answer» SOLUTION :`vecA.vecB=2+12+30=44` Magnitude `A=sqrt(4+16+25)=sqrt45` UNITS Magnitude `B=sqrt(1+9+36)=sqrt46` units The ANGLE between the two VECTORS is given by `theta=cos^(-1)((vecA.vecB)/(AB))=cos^(-1)((44)/(sqrt(45)xxsqrt46))=cos^(-1)((4)/(45.49))=cos^(-1)(0.967)` `:.theta~=15^(@)`.