Question

What is the angle between `<-2,-5,6>` and `< -1,-1,2 >`?

Answer 1

`alpha~=15^o`

Explanation:

`1)" Find the magnitude of<-2,-5,6>"`
`||A||=sqrt((-2)^2+(-5)^2+6^2)" "A=sqrt65`
`2)" Find the magnitude of <-1,-1,2>"`
`||B||=sqrt((-1)^2+(-1)^2+2^2)" "B=sqrt6`
`3)" Find dot product of <-2,-5,6> and <-1,-1,2>"`
`A * B =(-2*-1)+(-5*-1)+6*2`
`A*B=+2+5+12=19`
`A.B=||A||*||B||*cos alpha`
`19=sqrt65*sqrt6*cos alpha`
`19=sqrt (65*6)*cos alpha`
`cos alpha=19/(19,749)`
`cos alpha=0,962`
`alpha~=15^o`