Find the acute angle between the planes `barr*(2hati + hatj - hatk) = 3` and `barr*(hati + 2hatj + hatk) =1`.

Explanation

The acute angle `theta` between the planes
`barr*bar(n_(1))=p_(1)andbarr*bar(n_(2))=p_(2)` is given by
`cos theta=|(|bar(n_(1))*bar(n_(2)))/(|bar(n_(1))|*|bar(n_(1))|)|" "...(1)` Here, `bar(n_(1)) =2hati+hatj-hatk and bar(n_(2))=hati+2hatj+hatk`
`therefore bar(n_(1))*bar(n_(2))=(2hati+hatj-hatk)*(hati+2hatj+hatk)`
` = 2(1) + 1(2) + (-1) (1)`
` = 2+ 2-1 =3`
Also, `|bar(n_(1))|=sqrt(2^(2)+1^(2)+(-1)^(2))=sqrt(4+1+1)=sqrt(6)`
and `|bar(n_(2))|=sqrt(1^(2)+2^(2)+1^(2))=sqrt(1+4+1)=sqrt(6)`
`therefore` from (1), we get
` cos theta = |(3)/(sqrt(6) * sqrt(6))| = (1)/(2)`
`therefore cos theta = cos ""(pi)/(3) " " therefore theta = (pi)/(3)`
Hence, the acute angle between the planes is `(pi)/(3)`.