Find the angle between the planes whose vector equations are `vecr.(2hati+2hatj-3hatk)=5` and `vecr.(3hati-3hatj+5hatk)=3`.

Explanation

We know that the acute angle `theta` between the planes `vecr.vecn_(1)=q_(1)` and `vecr.vecn_(2)=q_(2)` is given by
`costheta=|vecn_(1).vecn_(2)|(|vecn_(1)||vecn_(2)|)`.
Here, `vecn_(1)=(hati+hatj+2hatk)` and `n_(2)=(2hati-hatj+hatk)`.
`therefore costheta= |(2hati+2hatj-3hatk).(3hati-3hatj+5hatk)|/|(2hati+2hatj-3hatk)(3hati-3hatj+5hatk}`
`=|2 xx 3 +2 xx (-3)+(-3) xx 5|/(sqrt(2^(2)+2^(2)+(-3)^(2))sqrt(3^(2)+(-3)^(2)+5^(2))`
=|-15|/(sqrt(17)sqrt(43))=15/sqrt(731)`.
`rArr theta=cos^(-1)(15/sqrt(731))`.
Hence, the angle between the given planes is `cos^(-1)(15/sqrt(731))`.