Find the angle between the planes `vecr.(3hati-4hatj+5hatk)=0` and `vecr.(2hati-hatj-2hatk)=7`.

Answer

`pi/2`

Explanation

The given planes are `3x-4y+5z=0` and `2x-y-2z=7`.
The d.r.'s of normals to these planes are 3,-4,5 and 2,-1,2.
`therefore cos theta =|(3 xx 2)+(-4) xx(-1)+5 xx (-2)|/(sqrt(3^(2)+(-4)^(2)+5^(2))sqrt(2^(2)+(-1)^(2)+(-2)^(2))=0 rArr theta=pi/2`.