Show that the line of intersection of the planes `vecr*(hati+2hatj+3hatk)=0 and vecr*(3hati+2hatj+hatk)=0` is equally inclined to `hati and hatk`. Also find the angleit makes with `hatj`.

Explanation

The line of intersection of the two planes will be perpendicular to the normals to the planes. Hence, it is parallel to the vector `(hati+2hatj+3hatk)xx(3hati+2hatj+hatk)=(-4hati+8hatj-4hatk)`. Now,
`" "(-4hati+8hatj-4hatk)*hati=-4 and (-4hati+8hatj-4hatk)*hatk=-4`
Hence, the line is equally inclined to `hati and hatk`. Also,
`" "((-4hati+8hatj-4hatk))/(sqrt(16+64+16))*hatj=(8)/(sqrt(96))=sqrt((2)/(3))`
If `theta` is the required angle, then `costheta=sqrt((2)/(3)) or theta = cos^(-1) sqrt((2)/(3))`.