Question

Prove that the planes `x+4y-3=1` and `-3x+6y+7z=0` are perpendicular to each other?

Answer 1

Planes are perpendicular to each other.

Explanation:

The normal to a plane `ax+by+cz+d=0` is `<a,b,c>`

hence normal to `x+4y-3=1` is `<1,4,-3>`

and normal to `-3x+6y+7z=0` is `<-3,6,7>`

and angle between planes is equal to angle between their normals.

As normals are `<1,4,,-3>` and `<-3,6,7>` and if angle between them is `alpha`

then `cosalpha={|n_1.n_2|)/(|n_1||n_2|}`

= `(|1xx(-3)+4xx6+(-3)xx7|)/(sqrt(1^2+4^2+(-3)^2)sqrt((-3)^2+6^2+7^2)`

=`(|-3+24-21|)/(sqrt(1+16+9)sqrt(9+36+49)|)`

= `0`

Hence.`alpha=90^@` and planes are perpendicular to each other.