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

1 Answer
May 27, 2017

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.