Question

Prove that the planes 7x +4y -4z +30 =0, 36x - 51y+12z + 17 = 0 14x +8y- 8z -12 = 0 and 12x -17y + 4z -3 =0 form the four faces of a cuboid

Answer 1

Let plane with equation $7x +4y -4z +30 =0$ be $P_1,$

plane with equation

$14x+8y- 8z -12 = 0 \ or\ dividing\ by\ 2:\ 7x+4y-4z-6=0$ be plane $P_2,$

plane with equation $36x - 51y+12z + 17 = 0$ be $P_3,$

and plane with equation

$12x -17y + 4z -3 =0\ multiplying\ by\ 3:\ 36x-51y+12z-9=0$ be plane $P_4.$

Let general equation of a plane be $Ax+By+Cz+D=0$

Since planes $P_1$ and $P_2$ have equal coefficients A, B and C, $P_1$ is parallel to $P_2.$

Planes $P_3$ and $P_4$ have equal coefficients A, B and C too, therefor $P_3$ is parallel to $P_4.$

Normal vector of planes $P_1$ and $P_2$ is $n_1=(7,4,-4)$

Normal vector of planes $P_3$ and $P_4$ is $n_2=(36,-51,12)$

Since scalar product $n_1\cdot n_2=7\cdot36+4\cdot(-51)-4\cdot 12=0$ ,plane P₁ is orthogonal to plane P₃.

Plane P₂ is parallel to P₁ and plane P₄ is parallel to P₃, therefor P₂ is orthogonal to P₄.

So, we have:

P₁ is parallel to P₂,

P₃ is parallel to P₄,

P₁ is orthogonal to P₃ and P₄,

and P₂ is orthogonal to P₃ and P₄,

therefor these four planes form four faces of cuboid.