Answer to Question #92451 in Analytic Geometry for Palash Ganguli

Question

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

Answer 1

1. 7x + 4y - 4z =-30
2. 14x + 8y -8z = 12 <=> 7x + 4y - 4z = 6
3. 36x - 51y + 12z = -17 <=> 12x - 17y +4z = -17/3
4. 12x -17y + 4z = 3

Both planes are parallel if condition suffices

$$\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}.$$

where a₁x + b₁y + c₁z = d₁ and a₂x + b₂y + c₂z = d₂ are planes.

Let's consider 1st and 2nd equation.

$$\frac{7}{7} = \frac{4}{4} = \frac{-4}{-4}$$

Let's consider 3th and 4th equation.

$$\frac{12}{12} = \frac{-17}{-17} = \frac{4}{4}$$

So, 1,2 and 3,4 planes are parallel. A cuboid is a figure bounded by six face parallel to each opposite one. So, the planes form cuboid.