Answer to Question #91560 in Analytic Geometry for Ra

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

Solution:

7x + 4y - 4z + 30 = 0 Plane ------- (1) 

36x- 51y + 12z + 17 = 0 Plane ------- (2)  

14x + 8y -8z - 12 = 0 Plane ------- (3)  

12x - 17y + 4z - 3 = 0 Plane ------- (4) 

To prove these four planes are fours face of a cuboid, we need to prove following things

(A) Plane (1) and Plane (3) are parallel

(B) Plane (2) and Plane (4) are parallel

(C) Plane (1) or Plane (3) is perpendicular to Plane (2) or Plane (4)

(A) Plane (1) and Plane (3) are parallel

Condition :

$\dfrac{a1}{a2}$ = $\dfrac{b1}{b2}$ = $\dfrac{c1}{c2}$ , $\dfrac{c1}{c2}$ $\neq$ $\dfrac{d1}{d2}$

$\dfrac{7}{14}$ $=$ $\dfrac{4}{8}$ $=$ $\dfrac{-4}{-8}$ , $\dfrac{-4}{-8}$ $\neq$ $\dfrac{30}{-12}$

So Plane (1) and Plane (3) are parallel

(B) Plane (2) and Plane (4) are parallel

$\dfrac{a1}{a2}$ $=$ $\dfrac{b1}{b2}$ $=$ $\dfrac{c1}{c2}$ , $\dfrac{c1}{c2}$ $\neq$ $\dfrac{d1}{d2}$

36/12 = (-51)/(-17) = 12/4 , 12/4 ≠ 17/(-3) 

Plane (2) and Plane (4) are parallel

(C) Plane (1) or Plane (3) is perpendicular to Plane (2) or Plane (4)

(i) Condition for Plane(1) is perpendicular to Plane(4)

(Normal of Plane 1) * (Normal of Plane 2) = 0 (here the dot product of vectors is applied)

<7, 4, -4> * <36, -51, 12>

252 - 205 -48

=0

So Plane(1) is perpendicular to Plane(2)

(ii) Condition for Plane(3) is perpendicular to Plane(4) (here the dot product of vectors is applied)

(Normal of Plane 3) * (Normal of Plane 4) = 0

<14, 8, -8>* <12, -17 , 4>

168 - 136 -32

=0

So Condition for Plane(3) is perpendicular to Plane(4)

Now we can say these four planes are four faces of cuboid.