Answer to Question #92451 in Analytic Geometry for Palash Ganguli

Question #92451
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.
1
Expert's answer
2019-08-09T10:15:46-0400
  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.



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