Q9 of 181 Page 1196

Find the distance between the parallel planes 2x + 3y + 4 = 4 and 4x + 6y + 8z = 12.

Formula : The distance between two parallel planes, say


Plane 1:ax + by + cz + d1 = 0 &


Plane 2:ax + by + cz + d2 = 0 is given by the formula



where () are constants of the planes


Therefore ,


First Plane 2x + 3y + 4 = 4


2x + 3y + 4 – 4 = 0 …… (1)


Second plane 4x + 6y + 8z = 12


4x + 6y + 8z - 12 = 0


2(2x + 3y + 4z – 6) = 0


2x + 3y + 4z – 6 = 0 …… (2)


Using equation (1) and (2)


Distance between both planes





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