Math, asked by stevenbablu, 2 months ago

The distance between two parallel planes 2x – 3y + 6z + 1 = 0 and 4x - 6y + 12 z – 5 = 0 is

Answers

Answered by Anonymous

1

Answer:

The given equations of the plane are 2x−3y+6z+12=0 and 2x−3y+6z+−2=0.

The distance between the planes is a2+b2+c2∣d1−d2∣

=22+32+62∣12−(−2)∣

= 714

= 2

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