What is the distance between the planes -6x+9y-9z=27 and 2x-3y+3z=12 ?

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

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To calculate a distance between two planes, the planes must be parallel. We'll verify if the given planes are parallel.

We'll divide the first equation -6x+9y-9z=27 by 3:

-2x + 3y - 3z = 9

We'll multiply by -!:

2x - 3y + 3z = -9

We notice that the coefficients of the first equation of the plane are the same with the coefficients of the second plane, therefore the planes are parallel.

The distance between the given planes is the distance from a point located in a plane to the other plane.

Now, we'll choose a point located in the first plane: M(0,0,2).

The distance formula is:

d(M,P2) = |2*0 - 3*0 + 3*2 + 9|/sqrt(2^2 + (-3)^2 + 3^2)

d(M,P2) = 9/sqrt22

d(M,P2) = 9sqrt22/22

The requested distance between the given planes is d(M,P2) = 9sqrt22/22.

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