# Find the distance between the planes 8y - 12x + 4z= 20 and 10y - 15x + 5z= 20.

lemjay | High School Teacher | (Level 3) Senior Educator

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To solve,  determine if the planes are parallel, coincident or if they intersect.

To do so, simplify each equation.

For plane 1,

`-12x + 8y +4z = 20`

divide both sides by the GCF which is 4.

`(-12x+8y+4z)/4=20/4`

`-3x+2y+z=5`

And for plane 2,

`-15x+10y+5z=20`

divide both sides by the GCF too.

`(-15x+10y+5z)/5=20/5`

`-3x+2y+z=4`

Since the simplified form of the two planes have same coefficients of x, y and z , but with different constants, this means that they are parallel.

So, to solve for the distance between parallel planes, apply the formula:

`d= |(D_2-D_1)/sqrt(A^2+B^2+C^2)|`

`d=|(4-5)/sqrt((-3)^2+2^2+1^2)|=|(-1)/sqrt15|=|-1/sqrt15|`

Rationalize the denominator.

`d=|-1/sqrt15*sqrt15/sqrt15|=|-sqrt15/15|`

Then, take the absolute value.

`d=sqrt15/15`

Hence, the distance between the  given planes is `sqrt15/15` units.