Find the distance between the planes x+2y-3z=8 and 3x+6y-9z=10
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Distance between two parallel planes , ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0 is |d₁ - d₂ |/√(a² + b² + c²)
Here given two planes x + 2y - 3z = 8
and 3x + 6y - 9z = 10 ⇒3(x + 2y - 3z) = 10
⇒x + 2y - 3z = 10/3
Now, you can both the given planes are parallel
So, distance between it is |8 - 10/3|/√{1² + (2)² + (-3)²} = |14/3|/√14 = √14/3 unit
Here given two planes x + 2y - 3z = 8
and 3x + 6y - 9z = 10 ⇒3(x + 2y - 3z) = 10
⇒x + 2y - 3z = 10/3
Now, you can both the given planes are parallel
So, distance between it is |8 - 10/3|/√{1² + (2)² + (-3)²} = |14/3|/√14 = √14/3 unit
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