Answer:
Therefore, the distance between the given planes π1: -2x + 2y - 6z + 5 = 0 and π2: x - y + 3z + 8 = 0 is 22 units.
Explanation:
To determine the distance between two planes, we need to find the perpendicular distance between them.
1. First, we need to find the normal vectors of both planes. The normal vector of a plane is the coefficients of x, y, and z in the plane's equation.
For π1: -2x + 2y - 6z + 5 = 0, the normal vector is (-2, 2, -6).
For π2: x - y + 3z + 8 = 0, the normal vector is (1, -1, 3).
2. Next, we will calculate the dot product of the two normal vectors. The dot product of two vectors is calculated by multiplying their corresponding components and summing them.
Dot Product = (-2)(1) + (2)(-1) + (-6)(3)
Dot Product = -2 - 2 - 18
Dot Product = -22
3. The magnitude (or length) of the dot product will give us the distance between the planes. The magnitude of a vector is calculated using the formula sqrt(x^2 + y^2 + z^2), where x, y, and z are the components of the vector.
Distance = magnitude of the dot product = sqrt((-22)^2)
Distance = sqrt(484)
Distance = 22