Final answer:
The distance between the planes 3x + y - 4z = 2 and 3x + y - 4z = 24 is found using the formula for distance between parallel planes and is calculated to be 22 / sqrt(26).
Step-by-step explanation:
The question asks for the distance between two parallel planes. The planes given by the equations 3x + y - 4z = 2 and 3x + y - 4z = 24 are parallel because they have the same normal vector, and the only difference between them is the constant term.
To find the distance (D) between these planes, we use the formula:
D = |c₂ - c₁| / sqrt(a² + b² + c²)
Where:
- a, b, and c are the coefficients of x, y, and z in the plane equation.
- c₁ and c₂ are the constant terms in the plane equations.
In this case:
- a = 3
- b = 1
- c = -4
- c₁ = 2
- c₂ = 24
Plugging these values into the formula gives:
D = |24 - 2| / sqrt(3² + 1² + (-4)²) = 22 / sqrt(9 + 1 + 16) = 22 / sqrt(26)
Thus, the distance between the planes is 22 / sqrt(26).