Final answer:
The distance between the point (-3, 3, -2) and the plane 2x + y - 3z = 5 is calculated using the point-to-plane distance formula, which yields approximately 0.53 units.
Step-by-step explanation:
To find the distance between a point and a plane in three-dimensional space, you can use the point-to-plane distance formula:
D = |Ax + By + Cz + D| / √(A² + B² + C²), where:
A, B, C are the coefficients of the plane equation.
D is the constant term in the plane equation.
(x, y, z) is the point in space.
For the plane 2x + y - 3z = 5 and the point (-3, 3, -2), the distance can be calculated as follows:
D = |(2)(-3) + (1)(3) - (3)(-2) - 5| / √(2² + 1² + (-3)²) = | -6 + 3 + 6 - 5 | / √(4 + 1 + 9) = |-2| / √14 = 2 / √14 = 2 / 3.74 = 0.53 (approximately)
Therefore, the distance between the point and the plane is approximately 0.53 units.