Question

Find the given distances. The distance between the parallel planes \(2(x-1)+2(y+1)+(z-2)=0\) and $$ 2(x-3)+2(y-1)+(z-3)=0 $$

Short answer

The distance between the planes is \(\frac{7}{3}\).

Step-by-step solution

Understand the General Form of a Plane Equation

The general form of the plane equation is \(Ax + By + Cz + D = 0\), where \(A\), \(B\), and \(C\) are the coefficients of the plane's normal vector \(\vec{n} = (A, B, C)\). For the first plane, it is \(2(x-1) + 2(y+1) + (z-2) = 0\), which can be expanded to form \(2x + 2y + z = 4\). Similarly, the second plane can be expanded from \(2(x-3) + 2(y-1) + (z-3) = 0\) to \(2x + 2y + z = 11\).

Confirm Planes are Parallel

Planes are parallel if their normal vectors are collinear (or the same). The normal vector in both expanded plane equations \(2x + 2y + z\) is \(\vec{n} = (2, 2, 1)\). Since the normal vectors are identical, the planes are parallel.

Use Distance Formula for Parallel Planes

The distance \(d\) between two parallel planes \(Ax + By + Cz + D_1 = 0\) and \(Ax + By + Cz + D_2 = 0\) is given by \(d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}}\). Here, \(D_1 = -4\) for the plane \(2x + 2y + z = 4\) and \(D_2 = -11\) for the plane \(2x + 2y + z = 11\).

Calculate the Distance

Plug the values into the formula: \[d = \frac{|-11 - (-4)|}{\sqrt{2^2 + 2^2 + 1^2}} = \frac{|-11 + 4|}{\sqrt{4 + 4 + 1}} = \frac{| -7|}{\sqrt{9}} = \frac{7}{3}\] Thus, the distance between the parallel planes is \(\frac{7}{3}\).

Key concepts

Normal Vector

Understanding the normal vector is crucial when dealing with plane equations. A normal vector is a vector that is perpendicular to a surface or a plane in 3D space. For a plane described by the equation \(Ax + By + Cz + D = 0\), the normal vector \((A, B, C)\) represents the direction that is orthogonal to any vector lying on the plane.
For the exercise, both planes have the same normal vector, \(\vec{n} = (2, 2, 1)\). This not only helps us identify that these planes are parallel, but it also simplifies the process if you need to perform calculations involving the plane, such as finding the distance between parallel planes, or determining angles with other geometric shapes.

Plane Equation

The equation of a plane in the general form \(Ax + By + Cz + D = 0\) is a simple yet powerful tool. Each component has a specific role:

• extbf{A, B,} and extbf{C} are coefficients that together form the normal vector.
• extbf{x, y,} and extbf{z} are the variables representing the coordinates in 3D space.
• extbf{D} is a scalar that adjusts the plane's position along its normal vector.

The given planes in the exercise are first presented in factored form: \(2(x-1)+2(y+1)+(z-2)=0\) and \(2(x-3)+2(y-1)+(z-3)=0\).
To understand and work with these equations efficiently, we convert them to the expanded form \(2x + 2y + z = 4\) and \(2x + 2y + z = 11\). This makes it easier to identify important features like the normal vector and allows for direct use of the distance formula.

Distance Formula

When you need to find the distance between two parallel planes, the distance formula is your go-to tool. The formula is given as \[d = \frac{|D_2 - D_1|}{\sqrt{A^2 + B^2 + C^2}}\]where \(Ax + By + Cz + D_1 = 0\) and \(Ax + By + Cz + D_2 = 0\) are the equations of the parallel planes, and \(\vec{n} = (A, B, C)\) is the common normal vector.
In the exercise, the values were:

• First Plane: \(D_1 = -4\)
• Second Plane: \(D_2 = -11\)
• Normal Vector Coefficients: \( (A, B, C) = (2, 2, 1) \)

The distance is calculated as:\[\frac{| -11 + 4 |}{\sqrt{2^2 + 2^2 + 1^2}} = \frac{7}{3}\]This formula offers a straightforward way to measure how far apart the parallel planes are.

Parallel Planes

Parallel planes are planar surfaces that never intersect, analogous to how parallel lines behave in 2D space. When two planes are parallel, their normal vectors are identical or scalar multiples of each other.
In our exercise, checking that both planes share a normal vector \(\vec{n} = (2, 2, 1)\) confirms their parallelism. This property is key because it allows us to apply the distance formula easily.
Planes always maintain a constant distance apart, which is why the normal vector and the position in the equation are crucial for calculations. Once you recognize the parallel nature via identical normal vectors, you can explore spatial relationships, effectively moving on to other calculations, like the distance, without complication.