Problem 54 - Essential Calculus Early Transcendentals

Question

Find equations of the planes that are parallel to the plane \(x+2 y-2 z=1\) and two units away from it.

Step-by-Step Solution

Answer

The planes are \(x + 2y - 2z = 7\) and \(x + 2y - 2z = -5\).

Step 1: Identify the Plane Equation

The given plane equation is \(x + 2y - 2z = 1\). This plane is in the standard form \(ax + by + cz = d\), where \(a = 1\), \(b = 2\), \(c = -2\), and \(d = 1\). The normal vector to this plane is \(\mathbf{n} = \langle 1, 2, -2 \rangle\).

Step 2: Find Distance Formula

The distance \(D\) between two parallel planes \(ax + by + cz = d_1\) and \(ax + by + cz = d_2\) is given by \(|d_2 - d_1| / \sqrt{a^2 + b^2 + c^2}\). Here, we know \(d = 1\), and we want the distance to be 2 units, so: \(\frac{|d_2 - 1|}{\sqrt{1^2 + 2^2 + (-2)^2}} = 2\).

Step 3: Compute the Normal Vector Magnitude

Calculate the magnitude of the normal vector \(\mathbf{n} = \langle 1, 2, -2 \rangle\). \(\sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1 + 4 + 4} = \sqrt{9} = 3\).

Step 4: Solve for the Distance

Plug the magnitude into the distance formula equation: \(\frac{|d_2 - 1|}{3} = 2\). Solve for \(|d_2 - 1|\), so \(|d_2 - 1| = 6\).

Step 5: Find Possible Values for \(d_2\)

From \(|d_2 - 1| = 6\), we have the two equations \(d_2 - 1 = 6\) and \(d_2 - 1 = -6\). Solving these, we get \(d_2 = 7\) or \(d_2 = -5\).

Step 6: Write the Equations of the Parallel Planes

The two equations for planes parallel and two units away are \(x + 2y - 2z = 7\) and \(x + 2y - 2z = -5\). These are the required planes.

Key Concepts

Parallel Planes

Parallel planes have the same normal vector, which means they have the same orientation in space, even if they are not right next to each other. A key feature of parallel planes is that they never intersect; they are equidistant anywhere due to having constant separation. In context here, we start with the given plane equation:

• Original Plane: \(x + 2y - 2z = 1\)

To find planes parallel to this original plane, we maintain the coefficients of \(x\), \(y\), and \(z\), so our goal is to manipulate only the constant term to achieve additional planes. This is because changing the constant \(d\) moves the plane parallel in space, without altering its overall orientation or direction.
For a problem like finding a parallel plane two units away, it is essential to understand that any plane \(ax + by + cz = d_2\) will be parallel if \(a\), \(b\), and \(c\) are unchanged.
Then, we adjust the \(d_2\) value, calculated as shown further below, to ensure it is precisely two units away.

Distance Formula

The distance formula is a critical calculation in geometry to determine the gap between parallel planes. When dealing with planes, particularly parallel planes defined by equations such as \(ax + by + cz = d_1\) and \(ax + by + cz = d_2\), the formula we use is:

• Distance \(D = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}}\)

This formula effectively gives us the perpendicular distance between two planes. In this case, we want this distance to be 2 units, so we need to solve for \(d_2\) given \(d_1 = 1\) from the original plane. The denominator \(\sqrt{a^2 + b^2 + c^2}\) represents the magnitude of the normal vector, which is consistent across both planes when they are parallel.

Normal Vector

The normal vector is a crucial component in understanding plane geometry. A normal vector is a vector that is perpendicular to a surface. In our context of plane equations like \(ax + by + cz = d\), the components of this normal vector are \(\langle a, b, c \rangle\). For the given plane \(x + 2y - 2z = 1\), the normal vector is:

• \(\mathbf{n} = \langle 1, 2, -2 \rangle\)

The magnitude of this vector is also significant when calculating distances between parallel planes. The magnitude provides the scale needed to compute how far apart the parallel planes are. To find this magnitude, calculate:

• \(\text{Magnitude} = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{9} = 3\)

This magnitude shows the strength or length of the normal vector in space and is necessary to correctly apply the distance formula. The normal vector helps ensure the planes remain parallel, locking their orientation in space.

Plane Equation

A plane equation is a mathematical expression that defines a flat, two-dimensional surface in three-dimensional space. The general form of a plane equation is \(ax + by + cz = d\). Each term serves a specific purpose:

• \(a\), \(b\), and \(c\) are coefficients that define the direction of the normal vector.
• \(d\) is a constant that shifts the plane parallel along its normal direction.

For our exercise with the plane \(x + 2y - 2z = 1\), we ask to find parallel planes two units away. This involves deciding new values for \(d\) such that the distance between the initial and new plane is maintained at 2 units.
Through calculation, adjusting \(d_2\) led us to two solutions: \(7\) and \(-5\). Thus, the plane equations parallel and two units apart from the original are:

• \(x + 2y - 2z = 7\)
• \(x + 2y - 2z = -5\)

Each equation reflects a plane that maintains its direction, as dictated by the normal vector, while the constant term \(d\) ensures their precise placement.