Problem 78 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the plane each of whose points is equidistant from \((-2,1,4)\) and \((6,1,-2)\).

Step-by-Step Solution

Answer

The equation of the plane is \( 8x - 6z = 10 \).

Step 1: Identify the Midpoint of Given Points

First, we need to find the midpoint of the two points \(P_1 = (-2, 1, 4)\) and \(P_2 = (6, 1, -2)\). This midpoint will be a point on the plane. The formula for the midpoint \(M\) is given by: \[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \] Substituting in our points, we have: \[ M = \left( \frac{-2 + 6}{2}, \frac{1 + 1}{2}, \frac{4 - 2}{2} \right) = (2, 1, 1) \] Thus, the midpoint is \(M = (2, 1, 1)\).

Step 2: Find the Direction Vector Between the Points

The vector from \(P_1 \) to \(P_2 \) can be found using the formula: \[ \vec{d} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) \] Substituting the points into this formula gives: \[ \vec{d} = (6 - (-2), 1 - 1, -2 - 4) = (8, 0, -6) \] Thus, the direction vector is \(\vec{d} = (8, 0, -6)\).

Step 3: Derive the Plane Equation

The equation of the plane can be derived using the normal vector of the plane, which is in the direction of \(\vec{d} = (8, 0, -6)\). The general form of the plane equation is: \[ Ax + By + Cz = D \] where \( (A, B, C) \) is the normal vector. In this case, the normal vector \(\vec{d} = (8, 0, -6)\) gives \(A = 8, B = 0, C = -6\). Using the midpoint \(M = (2, 1, 1)\), substitute these into the plane equation \((8)(x - 2) + (0)(y - 1) + (-6)(z - 1) = 0\). Simplifying, we get: \[ 8(x - 2) - 6(z - 1) = 0 \] \[ 8x - 16 - 6z + 6 = 0 \] Thus, \[ 8x - 6z = 10 \] is the equation of the plane.

Key Concepts

Midpoint Calculation

When you have two points in a 3D space and you want to find a point that lies exactly halfway between them, you calculate the midpoint. The midpoint gives you a point that is equidistant from each of the original points. The formula to find the midpoint \( M \) between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \) is:

• \( M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2} \right) \)

To help visualize this, imagine you are drawing a line directly connecting the two points. The midpoint would be right in the center of this line. For example, to find the midpoint of points \( P_1 = (-2, 1, 4) \) and \( P_2 = (6, 1, -2) \), you simply plug the respective coordinates into the midpoint formula. Each coordinate is calculated by adding the coordinates of the two points together and dividing by two:

• The x-coordinate: \( \frac{-2 + 6}{2} = 2 \)
• The y-coordinate: \( \frac{1 + 1}{2} = 1 \)
• The z-coordinate: \( \frac{4 - 2}{2} = 1 \)

Hence, the midpoint \( M \) is \((2, 1, 1)\).

Direction Vector

A direction vector—a useful tool in geometry—helps indicate the path or direction between two points in space. It essentially captures the difference in position from the starting point to the ending point. The formula for the direction vector \( \vec{d} \) from point \( P_1 = (x_1, y_1, z_1) \) to point \( P_2 = (x_2, y_2, z_2) \) is:

• \( \vec{d} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) \)

Think of it as an arrow pointing from the starting location to the destination. For example, if you calculate the direction vector from point \( P_1 = (-2, 1, 4) \) to point \( P_2 = (6, 1, -2) \), you subtract each corresponding coordinate like so:

• \( x_2 - x_1 = 6 - (-2) = 8 \)
• \( y_2 - y_1 = 1 - 1 = 0 \)
• \( z_2 - z_1 = -2 - 4 = -6 \)

The vector is then \( \vec{d} = (8, 0, -6) \). It's a fascinating way to convey which direction and how far one point is from another in three-dimensional space.

Plane Equation Derivation

The equation of a plane is derived using a normal vector and a point on the plane. The normal vector is perpendicular to the plane and in this problem, it arises from the direction vector between the given points. For a plane with a normal vector \( (A, B, C) \), the general formula is:

• \( Ax + By + Cz = D \)

Here, the given direction vector \( (8, 0, -6) \) serves as the normal vector with \( A = 8, B = 0, C = -6 \). To find the constant \( D \), substitute in a point on the plane, such as the midpoint \( M = (2, 1, 1) \):

• Substitute into the plane equation: \( 8(x - 2) + 0(y - 1) - 6(z - 1) = 0 \)
• This simplifies to \( 8x - 16 - 6z + 6 = 0 \)

Thus, simplifying further gives us:

• \( 8x - 6z = 10 \)

This represents the final equation of the plane—showcasing how each point on it is equidistant from the two original points. This smooth process elegantly illustrates the power of geometric properties in three-dimensional analysis.