Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the plane through the origin that is perpendicular to the \(x y\)-plane and the plane \(3 x-2 y+z=4\).

Step-by-Step Solution

Answer

The equation of the plane is \(-2x - 3y = 0\).

Step 1: Understand the Problem

We need to find the equation of a plane that passes through the origin and is perpendicular to two given planes: the xy-plane and the plane described by the equation \(3x - 2y + z = 4\).

Step 2: Identify the Normal Vectors

A plane is defined by its normal vector. The normal vector to the \(xy\)-plane is \(\mathbf{n}_1 = \langle 0, 0, 1 \rangle\). The normal vector to the plane \(3x - 2y + z = 4\) is \(\mathbf{n}_2 = \langle 3, -2, 1 \rangle\).

Step 3: Find the Normal Vector of the Desired Plane

Since the desired plane is perpendicular to both the \(xy\)-plane and the plane \(3x - 2y + z = 4\), its normal vector \(\mathbf{n}\) should be the cross product of \(\mathbf{n}_1\) and \(\mathbf{n}_2\): \[\mathbf{n} = \mathbf{n}_1 \times \mathbf{n}_2\]

Step 4: Calculate the Cross Product

The cross product \(\mathbf{n} = \langle 0, 0, 1 \rangle \times \langle 3, -2, 1 \rangle\) can be calculated as follows: \[ \mathbf{n} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 0 & 0 & 1 \ 3 & -2 & 1 \end{array} \right| = \langle 1 \cdot (-2) - 0 \cdot 1, -(1 \cdot 3 - 0 \cdot 1), 0 \cdot (-2) - 0 \cdot 3 \rangle = \langle -2, -3, 0 \rangle \]

Step 5: Write the Equation of the Plane

The equation of the plane can be written as \(-2x - 3y + 0z = D\), where \(D\) is a constant. Since the plane passes through the origin, \((0, 0, 0)\), substituting these coordinates into the equation gives us \(-2(0) - 3(0) + 0(0) = 0\), which implies \(D = 0\).

Step 6: Formulate the Final Equation

The equation of the plane is: \[-2x - 3y = 0\]. This represents a plane through the origin that is perpendicular to both the given planes as required.

Key Concepts

Cross Product

The cross product is a mathematical operation used primarily with vectors in three-dimensional space. It takes two vectors as input and produces another vector that is perpendicular to both of the input vectors. This output vector is often referred to as a normal vector because it stands at a right angle to a plane formed by the initial two vectors.

To calculate the cross product, consider two vectors represented in 3D as \(\mathbf{a} = \langle a_1, a_2, a_3 \rangle\) and \(\mathbf{b} = \langle b_1, b_2, b_3 \rangle\). Their cross product, \(\mathbf{a} \times \mathbf{b}\), is given by the determinant of a matrix:

• The first row contains the unit vectors, \(\mathbf{i}, \mathbf{j}, \mathbf{k}\).
• The second row contains the components of \(\mathbf{a}\).
• The third row contains the components of \(\mathbf{b}\).

Thus,\[\mathbf{a} \times \mathbf{b} = \left| \begin{array}{ccc} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{array} \right|\]The cross product is essential in finding a normal to a plane when given two non-parallel vectors lying on the plane.

Normal Vector

In plane geometry and three-dimensional space, a normal vector is a vector that is perpendicular to a surface or a plane. Understanding the normal vector is crucial for defining the orientation of a plane.

Given the equation of a plane in the form \(ax + by + cz = d\), \(\langle a, b, c \rangle\) is the normal vector. This vector is significant because it establishes the plane's direction throughout the space. For instance:

• The normal vector \(\langle 0, 0, 1 \rangle\) is perpendicular to the xy-plane, meaning it points directly upwards along the z-axis.
• For the plane equation \(3x - 2y + z = 4\), its normal vector is \(\langle 3, -2, 1 \rangle\).

Normal vectors are pivotal in calculating the cross product, for they provide the direction perpendicular to both vectors forming a plane.

3D Coordinates

Working in three-dimensional coordinates involves points and vectors that use three values to describe their location in space. Each point in 3D space is expressed as \(\langle x, y, z \rangle\), representing its position along the axes:

• \(x\) is the point's position along the x-axis.
• \(y\) is the position along the y-axis.
• \(z\) is the position along the z-axis.

Understanding 3D coordinates is vital because many problems, such as the intersection of geometrical shapes and finding planes, require manipulation in this multi-dimensional space.

When describing vectors, like a normal vector, 3D coordinates provide the direction and magnitude. They help visualize complex spatial relationships and perform operations like the cross product.

Equation of a Plane

The equation of a plane in three-dimensional space is expressed typically in the form \(ax + by + cz = d\), where \(a, b,\) and \(c\) are the coefficients representing the plane's orientation via its normal vector, and \(d\) represents the plane's specific placement in space.

To determine the equation of a plane:

• You need a point that lies on the plane (at least one set of coordinates like \(x, y, z\)).
• And a normal vector \(\langle a, b, c \rangle\) that points perpendicularly to the plane.

For example, if a plane passes through the origin, \(d\) becomes zero because substituting the origin into the plane equation results in \(0 = d\).

In the exercise, finding the equation of the desired plane involved using the cross product to determine the correct normal vector. This ensured the plane was perpendicular to both given planes, verifying the understanding of plane orientation and its mathematical expression.