Problem 32 - Essential Calculus Early Transcendentals

Question

\(21-32=\) Find an equation of the plane. The plane that passes through the line of intersection of the planes \(x-z=1\) and \(y+2 z=3\) and is perpendicular to the plane \(x+y-2 z=1\)

Step-by-Step Solution

Answer

The equation of the plane is \(2x - y - 4z + 1 = 0\).

Step 1: Understand the line of intersection

The line of intersection that the plane passes through is found by solving the system of equations given by the two planes:x - z = 1 \(\rightarrow\) Plane 1 y + 2z = 3 \(\rightarrow\) Plane 2.This represents a line and any plane through this line will be a linear combination of Plane 1 and Plane 2.

Step 2: Find the normal vector to the new plane

The new plane is perpendicular to the plane with equation \(x + y - 2z = 1\). Thus, its normal vector must be parallel to the normal vector of this plane, \( \langle 1, 1, -2 \rangle\).

Step 3: Form the equation of the plane

Since the plane should pass through the intersection line of the planes \(x - z = 1\) and \(y + 2z = 3\), and is perpendicular to the plane \(x + y - 2z = 1\), the equation of the new plane can be written as:\[(x - z - 1) + k(y + 2z - 3) = 0\]where \(k\) is a constant. This ensures the plane passes through the line of intersection.

Step 4: Use normal vector condition

Since the plane is perpendicular to \(x + y - 2z = 1\), its normal vector must be \( \langle 1, 1, -2 \rangle \). By equating the coefficients of the normal vector of the form \(\langle 1, k, -1+2k \rangle\) to \( \langle 1, 1, -2 \rangle \), we find that:\(1 = 1\) (satisfied)\(k = 1\)\(-1 + 2k = -2\) \(\Rightarrow k = \frac{-1}{2}\).Thus, the equation becomes:\[ (x - z - 1) - \frac{1}{2}(y + 2z - 3) = 0 \]

Step 5: Simplify the equation

From the last step, simplify the equation:\[(x - z - 1) - \frac{1}{2}(y + 2z - 3) = 0\]Multiply through by 2 for simplicity and clarity:\[ 2(x - z - 1) - (y + 2z - 3) = 0 \]\[ 2x - 2z - 2 - y - 2z + 3 = 0 \]\[ 2x - y - 4z + 1 = 0 \]This is the equation of the desired plane.

Key Concepts

Line of Intersection

The line of intersection of two planes can be thought of as the seam where the surfaces of two planes meet. When two planes intersect in three-dimensional space, they usually intersect along a line. To find this intersection, we solve the equations of the two planes simultaneously. For instance, if we have two plane equations, like:

• Plane 1: \( x - z = 1 \)
• Plane 2: \( y + 2z = 3 \)

We solve this system to get a common line, which represents every point along the intersection of the two planes.
This line can be expressed as a vector equation. By substituting one variable in terms of another, you determine how points align along the line. Through this process, we can find a point on the line and use direction vectors to represent the line algebraically. This understanding gives us the ability to analyze how other planes, such as the new plane in the exercise, can pass through this line of intersection.

Normal Vector

The normal vector is an essential concept when dealing with plane equations. It acts as a directional guide and depicts the orientation of a plane in 3D space. Specifically, a normal vector is a vector that is perpendicular to every line lying on the plane.

• For the plane equation of the form \( ax + by + cz = d \), the vector \( \langle a, b, c \rangle \) is the normal vector.

In our exercise, the plane that the new plane is perpendicular to has the equation \( x + y - 2z = 1 \). Therefore, its normal vector is \( \langle 1, 1, -2 \rangle \). For the new plane to be perpendicular to this one, its own normal vector must be parallel to \( \langle 1, 1, -2 \rangle \).
Using this vector ensures the geometric relationship between the planes is maintained. By equating this new plane's equation's coefficients with those of its required normal vector, the overall orientation and perpendicular condition are satisfied.

Perpendicular Planes

In geometry, when two planes are perpendicular, their normal vectors are parallel but with opposite orientations. This means one plane runs at a 90-degree angle to the other in three-dimensional space. A major characteristic of perpendicular planes is that their dot product of normal vectors equals zero, a concept vital for verifying perpendicularity.
In our given problem, the desired plane must also be perpendicular to the established plane, \( x + y - 2z = 1 \). To ensure this perpendicularity, the normal vector of the new plane, \( \langle 1, k, -1+2k \rangle \), is determined to align with the already established normal vector \( \langle 1, 1, -2 \rangle \). This alignment sets the equation's coefficients in a way that when solved, satisfies the condition for the planes being perpendicular.
Thus, understanding how normal vectors serve as the bridge between plane orientation and perpendicularity advances our ability to solve more complex spatial problems while assuring exact alignments based on geometric constraints.