Problem 38 - Essential Calculus Early Transcendentals

Question

\(35-38=\) Determine whether the planes are parallel, perpendicular, or neither. If neither, find the angle between them. $$x+2 y+2 z=1, \quad 2 x-y+2 z=1$$

Step-by-Step Solution

Answer

The planes are neither parallel nor perpendicular; the angle between them is \( \cos^{-1}\left(\frac{4}{9}\right) \).

Step 1: Identify Normal Vectors

The first step is to identify the normal vectors of the given planes. A plane of the form \( ax + by + cz = d \) has a normal vector \( \mathbf{n} = \langle a, b, c \rangle \). Thus, the normal vector for the plane \( x + 2y + 2z = 1 \) is \( \mathbf{n}_1 = \langle 1, 2, 2 \rangle \) and for the plane \( 2x - y + 2z = 1 \), it is \( \mathbf{n}_2 = \langle 2, -1, 2 \rangle \).

Step 2: Determine if Planes are Parallel

Parallel planes have normal vectors that are scalar multiples of each other. Check if \( \mathbf{n}_1 \) is a scalar multiple of \( \mathbf{n}_2 \). Since \( \mathbf{n}_1 = \langle 1, 2, 2 \rangle \) and \( \mathbf{n}_2 = \langle 2, -1, 2 \rangle \) are not scalar multiples (no single scalar can equate all corresponding components), the planes are not parallel.

Step 3: Determine if Planes are Perpendicular

Perpendicular planes have normal vectors that are orthogonal, meaning their dot product is zero. Compute the dot product: \( \mathbf{n}_1 \cdot \mathbf{n}_2 = 1\cdot2 + 2\cdot(-1) + 2\cdot2 = 2 - 2 + 4 = 4 \). Since the dot product is not zero, the planes are not perpendicular.

Step 4: Calculate Angle Between Planes

Since the planes are neither parallel nor perpendicular, calculate the angle \( \theta \) between them using the formula: \( \cos \theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|} \). First, find the magnitudes: \( \|\mathbf{n}_1\| = \sqrt{1^2 + 2^2 + 2^2} = \sqrt{9} = 3 \) and \( \|\mathbf{n}_2\| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{9} = 3 \). Substitute into the formula: \( \cos \theta = \frac{4}{3 \times 3} = \frac{4}{9} \). Thus, \( \theta = \cos^{-1}\left(\frac{4}{9}\right) \).

Step 5: Interpret Results

Since we've found \( \theta \), the angle between the planes is \( \cos^{-1}\left(\frac{4}{9}\right) \). This confirms that the planes are neither parallel nor perpendicular but intersect at an angle.

Key Concepts

Normal Vectors

When dealing with plane geometry, understanding normal vectors is crucial. A normal vector is a vector that is perpendicular to a plane. It plays a key role in defining the orientation of that plane in space.
To identify a normal vector for a plane given by the equation \( ax + by + cz = d \), you simply take the coefficients \( a, b, \) and \( c \) to form the vector \( \mathbf{n} = \langle a, b, c \rangle \).
For example:

• The plane \( x + 2y + 2z = 1 \) has a normal vector \( \mathbf{n}_1 = \langle 1, 2, 2 \rangle \).
• The plane \( 2x - y + 2z = 1 \) has a normal vector \( \mathbf{n}_2 = \langle 2, -1, 2 \rangle \).

Understanding these vectors helps in analyzing the geometric relationships between the planes.
In the context of the problem, comparing normal vectors can determine if two planes are parallel, perpendicular, or neither.

Angle Between Planes

To determine the angle between two planes, you first examine their normal vectors. If the planes are neither parallel nor perpendicular, you calculate the angle between their normal vectors using a specific formula.
The angle \( \theta \) between two planes with normal vectors \( \mathbf{n}_1 \) and \( \mathbf{n}_2 \) is found using this relationship:

• First, compute the cosine of the angle: \( \cos \theta = \frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{\|\mathbf{n}_1\|\|\mathbf{n}_2\|} \).
• The dot product \( \mathbf{n}_1 \cdot \mathbf{n}_2 \) and the magnitudes of the vectors \( \|\mathbf{n}_1\| \) and \( \|\mathbf{n}_2\| \) are used.

In our example, the normal vectors are \( \mathbf{n}_1 = \langle 1, 2, 2 \rangle \) and \( \mathbf{n}_2 = \langle 2, -1, 2 \rangle \). This method illustrates how planes can intersect at specific angles, providing a deeper understanding of their spatial relation.

Dot Product

The dot product, also known as the scalar product, is a fundamental algebraic operation applicable in plane geometry. It measures the degree of parallelism between two vectors.
Mathematically, for two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), the dot product is given by:

• \( \mathbf{a} \cdot \mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 \)
• This result is a scalar, not a vector.

The significance in plane geometry is:

• If the dot product is zero, the vectors (and thus the planes) are perpendicular.
• If not, as in our exercise, there is a specific angle, and further calculations can determine how the planes interact spatially.

In this exercise, the dot product of \( \mathbf{n}_1 = \langle 1, 2, 2 \rangle \) and \( \mathbf{n}_2 = \langle 2, -1, 2 \rangle \) was calculated to be 4, meaning the planes are neither parallel nor perpendicular.