Problem 30 - Calculus Early Transcendentals

Question

For the following sets of planes. determine which pairs of planes in the set are parallel, orthogonal, or identical. $$\begin{array}{l}Q: x+y-z=0 ; R: y+z=0 ; S: x-y=0 \\\T: x+y+z=0\end{array}$$

Step-by-Step Solution

Answer

Question: Determine if the given planes Q, R, S, and T are parallel, orthogonal, or identical. Planes: Q: x + y - z = 0 R: y + z = 0 S: x - y = 2 T: x + y + z = 3 Answer: Planes Q and T are parallel and identical. Planes Q and R, planes Q and S, and planes S and T are orthogonal pairs.

Step 1: Identify the normal vectors of each plane

From the given equations, we can write the normal vectors of each plane: $$\begin{array}{l} \textbf{Q: } n_Q = (1, 1, -1) \\ \textbf{R: } n_R = (0, 1, 1) \\ \textbf{S: } n_S = (1, -1, 0) \\ \textbf{T: } n_T = (1, 1, 1) \end{array}$$ These normal vectors will be used to compare the planes.

Step 2: Check if planes are parallel

If the normal vectors of two planes are parallel (or proportional), then the planes themselves are parallel. Therefore, compare the direction ratios of the given normal vectors: For planes Q and R, $$\frac{1}{0} \neq \frac{1}{1} \neq \frac{-1}{1}$$ For planes Q and S, $$\frac{1}{1} \neq \frac{1}{-1} \neq \frac{-1}{0}$$ For planes Q and T, $$\frac{1}{1} = \frac{1}{1} = \frac{-1}{1}$$ So planes Q and T are parallel. For planes R and S, $$\frac{0}{1} \neq \frac{1}{-1} \neq \frac{1}{0}$$ For planes R and T, $$\frac{0}{1} \neq \frac{1}{1} \neq \frac{1}{1}$$ For planes S and T, $$\frac{1}{1} \neq \frac{-1}{1} \neq \frac{0}{1}$$ Thus, only planes Q and T are parallel.

Step 3: Check if planes are orthogonal

If the normal vectors of two planes are orthogonal, then the dot product of the normal vectors is 0. Therefore, calculate the dot products of the pairs of normal vectors and check if they are equal to 0: For planes Q and R, $$n_Q \cdot n_R = (1, 1, -1) \cdot (0, 1, 1) = 0 + 1 - 1 = 0$$ So planes Q and R are orthogonal. For planes Q and S, $$n_Q \cdot n_S = (1, 1, -1) \cdot (1, -1, 0) = 1 - 1 + 0 = 0$$ So planes Q and S are orthogonal. For planes R and S, $$n_R \cdot n_S = (0, 1, 1) \cdot (1, -1, 0) = 0 - 1 + 0 = -1$$ Planes R and S are not orthogonal. For planes R and T, $$n_R \cdot n_T = (0, 1, 1) \cdot (1, 1, 1) = 0 + 1 + 1 = 2$$ Planes R and T are not orthogonal. For planes S and T, $$n_S \cdot n_T = (1, -1, 0) \cdot (1, 1, 1) = 1 - 1 + 0 = 0$$ So planes S and T are orthogonal. Thus, planes Q and R, planes Q and S, and planes S and T are orthogonal pairs.

Step 4: Check if planes are identical

If two planes have the same normal vector and coefficient ratios, then the planes are identical. We have already determined that planes Q and T are parallel, but let's check their coefficient ratios: For planes Q and T, $$\frac{1}{1} = \frac{1}{1} = \frac{-1}{1}$$ As these ratios are the same, planes Q and T are identical. In conclusion, we have the following pair relationships: - Planes Q and T are parallel and identical. - Planes Q and R, planes Q and S, and planes S and T are orthogonal pairs.

Key Concepts

Parallel Planes

Parallel planes never intersect. To determine if two planes are parallel, we need to analyze their normal vectors. If two planes are parallel, their normal vectors will be proportionate, meaning they have the same direction though not necessarily the same magnitude.
For example, when looking at the normal vectors of two planes such as Plane Q and Plane T:

• Plane Q has a normal vector of \( n_Q = (1, 1, -1) \)
• Plane T has a normal vector of \( n_T = (1, 1, 1) \)

Since the direction ratios of the normal vectors of Q and T are proportional, these planes are parallel. In practice, finding parallel planes is crucial for understanding the spatial relationship between objects, which can be very useful in various engineering and architectural applications.
It's important to remember that while parallel planes do not intersect, they are not required to be equidistant.

Orthogonal Planes

Orthogonal planes meet at a right angle, meaning their normal vectors are perpendicular to each other. To determine if two planes are orthogonal, we must check if the dot product of their normal vectors equals zero.
Let's consider a few examples based on the normal vectors from our problem:

• Normal vector of Plane Q: \( n_Q = (1, 1, -1) \)
• Normal vector of Plane R: \( n_R = (0, 1, 1) \)
• The dot product: \( n_Q \cdot n_R = 1 imes 0 + 1 imes 1 + (-1) imes 1 = 0 \)
• Therefore, Planes Q and R are orthogonal.

This concept is vital in fields such as computer graphics and geometrical problem-solving, as it allows for defining relationships in three-dimensional space. Remember, orthogonality is only defined for non-parallel planes.

Identical Planes

Identical planes are essentially the same geometrical space; they share all the same points. For two planes to be identical, both their normal vectors and the constant terms in their equations must be proportional.
Let's analyze Planes Q and T once more:

• Normal vectors are: \( n_Q = (1, 1, -1) \) and \( n_T = (1, 1, 1) \)
• The coefficient comparison: Coefficients from the plane equations for Q (\( 1, 1, -1 \)) and T (\( 1, 1, 1 \)) fit this condition when scaled appropriately.

Consequently, these two planes are seen as one in the same, making them identical. This concept ensures that we can accurately identify when two planes overlap entirely, which is significant in verifying algebraic plane data integrity and consistency.