Problem 26 - Calculus Early Transcendentals

Question

Determine if the following pairs of planes are parallel, orthogonal, or neither parallel nor orthogonal. $$2 x+2 y-3 z=10 \text { and }-10 x-10 y+15 z=10$$

Step-by-Step Solution

Answer

Plane 1: $$2x + 2y - 3z = 10$$ Plane 2: $$-10x - 10y + 15z = 10$$ Answer: The given planes are __parallel__.

Step 1: Write down the normal vectors of the given planes.

The normal vector of a plane in the form of ax + by + cz = d is the coefficient vector (a, b, c). So, for the given planes: Plane 1: $$2x + 2y - 3z = 10$$ Normal vector, $$\vec{n_1} = (2, 2, -3)$$ Plane 2: $$-10x - 10y + 15z = 10$$ Normal vector, $$\vec{n_2} = (-10, -10, 15)$$

Step 2: Check if the normal vectors are scalar multiples within a tolerance.

If the normal vectors are scalar multiples, the planes are parallel. Let's check whether a scalar multiple relationship exists: $$\frac{-10}{2} = \frac{-10}{2} = \frac{15}{-3} = 5$$ Since the relationship between corresponding components exists and is consistent for all components, the normal vectors are scalar multiples. Therefore, the planes are parallel.

Step 3: Conclusion

The given planes are parallel since their normal vectors are scalar multiples.

Key Concepts

Parallel Planes

Parallel planes in geometry are planes that never intersect, no matter how far they extend. This means they are always the same distance apart. Identifying parallel planes can be done by looking at their normal vectors.

Here's a simple way to check for parallel planes: compare the normal vectors of both planes. If one normal vector is a scalar multiple of the other, the planes are parallel. For example, if you have one plane with a normal vector \( \vec{n_1} = (a, b, c) \) and another with \( \vec{n_2} = (ka, kb, kc) \), where \( k \) is a scalar, then those planes are parallel.

In our exercise:

• Plane 1 has the normal vector \( \vec{n_1} = (2, 2, -3) \)
• Plane 2 has the normal vector \( \vec{n_2} = (-10, -10, 15) \)

The normal vectors are scalar multiples: \( \frac{-10}{2} = \frac{-10}{2} = \frac{15}{-3} = 5 \). Thus, the planes are parallel.

Orthogonal Planes

Orthogonal planes are planes that intersect at a right angle, which means they are perpendicular to each other. To determine if planes are orthogonal, you should consider their normal vectors.

If the dot product of two planes' normal vectors is zero, then those planes are orthogonal. Here's how it works: suppose you have normal vectors \( \vec{n_1} = (a, b, c) \) and \( \vec{n_2} = (d, e, f) \). The dot product is calculated as \( a \cdot d + b \cdot e + c \cdot f \). If this sum equals zero, then the planes are orthogonal.

Using the vectors from our example:

• \( \vec{n_1} = (2, 2, -3) \)
• \( \vec{n_2} = (-10, -10, 15) \)

Let's calculate the dot product: \( 2 \cdot (-10) + 2 \cdot (-10) + (-3) \cdot 15 = -20 - 20 - 45 = -85 \). This is not zero, meaning the planes are not orthogonal.

Normal Vector

The normal vector of a plane is a fundamental concept in geometry. It is a vector that is perpendicular to the plane, allowing us to define the plane's orientation in space.

Finding a normal vector is straightforward if we have the plane equation in the form \( ax + by + cz = d \). The normal vector is simply \( (a, b, c) \). This makes analyzing the plane much easier, as it helps determine properties like parallelism and orthogonality with other planes.

In our given exercise, the normal vectors were:

• For Plane 1: \( \vec{n_1} = (2, 2, -3) \)
• For Plane 2: \( \vec{n_2} = (-10, -10, 15) \)

Understanding the importance of normal vectors allows you to easily check conditions for parallelism (by scalar multiplication) and orthogonality (via the dot product) of planes.