Problem 16 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the plane through \((0,0,2)\) that is parallel to the plane \(x+y+z=1\).

Step-by-Step Solution

Answer

The equation is \(x + y + z = 2\).

Step 1: Identify the Normal Vector

The normal vector of a plane is derived from its equation in the form \(ax + by + cz = d\). For the given plane \(x + y + z = 1\), the normal vector is \((1, 1, 1)\).

Step 2: Use the Given Point

The equation of a plane in normal form is \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\), where \((x_0, y_0, z_0)\) is a point on the plane. Here, \((x_0, y_0, z_0) = (0,0,2)\).

Step 3: Substitute into the Plane Equation

Substitute the point \((0,0,2)\) and the normal vector \((1,1,1)\) into the equation: \(1(x - 0) + 1(y - 0) + 1(z - 2) = 0\). Simplifying this gives \(x + y + z - 2 = 0\).

Step 4: Write the Final Equation

The simplified equation \(x + y + z - 2 = 0\) represents the plane that passes through \((0, 0, 2)\) and is parallel to \(x + y + z = 1\).

Key Concepts

Normal Vector

In the world of geometry, a normal vector is essential to defining a plane. It acts as a perpendicular, or normal, force to the plane's surface. When it comes to equations of planes in 3D space, they are often given in the form \(ax + by + cz = d\). Here, the coefficients \(a\), \(b\), and \(c\) create the normal vector \((a, b, c)\).
For example, in the equation \(x + y + z = 1\), the normal vector is \((1, 1, 1)\). This vector is perpendicular to any point on the plane.

• The normal vector helps in defining orientation of the plane.
• It is crucial for understanding the relationships between planes, like whether they are parallel.

Without the normal vector, we would struggle to articulate the spatial direction of planes effectively. So, whenever you encounter a plane equation, identifying its normal vector should be your first step.

Parallel Planes

Parallel planes are fascinating because they never meet, much like parallel lines. For two planes to be parallel, their normal vectors must be identical or scalar multiples of each other.
This means they share the same orientation, preventing them from ever intersecting. Think of it like having two floors in a building - the floors are parallel to each other.
The given problem involves finding a plane parallel to \(x + y + z = 1\). Since the normal vector is \((1, 1, 1)\), any plane with this same normal vector will be parallel.

• Parallel planes have equal normal vectors.
• The only difference is the constant \(d\) in their equation \(ax + by + cz = d\).

Finding parallel planes often involves adjusting this constant while keeping the normal vector unchanged.

Point-Plane Equation

The point-plane equation provides a straightforward method to determine a plane's equation, given a point on the plane and a normal vector. To utilize this form, the formula is \(a(x - x_0) + b(y - y_0) + c(z - z_0) = 0\), where \((x_0, y_0, z_0)\) is the point, and \((a, b, c)\) is the plane's normal vector.
In our example, the point \((0, 0, 2)\) is used with the normal vector \((1, 1, 1)\) to craft the plane's equation.
This method revolves around substituting the known point into the formula, simplifying to find an equation such as \(x + y + z - 2 = 0\).

• This equation tells us that all points \((x, y, z)\) on the plane satisfy this relation.
• The point-plane equation simplifies the process of defining planes in space.

By using a known point, it grounds the abstract concept of a plane in a tangible location, ensuring precision and clarity.