Problem 21 - University Calculus: Early Transcendentals

Question

Find equations for the planes. The plane through \(P_{0}(0,2,-1)\) normal to \(\mathbf{n}=3 \mathbf{i}-2 \mathbf{j}-\mathbf{k}\)

Step-by-Step Solution

Answer

The plane equation is \(3x - 2y - z = -3\).

Step 1: Plane Equation Form

The general equation for a plane in 3D space is given by: \( ax + by + cz = d \), where \( \mathbf{n} = \langle a, b, c \rangle \) is a normal vector to the plane. Here, \( \mathbf{n} = 3 \mathbf{i} - 2 \mathbf{j} - \mathbf{k} \) gives us \( a = 3 \), \( b = -2 \), \( c = -1 \).

Step 2: Use Point in Plane Equation

Substitute point \( P_{0}(0,2,-1) \) into the plane equation. Therefore, \( 3 \times 0 - 2 \times 2 - 1 \times (-1) = d \).

Step 3: Solve for d

Calculate \( d \) using the values from Step 2. \( d = 0 - 4 + 1 = -3 \).

Step 4: Write Final Plane Equation

Substitute the values back into the plane equation: \( 3x - 2y - z = -3 \).

Key Concepts

Understanding the Normal Vector

A normal vector is an essential concept when working with planes in geometry. It defines the orientation of a plane within a three-dimensional space. Think of it as an arrow that points directly out of the plane.

• The normal vector is perpendicular to the plane, making it a handy tool for defining and manipulating the plane's position.
• In mathematical terms, it involves components that relate to each axis in 3D space, typically denoted as \(\mathbf{n} = \langle a, b, c \rangle\).
• For the given problem, the normal vector is \(3 \mathbf{i} - 2 \mathbf{j} - \mathbf{k}\), translating to the vector \((3, -2, -1)\).

Knowing the normal vector allows us to build an equation for the plane by providing the necessary coefficients, leading us to systematize other crucial calculations.

What is 3D Space?

Three-dimensional (3D) space is the environment where all objects have length, width, and height. It's the common understanding of space around us, involving three coordinate axes:

• The x-axis generally represents left to right movements.
• The y-axis indicates upward or downward movements.
• The z-axis shows the depth, moving forward and backward.

In mathematics, these axes allow any point within this space to be specified by coordinates (\(x, y, z\)). This framework is vitally important for defining geometric shapes, such as lines and planes.3D space enables us to use vectors and coordinate planes to model real-world scenarios, making mathematical concepts more practical and applicable.

Deriving the Plane Equation

The equation of a plane in 3D space can be neatly expressed as \(ax + by + cz = d\). Here's how it works:

• The values \(a, b, c\) are derived from the plane's normal vector \(\mathbf{n} = \langle a, b, c \rangle\).
• In our problem, we substitute the normal vector values: \(3, -2, -1\).
• \(d\) represents the plane's displacement and is determined by plugging a known point from the plane into the equation.

For the point \(P_{0}(0, 2, -1)\), substitute into the equation: \(3 \times 0 - 2 \times 2 - (-1) = d\), leading to \(d = -3\).So, the final derived plane equation is: \(3x - 2y - z = -3\). This showcases the interaction between the normal vector and the point in forming the plane's equation, a crucial concept for defining 3D surfaces.