Problem 68 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the equation of the plane having the given normal vector \(\mathbf{n}\) and passing through the given point \(P .\) $$ \mathbf{n}=\langle 0,0,1\rangle ; P(1,2,-3) $$

Step-by-Step Solution

Answer

The equation of the plane is \( z + 3 = 0 \).

Step 1: Understanding the Plane Equation

The general equation of a plane given a normal vector \( \mathbf{n} = \langle a, b, c \rangle \) and a point \( P(x_0, y_0, z_0) \) is given by the formula: \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \] This equation uses the components of the normal vector \( \mathbf{n} \) and the coordinates of the point \( P \) on the plane.

Step 2: Apply the Normal Vector

Substitute the components of the normal vector \( \mathbf{n} = \langle 0, 0, 1 \rangle \) into the plane equation. The equation becomes: \[ 0(x - x_0) + 0(y - y_0) + 1(z - z_0) = 0 \] This simplifies to \( z - z_0 = 0 \).

Step 3: Substitute the Given Point

Substitute the coordinates of the point \( P = (1, 2, -3) \) into the equation \( z - z_0 = 0 \). Here, \( z_0 = -3 \), so the equation becomes: \[ z - (-3) = 0 \] Simplifying, we get: \[ z + 3 = 0 \].

Step 4: Finalize the Plane Equation

The equation \( z + 3 = 0 \) represents the plane with the given normal vector and passing through the point \( P \). This equation describes a plane parallel to the xy-plane at a height of \(-3\) in the z-direction.

Key Concepts

Normal Vector

A normal vector is a fundamental concept in plane and solid geometry. It’s a vector that is perpendicular to a given surface, or in this case, a plane. This means that the normal vector \( \mathbf{n} \) is oriented at a right angle to the plane.In our problem, we have the normal vector \( \mathbf{n} = \langle 0, 0, 1 \rangle \). This vector points straight in the positive z-direction.

• The components of the vector indicate how it affects each axis.
• Here, the x and y components are zero, simplifying the equation for the plane because it is unaffected by changes along the x and y directions.

Understanding the role of a normal vector helps you determine the orientation of the plane in space. It tells you that movement along the plane does not change elevation, while moving perpendicular to it does.

Plane Geometry

Plane geometry involves understanding and working with flat surfaces that extend infinitely in two dimensions. These surfaces are described using equations in three-dimensional space. A plane can be uniquely defined by a point and a direction, typically given by a normal vector.The standard equation of a plane using a normal vector \( \mathbf{n} = \langle a, b, c \rangle \) and a point \( P(x_0, y_0, z_0) \) is \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]

• The parameters \( a, b, \) and \( c \) are derived from the normal vector.
• The point \( (x_0, y_0, z_0) \) provides a specific location on the plane.

Understanding plane geometry helps in visualizing how the plane cuts across the three-dimensional space and defining its interactions with other geometrical forms.

Coordinate Geometry

Coordinate geometry, also referred to as analytic geometry, allows geometric problems to be solved using a coordinate system. In dealing with planes, we often use the Cartesian coordinates (x, y, z) to position points in three-dimensional space.When you have a normal vector and a point, you can quickly derive the equation of a plane in coordinate geometry. This equation describes locations in space that belong to the plane. For our specific example:

• The normal vector \( \mathbf{n} = \langle 0, 0, 1 \rangle \) suggests the plane would be parallel to the xy-plane.
• The point \( P(1, 2, -3) \) indicates a specific location on the plane, and plugging it into the equation gives \( z + 3 = 0 \).

This equation represents a flat surface at a constant z-value of \( -3 \), demonstrating how coordinate geometry translates geometric concepts into algebraic expressions.