Problem 18 - Precalculus Mathematics for Calculus

Question

A plane has normal vector \(n\) and passes through the point \(P\). (a) Find an equation for the plane. (b) Find the intercepts and sketch a graph of the plane. $$\mathbf{n}=\left\langle-\frac{2}{3},-\frac{1}{3}, 1\right\rangle, \quad P(-6,0,-3)$$

Step-by-Step Solution

Answer

Plane equation: \(-\frac{2}{3}x - \frac{1}{3}y + z = 1\). Intercepts: \((-\frac{3}{2}, 0, 0)\), \((0, -3, 0)\), \((0, 0, 1)\).

Step 1: Understand the Plane Equation

The general equation of a plane in 3D space is given by \( Ax + By + Cz = D \), where \( \vec{n} = \langle A, B, C \rangle \) is the normal vector to the plane. This can be expanded to find \( D \) using a point the plane passes through.

Step 2: Plug in the Normal Vector

We have the normal vector \( \vec{n} = \left\langle -\frac{2}{3}, -\frac{1}{3}, 1 \right\rangle \), so we know \( A = -\frac{2}{3} \), \( B = -\frac{1}{3} \), \( C = 1 \). The equation becomes \( -\frac{2}{3}x - \frac{1}{3}y + z = D \).

Step 3: Use the Point to Find D

Since the plane passes through \( P(-6, 0, -3) \), substitute these values into the equation: \( -\frac{2}{3}(-6) - \frac{1}{3}(0) + (-3) = D \). Calculate to get \( 4 - 3 = D \). Therefore, \( D = 1 \).

Step 4: Write the Plane Equation

Substitute \( D = 1 \) back into the plane equation to obtain \( -\frac{2}{3}x - \frac{1}{3}y + z = 1 \).

Step 5: Find Intercepts

To find the intercepts, set two of the variables to 0 and solve for the third. - **x-intercept**: Set \( y = 0 \), \( z = 0 \). Solve \( -\frac{2}{3}x = 1 \) to get \( x = -\frac{3}{2} \).- **y-intercept**: Set \( x = 0 \), \( z = 0 \). Solve \( -\frac{1}{3}y = 1 \) to get \( y = -3 \).- **z-intercept**: Set \( x = 0 \), \( y = 0 \). Solve \( z = 1 \) to get \( z = 1 \).

Step 6: Sketch the Plane

Using the intercepts found (-3/2, 0, 0), (0, -3, 0), and (0, 0, 1), draw the graph in a 3D coordinate system. Label each intercept and draw a triangular section of the plane touching each intercept point.

Key Concepts

Understanding the Normal Vector

A normal vector is a crucial concept when dealing with the equation of a plane. In the context of a plane in a 3D coordinate system, the normal vector is a vector that is perpendicular to the plane. It's like having a stick pointing directly away from a piece of paper. This vector helps define the orientation of the plane but not its position.
In mathematical terms, the normal vector for a plane can be represented as \( \vec{n} = \langle A, B, C \rangle \), where each component corresponds to a coefficient in the plane's equation \( Ax + By + Cz = D \).
Knowing the normal vector allows us to explain how the plane is tilted in space. If the normal vector points up, the plane faces upwards; if sideways, the plane lies horizontally. This orientation does not change with the position of the plane because the normal vector remains consistent.
Think of the normal vector as the backbone of the plane equation. By knowing it, we can determine how the plane slices through the 3D space and, ultimately, how it will interact with other objects, like lines or other planes.

Intercepts of the Plane

Intercepts are the points where the plane crosses the axes in a 3D coordinate system. Finding these intercepts helps us visualize and graph how the plane interacts with each axis.
For each axis, do the following to find the intercept:

• **x-intercept**: Set \( y = 0 \) and \( z = 0 \). Use the plane equation to solve for \( x \).
• **y-intercept**: Set \( x = 0 \) and \( z = 0 \). Solve the equation for \( y \).
• **z-intercept**: Set \( x = 0 \) and \( y = 0 \). Find the solution for \( z \).

These intercepts provide reference points that show where the plane intersects the three main directional axes in 3D space. For a clearer picture, imagine the plane as a flat surface that will cut through some part of each axis, revealing these points as its footprints on the axes. For example, if the equations for each are solved, the plane might touch points like \( (x, 0, 0) \), \( (0, y, 0) \), and \( (0, 0, z) \). These points are used to sketch and understand how the plane is oriented in the 3D space.

Navigating a 3D Coordinate System

The 3D coordinate system allows us to describe space using three axes: the x-axis, y-axis, and z-axis. This system is useful in modeling real-world phenomena and performing complex calculations that span multiple directions.
Think of 3D space like a room where the corners meet at a single point called the origin. In this room:

• The **x-axis** might run along the floor from left to right.
• The **y-axis** could extend from front to back.
• The **z-axis** goes straight up to the ceiling.

The equations of planes in a 3D system are represented as \( Ax + By + Cz = D \). These equations describe how planes slice or position themselves within that room. They tell us how high the plane is, how tilted it might be, or how far from the origin it stretches.
This coordinate system is essential for visualizing spatial relationships and performing mathematical operations across multiple dimensions. Understanding how to navigate this space helps us solve real-world 3D problems, whether it's designing a building, tracking an aircraft, or exploring virtual environments in computer graphics.