/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 17 A plane has normal vector \(n\) ... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 17

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. $$\mathbf{n}=\left\langle 3,0,-\frac{1}{2}\right\rangle, \quad P(2,4,8)$$

Short Answer

Expert verified
(a) Plane: \( 3x - \frac{1}{2}z = 2 \); (b) Intercepts: x = \( \frac{2}{3} \), y undefined, z = -4.

Step by step solution

01

Understand the Plane Equation

The general equation for a plane in 3D space can be written as \( ax + by + cz = d \), where \( \mathbf{n} = \langle a, b, c \rangle \) is the normal vector to the plane.
02

Substitute the Normal Vector

Given the normal vector \( \mathbf{n} = \langle 3, 0, -\frac{1}{2} \rangle \), substitute \( a = 3 \), \( b = 0 \), and \( c = -\frac{1}{2} \) into the plane equation, so we have \( 3x + 0y - \frac{1}{2}z = d \) which simplifies to \( 3x - \frac{1}{2}z = d \).
03

Use the Given Point to Find d

Substitute the coordinates of point \( P(2, 4, 8) \) into the equation \( 3x - \frac{1}{2}z = d \) to find \( d \). Using \( x = 2 \), \( y = 4 \), and \( z = 8 \), we substitute to get \( 3(2) + 0(4) - \frac{1}{2}(8) = d \), which simplifies to \( 6 - 4 = d \), thus \( d = 2 \).
04

Write the Plane Equation

Now substitute \( d = 2 \) back into the equation \( 3x - \frac{1}{2}z = d \), yielding the plane equation: \( 3x - \frac{1}{2}z = 2 \).
05

Find the Intercepts

To find the x-intercept, set \( y = 0 \) and \( z = 0 \). Equation becomes \( 3x = 2 \), hence \( x = \frac{2}{3} \). For the y-intercept, set \( x = 0 \) and \( z = 0 \). Since \( b = 0 \), the y-intercept is not defined in this equation. For the z-intercept, set \( x = 0 \) and \( y = 0 \), resulting in \( -\frac{1}{2}z = 2 \), so \( z = -4 \).
06

Sketch the Plane

Sketch the plane using the x-intercept (\( \frac{2}{3}, 0, 0 \)) and the z-intercept (\( 0, 0, -4 \)). The line through these points will lie in the plane alongside other potential points if needed. The plane is parallel to the y-axis due to \( b = 0 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Vector
The normal vector is a fundamental concept in 3D geometry, particularly when we're dealing with planes. A normal vector to a plane is one that is perpendicular to the plane itself. This is what makes it "normal" or orthogonal. For the given problem, the normal vector is \( \mathbf{n} = \langle 3, 0, -\frac{1}{2} \rangle \).

  • Orientation: The normal vector gives us the orientation of the plane. In this problem, \( a = 3 \), \( b = 0 \), and \( c = -\frac{1}{2} \) determine how the plane is tilted in 3D space.
  • Equation Contribution: These components correspond directly to the coefficients in the plane equation \( ax + by + cz = d \), which fundamentally defines the plane.
By interacting with a known point in the plane, the normal vector helps us find the specific equation of the plane in question.
Intercepts
Intercepts are points where a plane or line crosses the axes in the coordinate system. For any plane equation, identifying intercepts is key because they provide tangible points to understand and sketch the plane.

  • x-intercept: In our problem, setting \( y = 0 \) and \( z = 0 \) results in \( 3x = 2 \). Solving for \( x \), we find \( x = \frac{2}{3} \). So, the x-intercept is \( \left( \frac{2}{3}, 0, 0 \right) \).
  • y-intercept: Normally, you find the y-intercept by setting both \( x = 0 \) and \( z = 0 \). However, in our case, since \( b = 0 \), the y-term is missing, indicating the plane never actually crosses the y-axis in the usual manner.
  • z-intercept: Setting \( x = 0 \) and \( y = 0 \), we find \( -\frac{1}{2}z = 2 \), leading to \( z = -4 \). Thus, the z-intercept is \( (0, 0, -4) \).
In practical terms, intercepts allow us to pin down key points that can be graphed to visualize the plane's placement within the 3D space.
3D Geometry
3D geometry involves the study of mathematical shapes in a three-dimensional space, where concepts such as planes, lines, and points interact. A plane in 3D geometry is a flat, two-dimensional surface that extends infinitely along its length and width but has zero thickness.

  • Plane Representation: The equation of a plane is typically given in the form \( ax + by + cz = d \). In simpler terms, this represents all the points (x, y, z) that lie on the plane.
  • Planes in Space: Because a plane is two-dimensional, it divides the 3D space into two halves. The position of the plane is uniquely determined by a normal vector and a point lying on the plane, allowing it to exist in specific orientations within the space.
  • Graphical Representation: Sketching a plane can be aided using its intercepts which act as reference points. By connecting intercept points, like the ones found in the exercise, we can simulate the infinite plane in a finite 2D sketch.
Understanding 3D geometry is crucial for visualizing and solving problems involving physical spaces, whether it's in architecture, computer graphics, or any field requiring spatial reasoning.

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