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Chapter 12: Problem 8

In Exercises \(1-16,\) give a geometric description of the set of points in space whose coordinates satisfy the given pairs of equations. $$y^{2}+z^{2}=1, \quad x=0$$

Short Answer

Expert verified
The solution represents a circle of radius 1 in the yz-plane centered at the origin.

Step by step solution

01

Understanding the First Equation

The first equation is \( y^2 + z^2 = 1 \). This represents a circle in the \( yz \)-plane with a center at \((0, 0)\) and a radius of 1.
02

Understanding the Second Equation

The second equation is \( x = 0 \), which represents the \( yz \)-plane in three-dimensional space. This means all points satisfying this condition lie on the plane where \( x \) coordinate is zero.
03

Combining the Conditions

Since \( x = 0 \) restricts us to the \( yz \)-plane, and \( y^2 + z^2 = 1 \) describes a circle within this plane, combining these conditions, we have a circle of radius 1 centered at \((0,0,0)\) on the \( yz \)-plane.
04

Geometric Description

The set of points satisfying both equations is a vertical circle in the \( yz \)-plane with a radius of 1. This circle is centered at the origin \((0, 0, 0)\) in three-dimensional space.

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Key Concepts

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

Circle in 3D space
In coordinate geometry, a circle can be understood as a set of all points that are equidistant from a single point, known as the center. When visualizing this in three-dimensional space, a circle can occupy any plane 鈥� the challenge is just to define it clearly in terms of the axes. Let鈥檚 break it down further:
  • The circle鈥檚 general equation is given by \[ (y - a)^2 + (z - b)^2 = r^2, \]representing a circle with center \((a, b)\) and radius \(r\) in the yz-plane.
  • When this circle is situated in 3D space, to anchor it, you must define a fixed plane 鈥� for example, the equation \(x = c\), holds all points of the circle in a plane parallel to the yz-plane.
  • In the given exercise, \(y^2 + z^2 = 1\) signifies that the circle has a center at the origin, \((0,0)\) in the yz-plane, and a radius 1.
  • Consequently, the circle sits vertically on the yz-plane in three-dimensional space, having the x-coordinate fixed at zero.
It's fascinating to see how the same conceptual circle exists in different planes by just changing its plane equation.
Geometric description
Understanding geometry in three-dimensional space often involves visualizing shapes that you've encountered in two dimensions. The trick is to imagine how these shapes, like circles, exist and align in 3D. Let's have a look at how you might describe this geometrically:
  • For any shape, the geometric description often begins with its defining equations. For the circle in question, \(y^2 + z^2 = 1\) tells us the shape of what we see in the yz-plane.
  • This equation reveals a circle in the yz-plane, meaning all its points meet the criteria of being equidistant from the center (0,0) with a common radius, here being 1.
  • Next, \(x=0\) situates this circle within the yz-plane at the origin (0,0,0) in 3D space.
  • Therefore, the geometric description, harmonizing both equations, is a vertical circle based in the yz-plane with its center situated at the 3-dimensional origin.
This clear understanding of geometric descriptions is crucial for effective spatial visualization and problem-solving in more complex scenarios.
Planes in space
Planes in 3D space are flat, two-dimensional surfaces that extend infinitely within three-dimensional coordinate systems. They are primarily defined using linear equations and can be aligned with or intersect the axes. Here鈥檚 how they function conceptually:
  • A plane can be represented as \(Ax + By + Cz = D\), where \(A, B,\) and \(C\) are the coefficients indicating the orientation of the plane.
  • When dealing with planes that are parallel to a particular axis, such as the yz-plane, the equation simplifies to \(x = c\) where \(c\) is a constant.
  • In our context, \(x = 0\) is such a plane, serving as the yz-plane itself. This implies all points on this plane have an x-coordinate of zero.
  • Being able to recognize these planes helps significantly when situating other geometries like circles within the larger three-dimensional space.
Comprehending planes and how they relate to geometrical figures is foundational in visualizing and understanding complex geometrical relationships in space.

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Most popular questions from this chapter

In Exercises \(1-8,\) find the length and direction (when defined) of \(\mathbf{u} \times \mathbf{v}\) and \(\mathbf{v} \times \mathbf{u} .\) $$ \mathbf{u}=2 \mathbf{i}+3 \mathbf{j}, \quad \mathbf{v}=-\mathbf{i}+\mathbf{j} $$

In Exercises \(53-56,\) find the point in which the line meets the plane. $$ x=1+2 t, \quad y=1+5 t, \quad z=3 t ; \quad x+y+z=2 $$

\begin{equation} \begin{array}{l}{\text { a. Cauchy-Schwartz inequality since } \mathbf{u} \cdot \mathbf{v}=|\mathbf{u} \| \mathbf{v}| \cos \theta} \\ {\text { show that the inequality }|\mathbf{u} \cdot \mathbf{v}| \leq|\mathbf{u}||\mathbf{v}| \text { holds for any vectors }} \\ {\mathbf{u} \text { and } \mathbf{v} .} \\ {\text { b. Under what circumstances, if any, does }|\mathbf{u} \cdot \mathbf{v}| \text { equal }|\mathbf{u}||\mathbf{v}| ?} \\\ {\text { Give reasons for your answer. }}\end{array} \end{equation}

In Exercises \(39-44,\) find the distance from the point to the plane. $$ (0,1,1), \quad 4 y+3 z=-12 $$

Find the areas of the parallelograms whose vertices are given in Exercises \(35-40 .\) $$ A(-1,2), \quad B(2,0), \quad C(7,1), \quad D(4,3) $$
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