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Chapter 12: Problem 69

Are the statements true or false? Give reasons for your answer. There is only one point in the \(y z\) -plane that is a distance 5 from the point (3,0,0)

Short Answer

Expert verified
False; there are infinitely many points forming a circle.

Step by step solution

01

Understand the Condition

We need to determine if there is only one point in the \(yz\)-plane that is exactly 5 units away from the point \((3,0,0)\). The \(yz\)-plane consists of all points \((0,y,z)\). We need to find out how many points are 5 units from \((3,0,0)\) among points of form \((0,y,z)\).
02

Distance Formula Application

According to the distance formula, the distance between two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\) is given by \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\). Applying this to the problem, we have the fixed point \((3,0,0)\) and a general point in the \(yz\)-plane as \((0, y, z)\).
03

Set the Distance Equation

Substituting into the distance formula gives: \[ \sqrt{(0-3)^2 + (y-0)^2 + (z-0)^2} = 5 \] which simplifies to \[ \sqrt{9 + y^2 + z^2} = 5 \].
04

Solve the Equation

Squaring both sides to eliminate the square root gives:\[ 9 + y^2 + z^2 = 25 \] This simplifies to \[ y^2 + z^2 = 16 \]. This is the equation of a circle centered at \((0,0)\) with radius 4 in the \(yz\)-plane.
05

Interpret the Solution

The equation \(y^2 + z^2 = 16\) describes an entire circle in the \(yz\)-plane, implying that there are infinitely many points on this circle, each 5 units away from \((3,0,0)\).

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Key Concepts

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

Distance Formula
The distance formula is a mathematical equation used to determine the distance between two points in a Cartesian coordinate system.
  • In a 3D space, you have points that each have three coordinates: \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\).
  • The formula to find the distance \(d\) between these points is:\[d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}\]
This formula is derived from the Pythagorean theorem, which helps us compute the straight-line distance, akin to finding the hypotenuse in a right triangle.
In the example exercise, the points in question are \((3,0,0)\), which is fixed, and any point \((0,y,z)\) from the yz-plane.
Equation of a Circle
A circle is a set of points equidistant from a center point in a plane. The equation of a circle is \((x - h)^2 + (y - k)^2 = r^2\), where:
  • \((h, k)\) is the center of the circle.
  • \(r\) is the radius of the circle.
In the problem, after applying the distance formula and simplifying, we obtain \(y^2 + z^2 = 16\).
This represents a circle in the \(yz\)-plane, centered at \((0,0)\) with a radius of \(4\).
This equation is derived by setting up the situation where each point on the circle is \(5\) units from \((3,0,0)\) but restricted to the \(yz\)-plane.
3D Coordinate Plane
The 3D coordinate plane, also known as the 3-dimensional space, includes three axes:
  • The \(x\)-axis
  • The \(y\)-axis
  • The \(z\)-axis
Together, these axes allow us to locate any point in space with a coordinate \((x, y, z)\).
This coordinate system extends the 2D concept by adding depth (\(z\)-axis), expanding flat mappings into a more real-space application.
In the exercise context, it particularly involves the \(yz\)-plane, which only considers values where the \(x\)-coordinate is zero.
This means any point on the \(yz\)-plane can be denoted by \((0, y, z)\).
Point Distance Problem
The point distance problem involves finding points at a certain distance from a given point.
In many cases, like the one in the example, this translates to identifying not a single point but a collection of points that fulfill those distance criteria.
  • In the given problem, we seek a point in the \(yz\)-plane 5 units from \((3,0,0)\).
  • This process involved setting up the distance equation, which, after simplification, led us to a circle, indicating numerous solutions.
Rather than a solitary solution, this problem highlights how certain constraints (e.g., being in the \(yz\)-plane) result in circular sets of solutions, demonstrating the complexity yet beauty of spatial analytics in multivariable calculus.

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Most popular questions from this chapter

Any surface which is the graph of a two-variable function \(f(x, y)\) can also be represented as the level surface of a three-variable function \(g(x, y, z)\)

Give an example of: A function \(f(x, y, z)\) whose level surfaces are equally spaced planes perpendicular to the \(y z\) -plane.

Is the statement true or false? Give reasons for your answer. A level surface of a function \(g(x, y, z)\) cannot be a single point.

Is the statement true or false? Give reasons for your answer. If the level surfaces \(g(x, y, z)=k_{1}\) and \(g(x, y, z)=k_{2}\) are the same surface, then \(k_{1}=k_{2}\)

Give an example of: A function \(f(x, y, z)\) whose level sets are paraboloids.
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