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Chapter 11: Problem 90

Find the distance between the point and the plane.\((3,2,1)\) \(x-y+2 z=4\)

Short Answer

Expert verified
The shortest distance from the point (3,2,1) to the plane \(x - y + 2z = 4\) is \(d = \frac{1}{\sqrt6}\) or approximately 0.408

Step by step solution

01

Identify the coefficients from the plane equation

From the given plane equation \(x - y + 2z = 4\), you can see that a=1, b=-1 and c=2 are the coefficients of x, y and z respectively. The constant d, which is on the right side of the equation, is equal to -4.
02

Identify the coordinates of the given point

The given point in the 3-dimensional space is (3,2,1). Therefore, \(x_0=3\), \(y_0=2\), and \(z_0=1\).
03

Substitute values into the distance formula

The distance \(d\) from the point to the plane can be calculated by substituting the identified values into the formula. Therefore, \(d = \frac{{\left| a*x0 + b*y0 + c*z0 + d \right|}}{{\sqrt{a^2 + b^2 + c^2}}}\) becomes \(d = \frac{{\left| 1*3 + -1*2 + 2*1 - 4 \right|}}{{\sqrt{1^2 + (-1)^2 + 2^2}}}\)
04

Simplify and calculate the distance

After substitution, the calculation becomes \(d = \frac{{\left| -1 \right|}}{{\sqrt{1 + 1 + 4}}}\). This simplifies to \(d = \frac{1}{\sqrt6}\) or approximately 0.408.

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Key Concepts

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

3-dimensional geometry
In the fascinating world of 3-dimensional geometry, we explore the spatial relationships and measurements in a three-dimensional space. This branch of geometry considers points, lines, and planes, allowing us to understand how objects interact in space. A point in 3D space is represented by coordinates \((x, y, z)\), which specify its position along three axes:
  • x-axis for horizontal position
  • y-axis for vertical position
  • z-axis for depth or height
A plane, meanwhile, is a flat, two-dimensional surface extending indefinitely in 3D space and is generally described by an equation of the form \(ax + by + cz = d\). Here:
  • a, b, c are the coefficients that dictate the plane's orientation
  • d is a constant that influences its position
      When determining the distance from a specific point within this 3-dimensional realm to a plane, understanding these spatial coordinates and how they relate through geometry is essential.To find the distance from a point \((3, 2, 1)\) to a plane \(x - y + 2z = 4\), we use fundamental concepts like vectors and the distance formula to calculate spatial separation.
formula derivation
Deriving the formula for the distance from a point to a plane requires a structured approach using the components given in the plane's equation and the point's coordinates.We start with the standard equation of a plane: \(ax + by + cz = d\). The distance \(d\) from a point \((x_0, y_0, z_0)\) to this plane is calculated using the formula:\[ d = \frac{{|ax_0 + by_0 + cz_0 + d|}}{{\sqrt{a^2 + b^2 + c^2}}} \]**Steps in Derivation:**
  • Identify coefficients: From \(x - y + 2z = 4\), coefficients \(a = 1\), \(b = -1\), and \(c = 2\) are recognized. Constant \(d\) becomes -4 since the equation is usually in the form \(ax + by + cz - d = 0\).
  • Substitute point's coordinates: For point \((3, 2, 1)\), substitute into the formula: \(1\cdot3 + (-1)\cdot2 + 2\cdot1 - 4\).
  • Simplify and solve: Calculate the numerator as \(|(-1)|\) and the denominator as \(\sqrt{1 + 1 + 4}\) to arrive at the result \(\frac{1}{\sqrt{6}}\).
This derivation is fundamental for students understanding how geometric principles translate into measurable quantities given specific conditions in space.
vector analysis
Vector analysis plays a crucial role in solving geometric problems, particularly when measuring distances in 3-dimensional geometry.Vectors are essentially quantities represented by both magnitude and direction. In our context, the plane equation \(x - y + 2z = 4\) defines a vector normal to the plane. This normal vector is \(\langle 1, -1, 2 \rangle\), extracted directly from the plane's coefficients.**Steps in Vector Analysis:**
  • Identifying the Normal Vector: The components \(1, -1, 2\) represent a direction perpendicular to the plane surface.
  • Relate Point to the Normal: The shortest distance from a point to a plane is along a line parallel to this normal vector. Understanding this geometry assures that the distance derived is indeed the minimal separation.
  • Apply in Calculations: By integrating vector concepts, the distance formula slots directly into regular geometric analysis, resulting in more profound insights into space relations.
Vector analysis, therefore, enhances the comprehension of spatial measurements and supports the derivation of distance formulas in a systematic, logical way.

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Most popular questions from this chapter

Use vectors to find the point that lies two-thirds of the way from \(P\) to \(Q\). \(P(1,2,5), Q(6,8,2)\)

A point in the three-dimensional coordinate system has coordinates \(\left(x_{0}, y_{0}, z_{0}\right) .\) Describe what each coordinate measures.

Find the standard equation of the sphere with center \((-3,2,4)\) that is tangent to the plane given by \(2 x+4 y-3 z=8\).

What can be said about the vectors \(\mathbf{u}\) and \(\mathbf{v}\) if (a) the projection of \(\mathbf{u}\) onto \(\mathbf{v}\) equals \(\mathbf{u}\) and \((\mathrm{b})\) the projection of \(\mathbf{u}\) onto \(\mathbf{v}\) equals \(\mathbf{0}\) ?

Find the magnitude of \(\mathrm{v}\). \(\mathbf{v}=\mathbf{i}-2 \mathbf{j}-3 \mathbf{k}\)
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