/*! 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 39 Find the shortest distance from ... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the shortest distance from the point \((2,1,-1)\) to the plane \(x+y-z=1 .\)

Short Answer

Expert verified
The shortest distance from the point \((2,1,-1)\) to the plane is \(\sqrt{3}\).

Step by step solution

01

Understand the Problem

We need to find the shortest distance from the point \((2,1,-1)\) to the plane given by the equation \(x+y-z=1\). The shortest distance from a point to a plane is the perpendicular distance.
02

Identify the Plane's Normal Vector

The given plane equation is \(x+y-z=1\). The coefficients of \(x\), \(y\), and \(z\) give us the normal vector \(\mathbf{n} = \langle 1, 1, -1 \rangle\) to the plane.
03

Use the Distance Formula

The formula to calculate the distance \(d\) from a point \((x_1, y_1, z_1)\) to a plane \(ax + by + cz + d = 0\) is:\[d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\]For the plane equation \(x + y - z = 1\), we rearrange it to \(x + y - z - 1 = 0\). Here, \(a = 1\), \(b = 1\), \(c = -1\), and \(d = -1\).
04

Substitute Point Coordinates into the Formula

Substitute the coordinates \((2, 1, -1)\) of the point and the plane parameters into the formula:\[d = \frac{|1\cdot2 + 1\cdot1 - 1\cdot(-1) - 1|}{\sqrt{1^2 + 1^2 + (-1)^2}}\]
05

Calculate the Numerator

Calculate the expression in the numerator:\[|1\cdot2 + 1\cdot1 - 1\cdot(-1) - 1| = |2 + 1 + 1 - 1| = |3|\]
06

Calculate the Denominator

Calculate the expression in the denominator:\[\sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{1 + 1 + 1} = \sqrt{3}\]
07

Complete the Distance Calculation

Plug the results from steps 5 and 6 into the distance formula:\[d = \frac{|3|}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3}\]
08

Simplify Result if Necessary

Since \( \sqrt{3} \) is already in its simplest form, no further simplification is needed. The shortest distance is \( \sqrt{3} \).

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Key Concepts

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

Normal Vector
In the context of geometry and vectors, a **normal vector** is a crucial concept. It’s a vector that is perpendicular to a given surface, such as a plane. When dealing with planes, the normal vector is pivotal because it defines the plane's orientation in space. To find the normal vector of a plane from its equation, you can simply look at the coefficients of its variables.

For example, consider the plane equation:
  • \(x + y - z = 1\).
Here, the coefficients of \(x\), \(y\), and \(z\) are 1, 1, and -1 respectively. These coefficients form the components of the normal vector:
  • \(\mathbf{n} = \langle 1, 1, -1 \rangle\).
This vector is perpendicular to every line that lies on the plane, making it a handy tool for numerous calculations, including finding distances.
Distance Formula
The **distance formula** is used in algebra and geometry to find the shortest path from a point to a plane. This formula involves the normal vector of the plane and the coordinates of the point in question. When dealing with a plane equation of the form \(ax + by + cz + d = 0\), the formula for finding the distance \(d\) from a point \((x_1, y_1, z_1)\) to the plane is:
  • \[d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}\]
This equation breaks down neatly:
  • The numerator is the absolute value of the plane equation with the point coordinates substituted in.
  • The denominator is the magnitude of the normal vector, which gives a scale of the steepness.
By applying this formula, one can calculate the perpendicular distance effectively.
Perpendicular Distance
The **perpendicular distance** in three-dimensional space is the shortest distance from a point to a plane. This concept is fundamental in understanding distances in a geometric context. The key characteristic is that it forms a right angle (90 degrees) with the surface it meets, i.e., the plane.
To find this distance, it's necessary to use the point-to-plane distance formula. This method ensures that the distance calculated is the smallest possible, as it travels along the plane's normal vector. Additionally, this perpendicular path is not only mathematically precise but also geometrically optimal.
Plane Equation
The **plane equation** is vital in specifying a plane in three-dimensional space. It’s typically given in the form \(ax + by + cz = d\), or rearranged for calculations as \(ax + by + cz + d = 0\).
  • Each term coefficient \(a\), \(b\), and \(c\) relate to the components of the normal vector to the plane.
  • \(d\) adjusts the plane's position relative to the origin.
The equation not only sets out how the plane appears in space but also is central to deriving other related properties, such as the normal vector. In practical problems, rearranging the terms can facilitate easier computations, like determining distances. The understanding of plane equations is therefore crucial for navigating and solving geometric problems in a 3D space.

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Most popular questions from this chapter

Describe how the graph of \(g\) is obtained from the graph of \(f .\) $$g(x, y)=f(x, y)+2 \quad \text { (b) } g(x, y)=2 f(x, y)$$ $$g(x, y)=-f(x, y) \quad \text { (d) } g(x, y)=2-f(x, y)$$

Use a graphing device as in Example 4 (or Newton's method or a rootfinder) to find the critical points of \(f\) correct to three decimal places. Then classify the critical points and find the highest or lowest points on the graph. $$f(x, y)=2 x+4 x^{2}-y^{2}+2 x y^{2}-x^{4}-y^{4}$$

The total production \(P\) of a certain product depends on the amount \(L\) of labor used and the amount \(K\) of capital investment. In Sections 14.1 and 14.3 we discussed how the CobbDouglas model \(P=b L^{\alpha} K^{1-\alpha}\) follows from certain economic assumptions, where \(b\) and \(\alpha\) are positive constants and \(\alpha<1\) If the cost of a unit of labor is \(m\) and the cost of a unit of capital is \(n,\) and the company can spend only \(p\) dollars as its total budget, then maximizing the production \(P\) is subject to the constraint \(m L+n K=p .\) Show that the maximum production occurs when $$L=\frac{\alpha p}{m} \quad \text { and } \quad K=\frac{(1-\alpha) p}{n}$$

Suppose that a scientist has reason to believe that two quan- tities \(x\) and \(y\) are related linearly, that is, \(y=m x+b,\) at least approximately, for some values of \(m\) and \(b\) . The scientist performs an experiment and collects data in the form of points \(\left(X_{1}, y_{1}\right),\left(x_{2}, y_{2}\right), \ldots,\left(x_{n}, y_{n}\right)\) and then plots these points. The points don't lie exactly on a straight line, so the scientist wants $$ m \text { and } b \text { so that the line } y=m x+b$$ points as well as possible. (See the figure.) Let \(d_{i}=y_{i}-\left(m x_{i}+b\right)\) be the vertical deviation of the point \(\left(x_{i}, y_{i}\right)\) from the line. The method of least squares determines \(m\) and \(b\) so as to minimize \(\Sigma_{1-1}^{n} d_{i}^{2}\) , the sum of the squares of these deviations. Show that, according to this method, the line of best fit is obtained when $$\begin{array}{c}{m \sum_{i=1}^{n} x_{i}+b n=\sum_{i=1}^{n} y_{i}} \\ {m \sum_{i=1}^{n} x_{i}^{2}+b \sum_{i=1}^{n} x_{i}=\sum_{i=1}^{n} x_{i} y_{i}}\end{array}$$ Thus the line is found by solving these two equations in the two unknowns \(m\) and \(b\) . See Section 1.2 for a further discus- sion and applications of the method of least squares.)

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