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Chapter 1: Problem 16

Find the distance between point \(P=(3,1,2)\) and the plane given by \(x-2 y+z=5\) (see the following figure).

Short Answer

Expert verified
The distance is \(\frac{\sqrt{6}}{3}\).

Step by step solution

01

Define the Distance Formula

The distance, \(d\), from a point \((x_1, y_1, z_1)\) to a plane given by \(ax + by + cz + d = 0\) is calculated using the formula:\[ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \]Where \(a\), \(b\), and \(c\) are the coefficients of \(x\), \(y\), and \(z\) in the plane equation.
02

Identify Coefficients and Substitutions

For the given plane equation \(x - 2y + z = 5\), rewrite it in the form \(ax + by + cz + d = 0\):\[ x - 2y + z - 5 = 0 \]From this, we identify \(a = 1\), \(b = -2\), \(c = 1\), and \(d = -5\). The point \(P = (3, 1, 2)\) will be \((x_1, y_1, z_1) = (3, 1, 2)\).
03

Calculate Numerator of the Formula

Substitute the point \(P = (3, 1, 2)\) into the plane equation:\[ 1(3) + (-2)(1) + 1(2) - 5 \]This simplifies to:\[ 3 - 2 + 2 - 5 = -2 \]The absolute value is: \[ |-2| = 2 \]
04

Compute Denominator of the Formula

Using the coefficients \(a = 1\), \(b = -2\), and \(c = 1\), compute the magnitude of the normal vector:\[ \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6} \]
05

Calculate the Distance

Plug the results from the numerator and denominator into the distance formula:\[ d = \frac{2}{\sqrt{6}} \]Simplify the expression:\[ d = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3} \]
06

Final Result

The distance from point \(P = (3,1,2)\) to the plane \(x-2y+z=5\) is \(\frac{\sqrt{6}}{3}\).

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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 crucial mathematical tool for finding the shortest distance between a point and a plane. The idea behind this formula is to measure how far you need to travel in a perpendicular direction from a given point to the nearest point on the plane.

To calculate this distance, you can use the formula: \[ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \] Here鈥檚 what each symbol represents:
  • \( a \), \( b \), and \( c \): the coefficients of the variables \( x \), \( y \), and \( z \), respectively, in the plane equation.
  • \( x_1 \), \( y_1 \), and \( z_1 \): the coordinates of the point from which you鈥檙e measuring the distance.
  • \( d \): the constant in the plane equation when written in the form \( ax + by + cz + d = 0 \).
  • \(|...|\): absolute value, ensuring the distance is always non-negative.
  • \(\sqrt{...}\): square root calculates the magnitude of the normal vector to the plane.
This formula efficiently transforms a seemingly 3D problem into a manageable calculation.
Plane Equation Coefficients
The coefficients in the equation of a plane play a vital role in determining the characteristics of the plane. When you see an equation like \( x - 2y + z = 5 \), these coefficients of \( x \), \( y \), and \( z \) (\( 1 \), \(-2\), and \( 1 \), respectively) are front and center.

These coefficients \((a, b, c)\) help identify the plane鈥檚 orientation in space. The plane can be rewritten as \( ax + by + cz - d = 0 \), where \( a = 1\), \( b = -2\), \( c = 1\), and \( d = -5\).

Recognizing and manipulating these coefficients is essential in:
  • Translating the plane equation into the standard form necessary for calculations.
  • Understanding how changes in \( a \), \( b \), and \( c \) affect the slope and tilt of the plane in three-dimensional space.
  • Substituting them directly in the distance formula to estimate the shortest distance from a point to the plane.
Grasping the role of each coefficient ensures that calculations based on plane equations are accurate and meaningful.
Normal Vector Magnitude
In the geometry of planes, the normal vector is key for understanding how the plane is oriented in space. The normal vector is a perpendicular vector to the plane and is defined by the coefficients \(a\), \(b\), and \(c\) from the plane equation \(ax + by + cz + d = 0\).

To find its magnitude, and hence understand the scaling factor for the vector, use: \[ \sqrt{a^2 + b^2 + c^2} \] This converts the vector into a single number, measuring its length.

Let鈥檚 break it down:
  • In the example equation \( x - 2y + z = 5 \), the coefficients \((1, -2, 1)\) describe the direction in which the normal vector points.
  • The magnitude \( \sqrt{1^2 + (-2)^2 + 1^2} = \sqrt{6} \) gives the length of the vector.
  • This magnitude is crucial in the distance formula because it scales the numerator to normalize the actual distance calculation.
Understanding the concept of a normal vector and its magnitude helps you visualize the plane鈥檚 orientation and make precise distance calculations.

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Most popular questions from this chapter

For the following exercises, points \(P, Q\), and \(R\) are given. Find the general equation of the plane passing through \(P, Q\), and \(R\). Write the vector equation \(\mathbf{n} \cdot \overrightarrow{P S}=0\) of the plane at \(a_{.}\), where \(S(x, y, z)\) is an arbitrary point of the plane. Find parametric equations of the line passing through the origin that is perpendicular to the plane passing through \(P, Q\), and \(R\). Find the distance from point \(P(1,5,-4)\) to the plane of equation \(3 x-y+2 z-6=0\).

Consider vectors \(\mathbf{a}=\langle 2,-4\rangle, \mathbf{b}=\langle-1,2\rangle\), and \(\mathrm{c}=0\) Determine the scalars \(\alpha\) and \(\beta\) such that \(\mathbf{c}=\alpha \mathbf{a}+\beta \mathbf{b}\).

John allocates \(d\) dollars to consume monthly three goods of prices \(a, b\), and \(c .\) In this context, the budget equation is defined as \(a x+b y+c z=d\), where \(x \geq 0, y \geq 0\), and \(z \geq 0\) represent the number of items bought from each of the goods. The budget set is given by \(\\{(x, y, z) \mid a x+b y+c z \leq d, x \geq 0, y \geq 0, z \geq 0\\}\), and the budget plane is the part of the plane of equation \(a x+b y+c z=d\) for which \(x \geq 0, y \geq 0\), and \(z \geq 0 .\) Consider \(a=\$ 8, b=\$ 5\), \(c=\$ 10\), and \(d=\$ 500 .\) a. Use a CAS to graph the budget set and budget plane. b. For \(z=25\), find the new budget equation and graph the budget set in the same system of coordinates.

The intersection between cylinder \((x-1)^{2}+y^{2}=1\) and sphere \(x^{2}+y^{2}+z^{2}=4\) is called a Viviani curve. a. Solve the system consisting of the equations of the surfaces to find the equation of the intersection curve. (Hint: Find \(x\) and \(y\) in terms of \(z .)\) b. Use a computer algebra system (CAS) to visualize the intersection curve on sphere \(x^{2}+y^{2}+z^{2}=4\).

For the following exercises, find the traces for the surfaces in planes \(x=k, y=k\), and \(z=k .\) Then, describe and draw the surfaces\(x^{2}=y^{2}+z^{2}\)
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