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Magazine Chapter 12: Problem 56
Use the formula of Problem 54 to find the distance between the given point and the given plane. The point \(P(5,12,-13)\) and the plane with equation \(3 x+4 y+5 z=12\)
Short Answer
Expert verified
The distance is \(\frac{7 \sqrt{2}}{5}\).
Step by step solution
01
Identify the Formula
The formula to find the distance between a point \(P(x_0, y_0, z_0)\) and a plane \(Ax + By + Cz + D = 0\) is given by: \[d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}.\]
02
Rearrange the Plane Equation
The given plane equation is \(3x + 4y + 5z = 12\). We rearrange it to standard form \(3x + 4y + 5z - 12 = 0\), where \(A = 3\), \(B = 4\), \(C = 5\), and \(D = -12\).
03
Substitute the Point Coordinates
The point coordinates are \(P(5, 12, -13)\), so \(x_0 = 5\), \(y_0 = 12\), and \(z_0 = -13\). Substitute these into the formula: \[d = \frac{|3(5) + 4(12) + 5(-13) - 12|}{\sqrt{3^2 + 4^2 + 5^2}}.\]
04
Calculate the Numerator
Calculate the value inside the absolute value: \[3(5) + 4(12) + 5(-13) - 12 = 15 + 48 - 65 - 12 = -14.\] Thus, the numerator is \(|-14| = 14\).
05
Calculate the Denominator
Calculate the denominator: \[\sqrt{3^2 + 4^2 + 5^2} = \sqrt{9 + 16 + 25} = \sqrt{50}.\]
06
Find the Distance
Substitute the results from the previous steps: \[d = \frac{14}{\sqrt{50}} = \frac{14}{5\sqrt{2}}.\] Simplify further if needed: \[d = \frac{14 \times \sqrt{2}}{10} = \frac{7 \sqrt{2}}{5}.\]
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Distance from Point to Plane
Finding the distance from a point to a plane in three-dimensional space is an interesting problem in vector calculus. The formula is quite simple and involves both the point's coordinates and the plane's coefficients.
- The distance is calculated using the formula: \( d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}} \), where \((x_0, y_0, z_0)\) are the point's coordinates and \(A, B, C, D\) are the plane's coefficients.
- This formula gives the shortest distance (or perpendicular distance) from the point to the plane.
Equation of a Plane
Understanding the equation of a plane is crucial for solving many problems in vector calculus. The general form of a plane's equation is \(Ax + By + Cz + D = 0\).
- Here, \(A, B,\) and \(C\) are coefficients describing the orientation of the plane. They can be viewed as components of a normal vector \(\mathbf{n} = \langle A, B, C \rangle\).
- The constant \(D\) shifts the plane in space relative to the origin.
Magnitude of a Vector
The magnitude of a vector, sometimes called the vector's length or norm, is a measure of how long the vector is. You can think of it as the distance from the origin to the point described by the vector in space.
- The magnitude of a vector \(\mathbf{v} = \langle a, b, c \rangle\) is calculated using \(\sqrt{a^2 + b^2 + c^2}\).
- In the context of our problem, this formula is used to calculate the denominator of the distance formula. Specifically, \(\sqrt{A^2 + B^2 + C^2}\) gives the magnitude of the normal vector to the plane.
Absolute Value
The absolute value is a mathematical concept that describes the non-negative value of a number. The absolute value of a number \(x\), denoted \(|x|\), is the distance of \(x\) from zero on a number line.
- It turns negative results into positive ones, ensuring that distances and lengths (which can't be negative) are properly calculated.
- In the distance from point to plane, the absolute value \(|Ax_0 + By_0 + Cz_0 + D|\) ensures the numerator is always non-negative.
