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Magazine Chapter 13: Problem 74
Find the shortest distance between the planes \(2 x-5 y+\) \(z=10\) and \(z=5 y-2 x\)
Short Answer
Expert verified
The shortest distance between the planes is \(\frac{\sqrt{30}}{3}\).
Step by step solution
01
Identify the Normal Vectors
For the plane equation of the form \(a x + b y + c z = d\), the normal vector is \(\langle a, b, c \rangle\). The normal vector for the first plane, \(2x - 5y + z = 10\), is \(\langle 2, -5, 1 \rangle\). For the second plane, \(-2x + 5y - z = 0\), the normal vector is \(\langle -2, 5, -1 \rangle\).
02
Verify if Planes are Parallel
Planes are parallel if their normal vectors are scalar multiples of each other. Check if \(\langle 2, -5, 1 \rangle\) is a scalar multiple of \(\langle -2, 5, -1 \rangle\). Notice that multiplying the vector \(\langle 2, -5, 1 \rangle\) by \(-1\) gives \(\langle -2, 5, -1 \rangle\), confirming that the planes are indeed parallel.
03
Find a Point on Each Plane
Choose simple values for \(x\) and \(y\) to find a point on each plane. For the first plane, let \(x = 0\), \(y = 0\), leading to \(z = 10\). So, a point on the first plane is \((0, 0, 10)\). For the second plane, let \(x = 0\), \(y = 0\), leading to \(z = 0\). So, a point on the second plane is \((0, 0, 0)\).
04
Compute the Shortest Distance
The shortest distance \(d\) between two parallel planes with common normal vector \(\langle a, b, c \rangle\) can be calculated using the formula \(d = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}}\). Substituting our values, where \(d_1 = 10\), \(d_2 = 0\), and the normal vector is \(\langle 2, -5, 1 \rangle\), we calculate \(d = \frac{|0 - 10|}{\sqrt{2^2 + (-5)^2 + 1^2}} = \frac{10}{\sqrt{30}} = \frac{10 \sqrt{30}}{30} = \frac{\sqrt{30}}{3}\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Normal Vector
A normal vector is an essential concept in understanding the geometry of planes. Imagine a plane in three-dimensional space鈥攁 mathematical flat surface. Every plane can be associated with a unique vector called the 'normal vector.' This vector is perpendicular to the plane, meaning it points directly out from the surface.
- The normal vector is represented by a triplet \(\langle a, b, c \rangle\), where \(a, b,\) and \(c\) are coefficients of the plane's equation \(ax + by + cz = d\).
- It determines the plane's orientation in space.
- The method to find it is simple: translate the coefficients of \(x, y,\) and \(z\) into vector form.
Plane Equation
A plane equation is a mathematical statement that defines all the points that lie on a plane. It is the expression of a flat surface in the three-dimensional coordinate system.
- The general form of a plane equation is \(ax + by + cz = d\).
- Here, \(a, b,\) and \(c\) represent the components of the normal vector to the plane.
- The variable \(d\) is called the constant or the distance from the plane to the origin along the normal vector.
Parallel Planes
Parallel planes are planes that maintain a constant distance from each other and never intersect. Understanding this concept requires examining the normal vectors derived from their equations.
- Two planes are parallel if their normal vectors are scalar multiples of each other.
- This means one vector can be multiplied by a constant to produce the other vector.
Distance Formula
Calculating the distance between parallel planes involves a specific formula that uses their normal vector and constant terms. This method finds the shortest distance directly along the direction of the normal vector.
- The formula to find this distance \(d\) is: \[d = \frac{|d_2 - d_1|}{\sqrt{a^2 + b^2 + c^2}}\] \ \ where \(\langle a, b, c \rangle\) is the common normal vector.
- \(d_1\) and \(d_2\) are the constant terms in the equations of each plane.
