# qr code vb.net open source Radius = r in .NET Creator Code 39 Full ASCII in .NET Radius = r

Code-39 Scanner In Visual Studio .NET
Using Barcode Control SDK for Visual Studio .NET Control to generate, create, read, scan barcode image in VS .NET applications.
Code 39 Creation In Visual Studio .NET
Using Barcode generator for Visual Studio .NET Control to generate, create ANSI/AIM Code 39 image in VS .NET applications.
Figure 17-4 A sphere of radius r in Cartesian xyz space, centered
Reading Code 3/9 In .NET
Using Barcode recognizer for .NET Control to read, scan read, scan image in Visual Studio .NET applications.
Encoding Bar Code In .NET
Using Barcode creation for VS .NET Control to generate, create bar code image in .NET applications.
away from the origin All points on the sphere s surface are at distance r from the center point (x0,y0,z0)
Decode Bar Code In Visual Studio .NET
Using Barcode decoder for .NET framework Control to read, scan read, scan image in VS .NET applications.
Code 39 Extended Encoder In Visual C#
Using Barcode generator for Visual Studio .NET Control to generate, create ANSI/AIM Code 39 image in VS .NET applications.
Surfaces in Three-Space
Code-39 Drawer In Visual Studio .NET
Using Barcode generation for ASP.NET Control to generate, create Code 39 Full ASCII image in ASP.NET applications.
Generating Code 3 Of 9 In Visual Basic .NET
Using Barcode generation for .NET Control to generate, create USS Code 39 image in VS .NET applications.
Squaring both sides of this equation and then transposing the left- and right-hand sides, we obtain (xp x0)2 + (yp y0)2 + (zp z0)2 = r2 Every point on the sphere s surface is the same distance from P as (x0,y0,z0), so we can generalize to get (x x0)2 + (y y0)2 + (z z0)2 = r2 This is the standard-form equation for a sphere of radius r, centered at the point (x0,y0,z0) in Cartesian xyz space
Bar Code Creator In .NET Framework
Using Barcode printer for Visual Studio .NET Control to generate, create barcode image in Visual Studio .NET applications.
DataMatrix Generation In .NET Framework
Using Barcode creation for VS .NET Control to generate, create Data Matrix image in .NET framework applications.
An example Suppose we have a sphere whose center is at the origin, and whose radius is 7 units If we let r = 7 in the general equation for a sphere centered at the origin, then we have
Code-128 Printer In .NET Framework
Using Barcode printer for .NET framework Control to generate, create Code-128 image in .NET framework applications.
Print USPS POSTal Numeric Encoding Technique Barcode In Visual Studio .NET
Using Barcode encoder for .NET Control to generate, create Delivery Point Barcode (DPBC) image in .NET applications.
x2 + y2 + z2 = 72 which can be simplified to x2 + y2 + z2 = 49
GS1 - 13 Maker In Java
Using Barcode drawer for Java Control to generate, create EAN / UCC - 13 image in Java applications.
Make GTIN - 12 In None
Using Barcode generation for Microsoft Word Control to generate, create Universal Product Code version A image in Word applications.
Another example Consider a sphere centered at the point ( 2,4, 1) with a radius of 5 units in Cartesian xyz space We can let
Encoding UPCA In Java
Using Barcode creation for Android Control to generate, create UPC-A Supplement 5 image in Android applications.
EAN 128 Generator In None
Using Barcode printer for Online Control to generate, create GTIN - 128 image in Online applications.
x0 = 2 y0 = 4 z0 = 1 r=5 in the general equation for a sphere centered at a point other than the origin When we plug in the numbers, we obtain [x ( 2)]2 + (y 4)2 + [z ( 1)]2 = 52 which simplifies to (x + 2)2 + (y 4)2 + (z + 1)2 = 25
EAN / UCC - 13 Scanner In None
Using Barcode decoder for Software Control to read, scan read, scan image in Software applications.
Code 128 Generation In .NET
Using Barcode generation for Reporting Service Control to generate, create Code128 image in Reporting Service applications.
Spheres
Recognizing Bar Code In C#
Using Barcode scanner for .NET Control to read, scan read, scan image in Visual Studio .NET applications.
Code 128 Code Set B Generator In Java
Using Barcode generation for BIRT reports Control to generate, create Code 128 image in Eclipse BIRT applications.
Are you confused
The radius of a sphere is usually defined as a positive real number If we define the radius of a particular sphere as a negative real number, we get the same equation as we would if we defined the radius as the absolute value of that number That s because we square the radius when we work out the formula For example, if we have a sphere centered at the origin with a radius of 4 units, then its equation is x2 + y2 + z2 = 42 which simplifies to x2 + y2 + z2 = 16 If we find a companion antisphere centered at the origin with radius 4 units, then its equation is x2 + y2 + z2 = ( 4)2 which also simplifies to x2 + y2 + z2 = 16 In physics and engineering, it s possible to come up with spheres having negative radii, as well as negative dimensions for other physical objects These results are usually mere artifacts of the calculation process, and don t have any significance in the real world However, if you ever encounter a sphere whose radius is represented by an imaginary number such as j4, then you have good reason to be confused until you know what sort of object or phenomenon the equation describes!
Here s a challenge!
Suppose we re told that the following equation represents a sphere in Cartesian xyz space: x2 6x + y2 2y + z2 + 4z = 86 We re also informed that the sphere has a radius of 10 units What are the coordinates of the center of the sphere
Solution
To solve this problem, we need some intuition We know that the radius r of the sphere is 10 units Therefore, r2 = 100 If we add 14 to both sides of the original equation, we get r2 on the right-hand side: x2 6x + y2 2y + z2 + 4z + 14 = 100 We can split the stand-alone constant 14 into the sum of 9, 1, and 4, getting x2 6x + y2 2y + z2 + 4z + 9 + 1 + 4 = 100
Surfaces in Three-Space
Rearranging the addends on the left-hand side produces the following equation: x2 6x + 9 + y2 2y + 1 + z2 + 4z + 4 = 100 When we group the terms on the left-hand side by threes, we obtain (x2 6x + 9) + (y2 2y + 1) + (z2 + 4z + 4) = 100 This is a sum of three perfect squares! Factoring them individually gives us (x 3)2 + (y 1)2 + (z + 2)2 = 100 The coordinates of the center point are therefore x0 = 3 y0 = 1 z0 = 2 Expressed as an ordered triple, it s (3,1, 2)