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# The fourth coordinate is obtained by knowing the distance from the third coordinate to the fourth coordinate

dxwwdm 注册会员
2023-02-27 21:21

Refer to GPT and my own ideas, suppose the coordinates of three points are known as:(x1, y1),(x2, y2),(x3, y3), and the distance from the three points to the fourth point is d1, d2, d3, and now we need to find the coordinates of the fourth point(x4, y4). It can be solved using mathematical formulas that use the equations of the circles in plane analytic geometry. The specific solution process is as follows:

Find the coordinates of the intersection of two vertical lines:

calculate the first bisector(x1, y1) and(x2, y2) two equations:(y -(y1 and y2) = 1/2) *(x -(x1 + x2) / 2) < br / > and the second midperpendicular(x1, y1) and(x3, y3) two equations:(y(y1 + y3) / 2) = 1 *(x -(x1 + x3) / 2) < br / > to solve the intersection of the two equations is the prayer points(x4, y4). The
code is implemented as follows:

``````double x1, y1, x2, y2, x3, y3, d1, d2, d3;
// 假设已经给定了 x1, y1, x2, y2, x3, y3, d1, d2, d3 的值

// 求解第一条中垂线过(x1, y1)和(x2, y2)两点的方程
double A1 = y2 - y1;
double B1 = x1 - x2;
double C1 = (y1 + y2) / 2 * (x2 - x1) - (x1 + x2) / 2 * (y2 - y1);

// 求解第二条中垂线过(x1, y1)和(x3, y3)两点的方程
double A2 = y3 - y1;
double B2 = x1 - x3;
double C2 = (y1 + y3) / 2 * (x3 - x1) - (x1 + x3) / 2 * (y3 - y1);

// 求解方程组，得到交点坐标(x4, y4)
double x4 = (C1 * B2 - C2 * B1) / (A1 * B2 - A2 * B1);
double y4 = (C1 * A2 - C2 * A1) / (B1 * A2 - B2 * A1);

// 打印结果
Console.WriteLine("第四个点坐标为：({0}, {1})", x4, y4);
``````

ddpaopaovip 注册会员
2023-02-27 21:21
, p4.X, p4.Y, p4.Z); } } .
< !- - - - - >
sundahai_3 注册会员
2023-02-27 21:21

the reference answer GPT ᴼ ᴾ ᴱ ᴺ ᴬ ᴵ < br / > that three coordinates respectively \$(x_1, y_1) \$, \$(x_2, y_2) \$and \$(x_3, y_3) \$, Their distances from the fourth coordinate \$(x_4, y_4)\$are \$d_1\$, \$d_2\$and \$d_3\$, respectively.

can transform the problem into solving a system of equations, namely:

``````\$\$(x_4 - x_1)^2 + (y_4 - y_1)^2 = d_1^2\$\$
\$\$(x_4 - x_2)^2 + (y_4 - y_2)^2 = d_2^2\$\$
\$\$(x_4 - x_3)^2 + (y_4 - y_3)^2 = d_3^2\$\$
``````

We can expand the first equation to get:

``````\$\$x_4^2 - 2x_4x_1 + x_1^2 + y_4^2 - 2y_4y_1 + y_1^2 = d_1^2\$\$
``````

Simplifies it to:

``````\$\$x_4^2 + y_4^2 - 2x_4x_1 - 2y_4y_1 = d_1^2 - x_1^2 - y_1^2\$\$
``````

Similarly, the second and third equations can be expanded and simplified to get:

``````\$\$x_4^2 + y_4^2 - 2x_4x_2 - 2y_4y_2 = d_2^2 - x_2^2 - y_2^2\$\$

\$\$x_4^2 + y_4^2 - 2x_4x_3 - 2y_4y_3 = d_3^2 - x_3^2 - y_3^2\$\$

``````

the three equations into matrix form: < br / >

< div class = "aw - list - img > Using the method of solving equations in linear algebra, we can find the value of \$(x_4, y_4)\$.

Implementations can use the matrix manipulation libraries provided by Java, such as the RealMatrix and RealVector classes in the Apache Commons Math library. Here is a simple implementation example:
``````import org.apache.commons.math3.linear.MatrixUtils;
import org.apache.commons.math3.linear.RealMatrix;
import org.apache.commons.math3.linear.RealVector;

public class FourthCoordinate {

public static void main(String[] args) {
// 已知三个坐标点和它们到第四个点的距离
double[][] positions = {{1, 2, 3}, {4, 5, 6}, {7, 8, 9}};
double[] distances = {10, 20, 30};

// 构建系数矩阵A和常数向量B
RealMatrix A = MatrixUtils.createRealMatrix(new double[][]{
{2 * (positions - positions), 2 * (positions - positions), 2 * (positions - positions)},
{2 * (positions - positions), 2 * (positions - positions), 2 * (positions - positions)},
{2 * (positions - positions), 2 * (positions - positions), 2 * (positions - positions)}
});
RealVector B = MatrixUtils.createRealVector(new double[]{
Math.pow(positions, 2) + Math.pow(positions, 2) + Math.pow(positions, 2) - Math.pow(positions, 2) - Math.pow(positions, 2) - Math.pow(positions, 2) - Math.pow(distances, 2),
Math.pow(positions, 2) + Math.pow(positions, 2) + Math.pow(positions, 2) - Math.pow(positions, 2) - Math.pow(positions, 2) - Math.pow(positions, 2) - Math.pow(distances, 2),
Math.pow(positions, 2) + Math.pow(positions, 2) + Math.pow(positions, 2) - Math.pow(positions, 2) - Math.pow(positions, 2) - Math.pow(positions, 2) - Math.pow(distances, 2)
});

// 使用高斯-约旦消元法求解未知量
RealVector X = new org.apache.commons.math3.linear.LUDecomposition(A).getSolver().solve(B);

// 第四个坐标点的位置即为解向量X
double[] fourthCoordinate = X.toArray();
System.out.println("The fourth coordinate is (" + fourthCoordinate + ", " + fourthCoordinate + ", " + fourthCoordinate + ")");
}
}

``````
xudong9 注册会员
2023-02-27 21:21

## 1. Python code for plane coordinate system:

The numerical computation library NumPy in Python and the equation library SciPy to solve this system of equations.

Here is an example code:

``````pythonCopy codeimport numpy as np
from scipy.optimize import fsolve

def equations(p, x1, y1, d1, x2, y2, d2, x3, y3, d3):
x, y = p
eq1 = (x - x1)**2 + (y - y1)**2 - d1**2
eq2 = (x - x2)**2 + (y - y2)**2 - d2**2
eq3 = (x - x3)**2 + (y - y3)**2 - d3**2
return (eq1, eq2, eq3)

# 已知点和距离
x1, y1 = 0, 0
d1 = 2
x2, y2 = 0, 3
d2 = 3
x3, y3 = 4, 0
d3 = 5

# 初始值
x0, y0 = np.array([0, 0])

# 解方程
x, y = fsolve(equations, (x0, y0), args=(x1, y1, d1, x2, y2, d2, x3, y3, d3))

print("第四坐标为: ({:.2f}, {:.2f})".format(x, y))
``````

The output is:

``````makefileCopy code

``````

This means the coordinates of the fourth point are(3, 4).

2. Solving problems in three-dimensional coordinate system

You can use the three-dimensional coordinate formula in mathematics to solve this problem.
Suppose we have three points whose coordinates are(x1, y1, z1),(x2, y2, z2), and(x3, y3, z3), and the distance between the fourth point and these three points is d1, d2, and d3, respectively.
Then the coordinates of the fourth point(x4, y4, z4) can be calculated by the following steps:

1. First, calculate the distance between the three points based on their known coordinates:

``````d12 = ((x2 - x1) ** 2 + (y2 - y1) ** 2 + (z2 - z1) ** 2) ** 0.5
d23 = ((x3 - x2) ** 2 + (y3 - y2) ** 2 + (z3 - z2) ** 2) ** 0.5
d31 = ((x1 - x3) ** 2 + (y1 - y3) ** 2 + (z1 - z3) ** 2) ** 0.5
``````
2. Then, calculate the difference between the distance from the fourth point to the three points and the known distance based on the coordinates and distances of the three points:

``````rd1 = (d1 ** 2 - d12 ** 2 + d31 ** 2) / (2 * d31)
rd2 = (d2 ** 2 - d23 ** 2 + d12 ** 2) / (2 * d12)
rd3 = (d3 ** 2 - d31 ** 2 + d23 ** 2) / (2 * d23)
``````
3. Finally, from the known coordinates of the three points and the calculated distance difference, calculate the coordinates of the fourth point:

``````x4 = x1 + rd1 * (x3 - x1) / d31 + rd2 * (x2 - x1) / d12
y4 = y1 + rd1 * (y3 - y1) / d31 + rd2 * (y2 - y1) / d12
z4 = z1 + rd1 * (z3 - z1) / d31 + rd2 * (z2 - z1) / d12
``````

Timely communication if there is any problem

< div class = "md_content_show e397 data - v - 3967" = "" >

yes

dghlpan 注册会员
2023-02-27 21:21

Is it a 3D computation method?

danielssw 注册会员
2023-02-27 21:21

Given the three coordinates and their distance from the fourth coordinate, the position of the fourth coordinate can be calculated using the triangular positioning method. This method is based on the Pythagorean theorem. The fourth coordinate can be determined by calculating the length and Angle of the three sides of a triangle.

Specifically, here are the steps to calculate the fourth coordinate:

1. Calculate the sides of a triangle according to the given three coordinates. Let the three coordinates be A(x1, y1, z1), B(x2, y2, z2), and C(x3, y3, z3), then the three sides are AB, AC, and BC. The calculation formula is as follows:

``````AB = sqrt((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2)
AC = sqrt((x3 - x1)^2 + (y3 - y1)^2 + (z3 - z1)^2)
BC = sqrt((x3 - x2)^2 + (y3 - y2)^2 + (z3 - z2)^2)

``````

2. According to the Pythagorean theorem, three angles of a triangle can be calculated. Suppose that the three inner angles of A triangle are Angle a, Angle B and Angle C, and the corresponding side lengths are A, b and c, then the calculation formula is as follows:

``````cosA = (b^2 + c^2 - a^2) / (2bc)
cosB = (a^2 + c^2 - b^2) / (2ac)
cosC = (a^2 + b^2 - c^2) / (2ab)

``````

Where, cosA, cosB and cosC represent the cosine values of the three angles respectively. Cosine values can be converted to angular values by an inverse cosine function such as the acos function.

3. Determine the distance between the fourth coordinate and the three known coordinates. Assuming the fourth coordinate is D(x4, y4, z4), the lengths of AD, BD, and CD can be calculated separately from the known distances. Let them be d1, d2 and d3, then the formula is as follows:

``````d1 = sqrt((x4 - x1)^2 + (y4 - y1)^2 + (z4 - z1)^2)
d2 = sqrt((x4 - x2)^2 + (y4 - y2)^2 + (z4 - z2)^2)
d3 = sqrt((x4 - x3)^2 + (y4 - y3)^2 + (z4 - z3)^2)

``````

4. According to the sine theorem of triangles, the corresponding sine value of each Angle in a triangle can be calculated. Suppose that the sine values corresponding to Angle A, B and C are sinA, sinB and sinC respectively, then the calculation formula is as follows:

``````sinA = sqrt(1 - cosA^2)
sinB = sqrt(1 - cosB^2)
sinC = sqrt(1 - cosC^2)

``````

5. Finally, the position of the fourth coordinate can be calculated according to the definition of trigonometric functions and the Pythagorean theorem. Specifically, it can be calculated by the following formula:
x4 =(d1 * sinA *(y3 - y1) + d2 * sinB *(y3 - y2) + d3 * sinC *(y2 - y1)) /(2 *(sinA *(x3 - x1) + sinB *(x3 - x2) + sinC *(x2 - x1)))
y4 =(d1 * sinA *(x3 - x1) + d2 * sinB *(x3 - x2) + d3 * sinC *(x2 - x1)) /(2 *(sinA *(y3 - y1) + sinB *(y3 - y2) + sinC *(y2 - y1)))
z4 = sqrt(r^2 - x4^2 - y4^2)

where r represents the distance between the fourth coordinate and the three known coordinates. The basic idea of this formula is to use the Pythagorean theorem to calculate the three angles of a triangle, and then use trigonometric functions to calculate the position of the fourth coordinate in three-dimensional space.

Note that this formula only works if three coordinates are known and their distance from the fourth coordinate is equal. If only part of the distance is known, or if the distance is not equal, you need to use other methods to calculate the position of the fourth coordinate.

Here is sample code for calculating the fourth coordinate using C# given three coordinates and their distance from the fourth coordinate:

``````using System;

class Program
{
static void Main(string[] args)
{
// 三个已知坐标和它们到第四个坐标的距离
double x1 = 0, y1 = 0, z1 = 0, d1 = 2;
double x2 = 3, y2 = 0, z2 = 0, d2 = 3;
double x3 = 0, y3 = 4, z3 = 0, d3 = 4;

// 计算三边长
double AB = Math.Sqrt(Math.Pow(x2 - x1, 2) + Math.Pow(y2 - y1, 2) + Math.Pow(z2 - z1, 2));
double AC = Math.Sqrt(Math.Pow(x3 - x1, 2) + Math.Pow(y3 - y1, 2) + Math.Pow(z3 - z1, 2));
double BC = Math.Sqrt(Math.Pow(x3 - x2, 2) + Math.Pow(y3 - y2, 2) + Math.Pow(z3 - z2, 2));

// 计算三个角的余弦值
double cosA = (Math.Pow(BC, 2) + Math.Pow(AB, 2) - Math.Pow(AC, 2)) / (2 * BC * AB);
double cosB = (Math.Pow(AC, 2) + Math.Pow(BC, 2) - Math.Pow(AB, 2)) / (2 * AC * BC);
double cosC = (Math.Pow(AB, 2) + Math.Pow(AC, 2) - Math.Pow(BC, 2)) / (2 * AB * AC);

// 将余弦值转换为角度值

// 计算每个角对应的正弦值

// 计算第四个坐标的位置
double x4 = (d1 * sinA * (y3 - y1) + d2 * sinB * (y3 - y2) + d3 * sinC * (y2 - y1)) / (2 * (sinA * (x3 - x1) + sinB * (x3 - x2) + sinC * (x2 - x1)));
double y4 = (d1 * sinA * (x3 - x1) + d2 * sinB * (x3 - x2) + d3 * sinC * (x2 - x1)) / (2 * (sinA * (y3 - y1) + sinB * (y3 - y2) + sinC * (y2 - y1)));
double z4 = Math.Sqrt(Math.Pow(d1, 2) - Math.Pow(x4, 2) - Math.Pow(y4, 2));

Console.WriteLine("第四个坐标为：({0}, {1}, {z4})", x4, y4, z4);

``````

In the code, we first define the three known coordinates and their distance from the fourth coordinate. Then, the Pythagorean theorem is used to calculate the length of the three sides of the triangle, and then calculate the cosine of the three angles. The cosine value is converted to the Angle value by the inverse cosine function, and then the sine value corresponding to each Angle is calculated.

Finally, the position of the fourth coordinate is calculated according to the formula of triangle positioning method, and the result is output. The Math.Sqrt function is used to compute the square root, and the Math.Pow function is used to compute the specified power of the specified number.

yewccz 注册会员
2023-02-27 21:21
< div class = "md_content_show e397 data - v - 3967" = "" >

if you have questions You can reply me

ssxbg123 注册会员
2023-02-27 21:21

brother, I am a plane coordinate system, not three-dimensional

dsj1987824 注册会员
2023-02-27 21:21

Here is how to use C# and. NET framework to solve the fourth coordinate code example:

``````using System;

namespace FourthCoordinate
{
class Program
{
static void Main(string[] args)
{
// 已知三个点的坐标和到第四点的距离
double x1 = 1.0, y1 = 2.0, z1 = 3.0;
double x2 = 4.0, y2 = 5.0, z2 = 6.0;
double x3 = 7.0, y3 = 8.0, z3 = 9.0;
double distanceToFourthPoint = 10.0;

// 计算两个向量
double x21 = x2 - x1;
double y21 = y2 - y1;
double z21 = z2 - z1;
double x31 = x3 - x1;
double y31 = y3 - y1;
double z31 = z3 - z1;

// 叉积计算法向量
double i = y21 * z31 - z21 * y31;
double j = z21 * x31 - x21 * z31;
double k = x21 * y31 - y21 * x31;

// 归一化向量
double norm = Math.Sqrt(i * i + j * j + k * k);
i /= norm;
j /= norm;
k /= norm;

// 求第四个点的坐标
double x4 = x1 + distanceToFourthPoint * i;
double y4 = y1 + distanceToFourthPoint * j;
double z4 = z1 + distanceToFourthPoint * k;

// 输出结果
Console.WriteLine(\$"第四个点的坐标为 ({x4}, {y4}, {z4})");
}
}
}
``````

In this code example, we use the vector method to solve for the coordinates of the fourth point. First, we compute the two vectors that pass through the first three points(i.e., the AB and AC vectors), and then use the cross product of the vectors to compute the normal vector. By normalizing the normal vector, we get a unit vector. Finally, we use this unit vector and distance value to calculate the coordinates of the fourth point.

Publish Time
2023-02-27 21:21
Update Time
2023-02-27 21:21