我在3D空间中有3个向量。我们称他们为xaxis，yaxis和zaxis。这些向量以3D空间中的任意point为中心。我有兴趣将关于xaxis向量的yaxis和zaxis向量转换为多个度θ。
对于以下代码，其值是任意且不重要的：
double xaxis[3], yaxis[3], zaxis[3], point[3], theta;
我如何围绕xaxis yaxis度转移zaxis和theta？
未来注意：这些尝试不起作用。请参阅我的答案以获得正确的解决方案，该解决方案是在BlueRaja-DannyPflughoeft的帮助下找到的
我尝试基于矩阵的旋转：
double rx[3][3];
double ry[3][3];
double rz[3][3];
double r[3][3];
rx[0][0] = 1;
rx[0][1] = 0;
rx[0][2] = 0;
rx[1][0] = 0;
rx[1][1] = cos(theta);
rx[1][2] = sin(theta);
rx[2][0] = 0;
rx[2][1] = -1.0 * sin(theta);
rx[2][2] = cos(theta);
ry[0][0] = cos(theta);
ry[0][1] = 0;
ry[0][2] = -1.0 * sin(theta);
ry[1][0] = 0;
ry[1][1] = 1;
ry[1][2] = 0;
ry[2][0] = sin(theta);
ry[2][1] = 0;
ry[2][2] = cos(theta);
//No rotation wanted on the zaxis
rz[0][0] = cos(0);
rz[0][1] = sin(0);
rz[0][2] = 0;
rz[1][0] = -1.0 * sin(0);
rz[1][1] = cos(0);
rz[1][2] = 0;
rz[2][0] = 0;
rz[2][1] = 0;
rz[2][2] = 1;
vtkMath::Multiply3x3(rx, ry, r); //Multiplies rx by ry and stores into r
vtkMath::Multiply3x3(r, rz, r); //Multiplies r by rz and stores into r
vtkMath::Multiply3x3(r, xaxis, xaxis);//multiplies a 3x3 by a 3x1
vtkMath::Multiply3x3(r, yaxis, yaxis);//multiplies a 3x3 by a 3x1
此尝试仅在飞机位于x-y平面时有效：
double x, y;
x = xaxis[0];
y = xaxis[1];
xaxis[0] = x * cos(theta) - y * sin(theta);
xaxis[1] = x * sin(theta) + y * cos(theta);
x = yaxis[0];
y = yaxis[1];
yaxis[0] = x * cos(theta) - y * sin(theta);
yaxis[1] = x * sin(theta) + y * cos(theta);
使用BlueRaja-DannyPflughoeft给出的轴角方法：
double c = cos(theta);
double s = sin(theta);
double C = 1.0 - c;
double Q[3][3];
Q[0][0] = xaxis[0] * xaxis[0] * C + c;
Q[0][1] = xaxis[1] * xaxis[0] * C + xaxis[2] * s;
Q[0][2] = xaxis[2] * xaxis[0] * C - xaxis[1] * s;
Q[1][0] = xaxis[1] * xaxis[0] * C - xaxis[2] * s;
Q[1][1] = xaxis[1] * xaxis[1] * C + c;
Q[1][2] = xaxis[2] * xaxis[1] * C + xaxis[0] * s;
Q[2][0] = xaxis[1] * xaxis[2] * C + xaxis[1] * s;
Q[2][1] = xaxis[2] * xaxis[1] * C - xaxis[0] * s;
Q[2][2] = xaxis[2] * xaxis[2] * C + c;
double x = Q[2][1] - Q[1][2], y = Q[0][2] - Q[2][0], z = Q[1][0] - Q[0][1];
double r = sqrt(x * x + y * y + z * z);
//xaxis[0] /= r;
//xaxis[1] /= r;
//xaxis[2] /= r;
xaxis[0] = x;// ?
xaxis[1] = y;
xaxis[2] = z;
答案 0 :(得分：2)
感谢BlueRaja - Danny Pflughoeft：
double c = cos(theta);
double s = sin(theta);
double C = 1.0 - c;
double Q[3][3];
Q[0][0] = zaxis[0] * zaxis[0] * C + c;
Q[0][1] = zaxis[1] * zaxis[0] * C + zaxis[2] * s;
Q[0][2] = zaxis[2] * zaxis[0] * C - zaxis[1] * s;
Q[1][0] = zaxis[1] * zaxis[0] * C - zaxis[2] * s;
Q[1][1] = zaxis[1] * zaxis[1] * C + c;
Q[1][2] = zaxis[2] * zaxis[1] * C + zaxis[0] * s;
Q[2][0] = zaxis[0] * zaxis[2] * C + zaxis[1] * s;
Q[2][1] = zaxis[2] * zaxis[1] * C - zaxis[0] * s;
Q[2][2] = zaxis[2] * zaxis[2] * C + c;
xaxis[0] = xaxis[0] * Q[0][0] + xaxis[0] * Q[0][1] + xaxis[0] * Q[0][2];
xaxis[1] = xaxis[1] * Q[1][0] + xaxis[1] * Q[1][1] + xaxis[1] * Q[1][2];
xaxis[2] = xaxis[2] * Q[2][0] + xaxis[2] * Q[2][1] + xaxis[2] * Q[2][2]; // Multiply a 3x3 by 3x1 and store it as the new rotated axis
yaxis[0] = yaxis[0] * Q[0][0] + yaxis[0] * Q[0][1] + yaxis[0] * Q[0][2];
yaxis[1] = yaxis[1] * Q[1][0] + yaxis[1] * Q[1][1] + yaxis[1] * Q[1][2];
yaxis[2] = yaxis[2] * Q[2][0] + yaxis[2] * Q[2][1] + yaxis[2] * Q[2][2]; // Multiply a 3x3 by 3x1 and store it as the new rotated axis
答案 1 :(得分：0)
我看到跟随矩阵乘法是错误的！
如上所述，可以使用xaxis[0]
xaxis[0] = xaxis[0] * Q[0][0] + xaxis[0] * Q[0][1] + xaxis[0] * Q[0][2];
xaxis[0] = xaxis[0] * (Q[0][0] + Q[0][1] + Q[0][2]);
这看起来不像矩阵乘法。它应该是：
xaxis1[0] = xaxis[0] * Q[0][0] + xaxis[1] * Q[0][1] + xaxis[2] * Q[0][2];
xaxis1[1] = xaxis[0] * Q[1][0] + xaxis[1] * Q[1][1] + xaxis[2] * Q[1][2];
xaxis1[2] = xaxis[0] * Q[2][0] + xaxis[1] * Q[2][1] + xaxis[2] * Q[2][2]; // Multiply a 3x3 by 3x1 and store it as the new rotated axis
yaxis1[0] = yaxis[0] * Q[0][0] + yaxis[1] * Q[0][1] + yaxis[2] * Q[0][2];
yaxis1[1] = yaxis[0] * Q[1][0] + yaxis[1] * Q[1][1] + yaxis[2] * Q[1][2];
yaxis1[2] = yaxis[0] * Q[2][0] + yaxis[1] * Q[2][1] + yaxis[2] * Q[2][2]; // Multiply a 3x3 by 3x1 and store it as the new rotated axis