计算两个向量垂直于在3D第三个向量 [英] Computing two vectors that are perpendicular to third vector in 3D

问题描述

这是在最好（最快）方式来计算两个向量垂直于所述第三矢量（X）和也彼此垂直？

What is the best (fastest) way to compute two vectors that are perpendicular to the third vector(X) and also perpendicular to each other?

这是我应该如何计算这个向量将现在：

This is how am I computing this vectors right now:

// HELPER - unit vector that is NOT parallel to X
x_axis = normalize(X);
y_axis = crossProduct(x_axis, HELPER);
z_axis = crossProduct(x_axis, y_axis);

我知道有解决这一无限多的，我不关心哪一个会是我的解决办法。

I know there is infinite number of solutions to this, and I don't care which one will be my solution.

这背后的问题：我需要构造变换矩阵，在那里我知道哪个方向应该X轴（矩阵第一列）被人指指点点。我需要计算Y和Z轴（第二和第三列）。正如我们所知，所有轴必须垂直于对方。

What is behind this question: I need to construct transformation matrix, where I know which direction should X axis (first column in matrix) be pointing. I need to calculate Y and Z axis (second and third column). As we know, all axes must be perpendicular to each other.

推荐答案

我做了什么，只要 X＆LT;＆GT; 0 或 Y'LT ;＆GT; 0 是

1. A = [-Y，X，0]
2. B = [-X * Z，Y * Z，X * X + Y * Y]

1. A = [-Y, X, 0]
2. B = [-X*Z, -Y*Z, X*X+Y*Y]

然后规范化载体。

[ X,Y,Z]·[-Y,X,0] = -X*Y+Y*X = 0
[ X,Y,Z]·[-X*Z,-Y*Z,X*X+Y*Y] = -X*X*Z-Y*Y*Z+Z*(X*X+Y*Y) = 0
[-Y,X,0]·[-X*Z,-Y*Z,X*X+Y*Y] = Y*X*Z+X*Y*Z = 0

这就是所谓的向量的零空间的。

如果 X = 0 和 Y = 0 然后 A = [1,0 ，0] ， B = [0,1,0] 。

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