给定一个任意单位向量，计算任意正交单位向量的最佳方法是什么? [英] Given a single arbitrary unit vector, what is the best method to compute an arbitrary orthogonal unit vector?

问题描述

基本上问了同样的问题here，但在非编程环境中.建议的解决方案是采用 { y, -x, 0 }.这适用于所有具有 x 或 y 分量的向量，但如果向量等于 + 或 - { 0, 0, 1 } 则失败.在这种情况下，我们将得到 { 0, 0, 0 }.

Essentially the same question was asked here, but in a non-programming context. A suggested solution is take { y, -x, 0 }. This would work for all vectors that have an x or y component, but fails if the vector is equal to + or - { 0, 0, 1 }. In this case we would get { 0, 0, 0 }.

我目前的解决方案(在 C++ 中):

My current solution (in c++):

// floating point comparison utilizing epsilon
bool is_equal(float, float);

// ...

vec3 v = /* some unit length vector */

// ...

// Set as a non-parallel vector which we will use to find the 
//   orthogonal vector. Here we choose either the x or y axis.
vec3 orthog;
if( is_equal(v.x, 1.0f) )
  orthog.set(1.0f, 0.0f, 0.0f);
else
  orthog.set(0.0f, 1.0f, 0.0f);

// Find orthogonal vector
orthog = cross(v, orthog);
orthog.normalize(); 

这个方法有效，但我觉得可能有更好的方法，我的搜索结果没有了.

This method works, but I feel that there may be a better method and my searches turn up nothing more.

只是为了好玩，我在 C++ 中快速编写了每个建议答案的幼稚实现代码，并验证它们都有效(尽管有些并不总是自然地返回单位向量，但我在需要的地方添加了一个 noramlize() 调用).

Just for fun I did a quick code up of naive implementations of each of the suggested answers in c++ and verified they all worked (though some don't always return a unit vector naturally, I added a noramlize() call where needed).

我最初的想法:

vec3 orthog_peter(vec3 const& v)
{
  vec3 arbitrary_non_parallel_vec = v.x != 1.0f ? vec3(1.0, 0.0f, 0.0f) : vec3(0.0f, 1.0f, 0.0f);
  vec3 orthog = cross(v, arbitrary_non_parallel_vec);

  return normalize( orthog );
}

https://stackoverflow.com/a/19650362/2507444

vec3 orthog_robert(vec3 const& v)
{
  vec3 orthog;
  if(v.x == 0.0f && v.y == 0.0f)
    orthog = vec3(1.0f, 1.0f, 0.0f);
  else if(v.x == 0.0f)
    orthog = vec3(1.0f, v.z / v.y, 1.0f);
  else if(v.y == 0.0f)
    orthog = vec3(-v.z / v.x, 1.0f, 1.0f);
  else
    orthog = vec3(-(v.z + v.y) / v.x, 1.0f, 1.0f);

  return normalize(orthog);
}

https://stackoverflow.com/a/19651668/2507444

// NOTE: u and v variable names are swapped from author's example
vec3 orthog_abhishek(vec3 const& v)
{
  vec3 u(1.0f, 0.0f, 0.0f);
  float u_dot_v = dot(u, v);

  if(abs(u_dot_v) != 1.0f)
    return normalize(u + (v * -u_dot_v));
  else
    return vec3(0.0f, 1.0f, 0.0f);
}

https://stackoverflow.com/a/19658055/2507444

vec3 orthog_dmuir(vec3 const& v)
{
  float length = hypotf( v.x, hypotf(v.y, v.z));
  float dir_scalar = (v.x > 0.0) ? length : -length;
  float xt = v.x + dir_scalar;
  float dot = -v.y / (dir_scalar * xt);

  return vec3(
    dot * xt, 
    1.0f + dot * v.y, 
    dot * v.z);
};

推荐答案

好吧，这是解决问题的一种方法.给定一个向量 (a, b, c).解方程 (a, b, c) dot (aa, bb, cc) = 0 for aa, bb, and cc(并确保 aa, bb, and cc 不全为零)，所以 (aa, bb, cc) 与 (a, b, c) 正交.我使用了 Maxima (http://maxima.sf.net) 来解决它.

Well, here's one way to go about it. Let a vector (a, b, c) be given. Solve the equation (a, b, c) dot (aa, bb, cc) = 0 for aa, bb, and cc (and ensuring that aa, bb, and cc are not all zero), so (aa, bb, cc) is orthogonal to (a, b, c). I've used Maxima (http://maxima.sf.net) to solve it.

(%i42) solve ([a, b, c] . [aa, bb, cc] = 0, [aa, bb, cc]), a=0, b=0;
(%o42)                 [[aa = %r19, bb = %r20, cc = 0]]
(%i43) solve ([a, b, c] . [aa, bb, cc] = 0, [aa, bb, cc]), a=0;
                                        %r21 c
(%o43)              [[aa = %r22, bb = - ------, cc = %r21]]
                                          b
(%i44) solve ([a, b, c] . [aa, bb, cc] = 0, [aa, bb, cc]), b=0;
                             %r23 c
(%o44)              [[aa = - ------, bb = %r24, cc = %r23]]
                               a
(%i45) solve ([a, b, c] . [aa, bb, cc] = 0, [aa, bb, cc]);
                        %r25 c + %r26 b
(%o45)         [[aa = - ---------------, bb = %r26, cc = %r25]]
                               a

请注意，我首先解决了特殊情况(a = 0 和 b = 0，或 a = 0，或 b = 0)，因为找到的解决方案对于某些等于 0 的分量并不都是有效的.出现的 %r 变量是任意常数.我会将它们设置为 1 以获得一些特定的解决方案.

Note that I've solved special cases first (a = 0 and b = 0, or a = 0, or b = 0) since the solutions found aren't all valid for some components equal to zero. The %r variables which appear are arbitrary constants. I'll set them equal to 1 to get some specific solutions.

(%i52) subst ([%r19 = 1, %r20 = 1], %o42);
(%o52)                    [[aa = 1, bb = 1, cc = 0]]
(%i53) subst ([%r21 = 1, %r22 = 1], %o43);
                                          c
(%o53)                   [[aa = 1, bb = - -, cc = 1]]
                                          b
(%i54) subst ([%r23 = 1, %r24 = 1], %o44);
                                  c
(%o54)                   [[aa = - -, bb = 1, cc = 1]]
                                  a
(%i55) subst ([%r25 = 1, %r26 = 1], %o45);
                                c + b
(%o55)                 [[aa = - -----, bb = 1, cc = 1]]
                                  a

希望这会有所帮助.祝你好运&继续努力.

Hope this helps. Good luck & keep up the good work.

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