问题描述
从本质上讲同一个问题被问此处,但在非编程环境。一个建议的解决方案是采取{Y,-x 0 }。这将工作针对具有x或y分量的所有矢量,但失败,如果该载体是等于+或 - {0,0,}。在这种情况下,我们会得到{0,0, 0 }
我目前的解决方案(在C ++中):
//浮点比较利用小量
布尔is_equal(浮动,浮动);
// ...
VEC3 V = / *一些单位长度向量* /
// ...
//设置作为非平行矢量我们将使用查找
//正交向量。这里我们选择其中的x或y轴。
VEC3 orthog;
如果(is_equal(V.X,1.0F))
orthog.set(1.0F,0.0,0.0);
其他
orthog.set(0.0,1.0F,0.0);
//查找正交向量
orthog =交(V,orthog);
orthog.normalize();
这个方法可行,但我觉得有可能是一个更好的方法,我的搜索露面罢了。
只是为了好玩我做了一个快速code可高达每一个在C ++中的参考答案的幼稚的实现并验证它们都工作(尽管有些并不总是自然地返回单位向量,我添加了一个noramlize()调用需要的地方)。
我最初的想法:
VEC3 orthog_peter(VEC 3常量和放大器; 5)
{
VEC3 arbitrary_non_parallel_vec = V.X!= 1.0F? VEC3(1.0,0.0,0.0):VEC3(0.0,1.0F,0.0);
VEC3 orthog =交(V,arbitrary_non_parallel_vec);
回归正常化(orthog);
}
http://stackoverflow.com/a/19650362/2507444
VEC3 orthog_robert(VEC 3常量和放大器; 5)
{
VEC3 orthog;
如果(V.X == 0.0&功放;&安培; V.Y == 0.0)
orthog = VEC3(1.0F,1.0F,0.0);
否则,如果(V.X == 0.0)
orthog = VEC3(1.0F,V.Z / V.Y,1.0F);
否则,如果(V.Y == 0.0)
orthog = VEC3(-v.z / V.X,1.0F,1.0F);
其他
orthog = VEC3( - (V.Z + V.Y)/ V.X,1.0F,1.0F);
回归正常化(orthog);
}
http://stackoverflow.com/a/19651668/2507444
//注:u和v变量名是从作者的例子交换
VEC3 orthog_abhishek(VEC 3常量和放大器; 5)
{
VEC3 U(1.0F,0.0,0.0);
浮u_dot_v =点(U,V);
如果(绝对(u_dot_v)!= 1.0F)
回归正常化(U +(V * -u_dot_v));
其他
返回VEC3(0.0,1.0F,0.0);
}
http://stackoverflow.com/a/19658055/2507444
VEC3 orthog_dmuir(VEC 3常量和放大器; 5)
{
浮长度= hypotf(V.X,hypotf(V.Y,V.Z));
浮动dir_scalar =(V.X> 0.0)?长度: - 长度;
浮XT = V.X + dir_scalar;
浮点= -v.y /(dir_scalar * XT);
返回VEC3(
点* XT,
1.0F +点* V.Y,
点* V.Z);
};
好了,这里有一个去了解它的方式。让矢量(A,B,C)进行说明。解方程(A,B,C)的点(AA,BB,CC)= 0为AA,BB,和CC(并确保AA,BB,而cc不都是零),所以(AA,BB,CC )正交于(A,B,C)。我用千里马( http://maxima.sf.net )来解决这个问题。
(%I42)解决([A,B,C]。[AA,BB,CC] = 0,[AA,BB,CC]),A = 0,B = 0;
(%O42)[AA =%R19,BB =%R20,CC = 0]
(%I43)解决([A,B,C] [AA,BB,CC] = 0,[AA,BB,CC])中,a = 0;
%R21Ç
(%o43)[AA =%R22,BB = - ------,CC =%R21]
b
(%I44)解决([A,B,C] [AA,BB,CC] = 0,[AA,BB,CC]。)时,b = 0;
%R23Ç
(%O44)[AA = - ------,BB =%R24,CC =%R23]
一个
(%I45)解决([A,B,C] [AA,BB,CC] = 0,[AA,BB,CC。);
R25%C +%R26 b
(%O45)[AA = - ---------------,BB =%R26,CC =%R25]
一个
请注意,我已经首次解决了特殊的情况(α= 0和b = 0,或a = 0时,或b = 0),因为发现了解决方案并不都是有效的某些组件等于零。这显示在%R变量是任意常数。我将它们设置为1,以得到一些具体的解决方案。
(%I52)SUBST([%R19 = 1%R20 = 1]%O42);
(%O52)[[AA = 1,BB = 1,CC = 0]]
(%I53)实名词([%R21 = 1,%R22 = 1],%o43);
C
(%O53)[AA = 1,BB = - - ,CC = 1]
b
(%I54)实名词([%R23 = 1,%R 24 = 1],%O44);
C
(%o54)[AA = - - ,BB = 1,CC = 1]
一个
(%I55)实名词([%R25 = 1,%R 26 = 1],%O45);
C + B
(%O55)[AA = - -----,BB = 1,CC = 1]
一个
希望这有助于。祝你好运和放大器;保持良好的工作。
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 }.
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();
This method works, but I feel that there may be a better method and my searches turn up nothing more.
[EDIT]
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).
My original idea:
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 );
}
http://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);
}
http://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);
}
http://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);
};
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
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
Hope this helps. Good luck & keep up the good work.
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