在3d中移动点的集合

我在3d中有一组点。我把这些点连接起来，形成一条直线。我必须得到另一条线，它是这条线的位移版本，使得最终的位移总是在原始线的右边。解决这个问题的方法是什么?如何每次都使上向量在正确的方向上?

假设这些点位于一个球体上。看看球体的顶视图，我想要这样写

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第一行是点的原始集合，第二行是移位的集合

好的，我正在添加代码

std::vector<osg::Vec3> vertArray; // The array containig the 3d points
std::vector<osg::Vec3> shiftVec; // The array to store the shifted vectors
osg::Vec3 line1, line2, result, upVec, p1, p2, cross1, cross2;
result = vertArray[1] - vertArray[0];
result.normalise();
result = result X upVec; // i am not sure how to get this upvec for general set of   points
shiftVec.push_back(result);
for(unsigned int i = 1; i < vertArray.size() - 1; i++)
{
 line 1 = vertArray[i-1] - vertArray[i];
 line 2 = vertArray[i+1] - vertArray[i];
 line1.normalise();
 line2.normalise();
 upVec = line1 X line2;
 line 1 = line1 X upVec;
 p1 = vertArray[i-1] + line1;
 line 2 = line2 X upVec;
 p2 = vertArray[i+1] + line2;
 cross1 = upVec;
 cross2 = (p2-p1)X line2
 float factor = cross2.lenght()/cross1.length();
 result = p1+line1*factor;
 result = result - vertArray[i];
 shiftVec.push_back(result);
}
result = vertArray[i] - vertArray[i-1];
result.normalise();
result = result X upVec; // i am not sure how to get this upvec for general set of points
shiftVec.push_back(result);