检查一个点是否在三维空间的直线上

本文关键字：三维空间 是否 一个 检查 | 更新日期: 2023-09-27 18:10:21

这个方法应该告诉你一个点是否在给定的行上。目前，只要方向矢量的分量不为0(我使用的是直线的参数表示)，它就可以正常工作。如果没有0，我应该得到xResult=yResult=zResult，如果点在直线上。如果方向向量上有一个0，这三个中至少有一个是0因此不等于其他的，但是这个点仍然可以在直线上。

找到一个点是否在给定的行上，处理零情况的最好方法是什么?

    /// <summary>
    /// Returns true if the passed point is on the line, false otherwise
    /// </summary>
    /// <param name="passedPoint"></param>
    /// <returns></returns>
    public Boolean IsOnLine(Line passedLine)
    {
        Boolean pointIsOnLine = false;
        //Get components of this point
        Dimension xPoint = new Dimension(DimensionType.Millimeter, X.Millimeters);
        Dimension yPoint = new Dimension(DimensionType.Millimeter, Y.Millimeters);
        Dimension zPoint = new Dimension(DimensionType.Millimeter, Z.Millimeters);
        //Get components of the base point of the line
        Dimension xBasePoint = new Dimension(DimensionType.Millimeter, passedLine.BasePoint.X.Millimeters);
        Dimension yBasePoint = new Dimension(DimensionType.Millimeter, passedLine.BasePoint.Y.Millimeters);
        Dimension zBasePoint = new Dimension(DimensionType.Millimeter, passedLine.BasePoint.Z.Millimeters);
        //Find difference between passed point and the base point
        Dimension xDifference = xPoint - xBasePoint;
        Dimension yDifference = yPoint - yBasePoint;
        Dimension zDifference = zPoint - zBasePoint;
        DimensionGenerator dg = new DimensionGenerator(DimensionType.Millimeter);
        //Instantiate the 3 result variables
        Dimension xResult = dg.MakeDimension(-1);
        Dimension yResult = dg.MakeDimension(-1);
        Dimension zResult = dg.MakeDimension(-1);

        //Solve for the multiplier using each direction and make sure they are all equal.
        //If any component of the direction vector is 0, the result will be zero and should therefore be directly assigned to 0 to avoid dividing by 0
        if(passedLine.XComponentOfDirection.Millimeters == 0)
        {
           xResult = dg.MakeDimension(0);
        }
        if(passedLine.YComponentOfDirection.Millimeters == 0)
        {
            yResult = dg.MakeDimension(0);
        }
        if(passedLine.ZComponentOfDirection.Millimeters == 0)
        {
            zResult = dg.MakeDimension(0);
        }
        else
        {
            xResult = dg.MakeDimension(xDifference.Millimeters / passedLine.XComponentOfDirection.Millimeters);
            yResult = dg.MakeDimension(yDifference.Millimeters / passedLine.YComponentOfDirection.Millimeters);
            zResult = dg.MakeDimension(zDifference.Millimeters / passedLine.ZComponentOfDirection.Millimeters);
        }

        //If the 3 results are equal, the point is on the line. If they are not, the point is not on the line.
        if (xResult == yResult && xResult == zResult)
        {
            pointIsOnLine = true;
        }
        else
        {
            pointIsOnLine = false;
        }
        return pointIsOnLine;

检查一个点是否在三维空间的直线上

如果处理的是不精确的数字，比较是否相等是个坏主意。我会尝试下面的方法:从直线的基点到该点取向量，然后取直线的方向向量。计算它们的外积。如果这个点在直线上，这两个向量就共线，它们的叉乘就是0向量。因此，计算向量的平方长度(以避免不必要的平方根)，如果它低于某个阈值，则点位于所讨论的直线上。

但是如果您想更接近您自己的代码，请注意您有三个if，后面跟着一个else。所以else语句块仍然会被执行如果其中一个first to条件适用。这可能会给你带来麻烦。还请注意，如果您以毫米为单位进行所有计算，则可以通过处理原始数字来节省大量代码。

我建议用另一种方式处理这件事。使用数学。如果任意两个向量之间的夹角为0，则这两个向量"共享一条直线"。

cos(θ)= (U * V)/(U V * | | | |)(积)。

应用于你的代码:

double inverse = ((xPoint*xBasePoint) + (yPoint*yBasePoint) + (zPoint*zBasePoint))/((sqrt(xPoint^2 +yPoint^2 + zPoint^2) + sqrt(xBasePoint^2 + yBasePoint^2 + zBasePoint^2));
  if(inverse == 1)
      //points are on the same line
   else
      //points are not on the same line or one point is (0,0,0)

另外，在分配xresult、yresult和zresult时，请查看if else语句。如果passsedline，最后还是会除以0。Z不为0(转到else语句)

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