In 3D geometry, skew lines L1 and L2, defined by their respective parametric equations, can have parallel vectors v1 and v2. If the dot product of these vectors, calculated as v1.v2 = ad + be + cf, equals zero, the vectors are perpendicular. However, this perpendicularity does not imply that the skew lines themselves intersect; they can be orthogonal without sharing a point in space. For example, while the x and y axes are perpendicular and intersect in the plane z=0, raising the x-axis to y=0, z=1 results in two perpendicular lines that do not intersect.
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do you mean that in the case that x-axis is in its original position the x and y axes are perpendicular and intersect, and in the case that the x-axis is in y = 0, z = 1 the x and y axes are perpendicular and doesn't intersect ?RUber said:The lines would be considered orthogonal, but that does not imply that they intersect.
Consider the x and y axes. The intersect in the plane z=0. But if you raise the x-axis to y=0, z=1, it will never intersect the y-axis defined by x=0, z=0.
mohamed el teir said:do you mean that in the case that x-axis is in its original position the x and y axes are perpendicular and intersect, and in the case that the x-axis is in y = 0, z = 1 the x and y axes are perpendicular and doesn't intersect ?
yes sort of that, if lines with perpendicular directions but the lines themselves are skew, are they considered perpendicular or not, so they are perpendicular, thank youRUber said:Yes. That was your question, right? If orthogonal lines must intersect or not.