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math: Clarify scalar multiples must be the same for D too

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eggy 2021-06-08 14:17:03 -04:00
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@ -605,7 +605,7 @@ $$d=\frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}$$
Two planes are parallel if their direction vectors are scalar multiples of each other: Two planes are parallel if their direction vectors are scalar multiples of each other:
$$\vec n_1 = k\vec n_2$$ $$\vec n_1 = k\vec n_2$$
If they are also coincident, the D-values will also be identical: If they are also coincident, the D-values will also be identical or equal to the same scalar multiple as the normal:
$$D_1=D_2$$ $$D_1=D_2$$
Otherwise, the planes intersect, the line along which is equal to the cross product between the two direction vectors. Otherwise, the planes intersect, the line along which is equal to the cross product between the two direction vectors.
@ -619,7 +619,7 @@ When looking at three planes:
If all three normals are scalar multiples: If all three normals are scalar multiples:
- If all three $D$-values are scalar multiples, the planes are parallel and coincident and they have infinite points of intersection along the plane equation. - If all three $D$-values are those same scalar multiples, the planes are parallel and coincident and they have infinite points of intersection along the plane equation.
- Otherwise, there are no solutions and the planes are parallel and distinct and/or parallel and coincident for two. - Otherwise, there are no solutions and the planes are parallel and distinct and/or parallel and coincident for two.
If two normals are scalar multiples: If two normals are scalar multiples: