Question 3.37: Determine the relative positions of two planes S21 = x0^→ + ...

Both S_{1}  and  S_{2} are two-dimensional vector subspaces of R³.

In case S_{1} = S_{2} :  if   \overrightarrow{x_{0} }-\overrightarrow{y_{0} } \in S_{1} ,  then  S^{2}_{1}=S^{2}_{2} is coincident; if \overrightarrow{x_{0} }-\overrightarrow{y_{0} } \notin S_{1} ,  then   S^{2}_{1}\parallel S^{2}_{2}.

In case S_{1} \cap S_{2} is one-dimensional: no matter \overrightarrow{x_{0} }-\overrightarrow{y_{0} } \in S_{1} \cap S_{2} or not, S^{2}_{1} intersects with S^{2}_{2} along a line, namely \overrightarrow{x_{0} }+S_{1} \cap S_{2} if \overrightarrow{y_{0} } is chosen to be equal to \overrightarrow{x_{0} }.

By dimension theorem for vector spaces: dim S_{1} + dim S_{2} = dim S_{1} \cap S_{2} + dim \left(S_{1} +S_{2}\right) (see Ex. <A> 11 of Sec. 3.2 or Sec. B.2), since
dim \left(S_{1} +S_{2}\right) \leq  3, therefore dim S_{1} \cap S_{2} \geq  1. Hence S_{1} \cap S_{2} = \left\{\overrightarrow{0 } \right\} never happens in R³ and S_{1}  and  S_{2} can not be skew to each other.

As a consequence of Examples 3.36 and 3.37, in a tetrahedron \Delta  \overrightarrow{a_{0} }\overrightarrow{a_{1} }\overrightarrow{a_{2} }\overrightarrow{a_{3} } (see Fig. 3.74), we have the following information:

1. Vertices are skew to each other.
2. Vertex \overrightarrow{a_{3} } is skew to the face \Delta  \overrightarrow{a_{0} }\overrightarrow{a_{1} }\overrightarrow{a_{2} }, etc.
3. The edge \overrightarrow{a_{0} }\overrightarrow{a_{1} } is skew to the edge \overrightarrow{a_{2} }\overrightarrow{a_{3} } but intersects with the face\overrightarrow{a_{1} }\overrightarrow{a_{2} }\overrightarrow{a_{3} }, etc.

4. Any two faces intersect along their common edge.

Do you know what happens to \Delta  \overrightarrow{a_{0} }\overrightarrow{a_{1} }\overrightarrow{a_{2} }\overrightarrow{a_{3} } \overrightarrow{a_{4} }  in  R^{4}?

To cultivate our geometric intuition, we give one more example.