Question 3.38: Determine the relative positions of (a) two straight lines i...
Determine the relative positions of
(a) two straight lines in R^{4},
(b) two (two-dimensional) planes in R^{4}, and
(c) one (two-dimensional) plane and one (three-dimensional) hyperplane in R^{4}.
(a) The answer is like Example 3.36. Why? Prove it.
(b) Let S^{2}_{1}=\overrightarrow{x_{0} }+S_{1} and S^{2}_{2}=\overrightarrow{y_{0} }+S_{2} be two planes in R^{4}.
Besides the known three relative positions in Example 3.37, namely, coincident, parallel and intersecting along a line, there does exist another two possibilities in R^{4}: intersecting at a point, and Case 4 but never skew to each other.
In case dim S_{1} \cap S_{2} =1: then
dim S_{1} = dim S_{2} =2
\Rightarrow dim \left(S_{1}+S_{2}\right)=2+2-1=3.
Hence, it is possible to choose \overrightarrow{x_{0} },\overrightarrow{y_{0} } \in R^{4} so that \overrightarrow{x_{0} }-\overrightarrow{y_{0} } \notin S_{1}+S_{2}.
According to (3.8.41), S^{2}_{1} \cap S^{2}_{2} = \pmb{\phi} holds. So Case 4 does happen.
In case S_{1} \cap S_{2} = \left\{\overrightarrow{0 }\right\}: then
dim S_{1} = dim S_{2} =2
\Rightarrow dim \left(S_{1}+S_{2}\right)= dim \left(S_{1}\oplus S_{2}\right) =2+2-0=4,
namely, S_{1}\oplus S_{2}=R^{4} holds. No matter how \overrightarrow{x_{0} } and \overrightarrow{y_{0} } are chosen, \overrightarrow{x_{0} }-\overrightarrow{y_{0} } \in R^{4} is always true. Therefore, S^{2}_{1} and S^{2}_{2} intersect at a point.
Since \overrightarrow{x_{0} }-\overrightarrow{y_{0} } \notin R^{4} does not happen in this case, so S^{2}_{1} and S^{2}_{2} are never skew to each other.
(c) Let S^{2} = \overrightarrow{x_{0} } +S_{1} and S^{3} = \overrightarrow{y_{0} } +S_{2} where dim S_{1} =2, dim S_{2} =3.
Since
dim S_{1} \cap S_{2} = dim S_{1}+ dim S_{2} -dim \left(S_{1}+S_{2}\right)
=2+3-dim \left(S_{1}+S_{2}\right)=5-dim \left(S_{1}+S_{2}\right)
and 3 \leq dim \left(S_{1}+S_{2}\right) \leq 4, therefore dim S_{1} \cap S_{2} could be 1 or 2 only.
In case dim S_{1} \cap S_{2} =1: then dim \left(S_{1}+S_{2}\right) =4. For any two points \overrightarrow{x_{0} },\overrightarrow{y_{0} } in R^{4} , \overrightarrow{x_{0} }-\overrightarrow{y_{0} } \in S_{1}+S_{2} =R^{4} always hold. Then S^{2} and S^{3} will intersect along the line \overrightarrow{x_{0} }+S_{1} \cap S_{2} if \overrightarrow{y_{0} }=\overrightarrow{x_{0} }.
In case dim S_{1} \cap S_{2} =2: then S_{1} \subseteq S_{2} and S_{1}+S_{2} =S_{2} holds. For any two points \overrightarrow{x_{0} },\overrightarrow{y_{0} } \in R^{4}, either
\overrightarrow{x_{0} }-\overrightarrow{y_{0} } \in S_{1}+S_{2} \Rightarrow S^{2} and S^{3} are coincident because S^{2} \subseteq S^{3},
or
\overrightarrow{x_{0} }-\overrightarrow{y_{0} } \notin S_{1}+S_{2} = S_{2} \Rightarrow S^{2} \parallel S^{3} , parallel to each other.
Remark In R^{4}
S^{1}_{1} and S^{2}_{2} can be only coincident, parallel or intersecting at a point.
How about S^{1} =\overrightarrow{x_{0} }+S_{1} and S^{3} = \overrightarrow{y_{0} }+S_{2} ? Since
dim S_{1} \cap S_{2} = 1+3-dim \left(S_{1}+S_{2}\right)=4-dim \left(S_{1}+S_{2}\right),
and 3 \leq dim \left(S_{1}+S_{2}\right)=4, dim S_{1} \cap S_{2} can be only 1 or 0.
- dim S_{1} \cap S_{2} = 1: then S_{1} \subseteq S_{2} and S_{1}+S_{2} =S_{2} hold. For \overrightarrow{x_{0} },\overrightarrow{y_{0} } in R^{4}, either
\overrightarrow{x_{0}}-\overrightarrow{y_{0} } \in S_{1}+S_{2} \Rightarrow S^{1} and S^{3} are coincident with S^{1} \subseteq S^{3},
or
\overrightarrow{x_{0}}-\overrightarrow{y_{0} } \notin S_{1}+S_{2} \Rightarrow S^{1} \parallel S^{3}.
2. dim S_{1} \cap S_{2} = 0: then S_{1}+S_{2} = S_{1}\oplus S_{2} = R^{4} . No matter how \overrightarrow{x_{0}} and \overrightarrow{y_{0} } are chosen in R^{4} , \overrightarrow{x_{0}}-\overrightarrow{y_{0} } \in S_{1}+S_{2} always holds. Then S^{1} and S^{3} will intersect at one point.
See Exs. <A> 7 and 8 for more practice.