To solve this problem, we need to analyze the geometric relationships between the given lines and planes. We will use vector and plane equations to determine if the statements are true or false.
We define the vectors and the plane normal as follows:
- \(\bar{\pi} = [1, 0, 0]\)
- \([M = [0, 1, 0]\)
- \([\bar{R} = [0, 0, 1]\)
- \(\bar{K} = [1, 1, 0]\)
- \(\bar{J} = [1, 0, 1]\)
- \([\bar{W} = [0, 1, 1]\)
- \(ITM\) plane normal vector = \([1, 1, 1]\)
To check if \(\bar{\pi}\) and \([M\) lie in the same plane, we verify if both vectors are orthogonal to the plane normal vector:
\[
\bar{\pi} \cdot [1, 1, 1] = 1 \neq 0
\]
\[
[M \cdot [1, 1, 1] = 1 \neq 0
\]
Since neither dot product is zero, \(\bar{\pi}\) and \([M\) do not lie in the same plane. Thus, statement A is false.
To check if \(\bar{\pi}\) and \([\bar{R}\) are parallel, we compute their cross product:
\[
\bar{\pi} \times [\bar{R} = [1, 0, 0] \times [0, 0, 1] = [0, -1, 0]
\]
Since the cross product is not zero, \(\bar{\pi}\) and \([\bar{R}\) are not parallel. Thus, statement B is false.
To check if \(\bar{K}\) and \([M\) are perpendicular, we compute their dot product:
\[
\bar{K} \cdot [M = [1, 1, 0] \cdot [0, 1, 0] = 1
\]
Since the dot product is not zero, \(\bar{K}\) and \([M\) are not perpendicular. Thus, statement C is false.
To check if \(\bar{R}\) and ITM do not lie in the same plane, we verify if \(\bar{R}\) is orthogonal to the plane normal vector:
\[
\bar{R} \cdot [1, 1, 1] = 1 \neq 0
\]
Since the dot product is not zero, \(\bar{R}\) and ITM do not lie in the same plane. Thus, statement D is true.
To check if \(\bar{J}\) and \([\bar{W}\) are skew, we verify that they are neither parallel nor in the same plane:
\[
\bar{J} \times [\bar{W} = [1, 0, 1] \times [0, 1, 1] = [-1, -1, 1]
\]
Since the cross product is not zero, they are not parallel. Next, we check if they lie in the same plane:
\[
\bar{J} \cdot [-1, -1, 1] = 0
\]
\[
[\bar{W} \cdot [-1, -1, 1] = 0
\]
Since both dot products are zero, \(\bar{J}\) and \([\bar{W}\) lie in the same plane. Thus, statement E is false.
To check if \(\bar{\pi}\) and \(I \pi\) do not intersect, we verify if they are not the same vector:
\[
\bar{\pi} \neq [1, 1, 1]
\]
Since they are not the same vector, \(\bar{\pi}\) and \(I \pi\) do not intersect. Thus, statement F is true.
Answered
