True/fasle statements about lines and planes 

*

*Line $l_1$ passes through point $M_1$ and is parallel to vector $\vec s_1$ which is not zero and  line $l_2$ passes through point $M_2$ and is parallel to vector $\vec s_2$ which is not zero as well. if the triple product $(\vec s_1 ,\vec s_2 ,\overrightarrow{M_1M2})=2$ then the lines do not intersect


*if two lines do not intersect then they are parallel


*if the line $l$ is parallel to the plane $P$ then the normal of the plane is a direction vector to the line


*if line $l$ is perpendicular to two other lines that are not parallel and are in the plane $P$ then the line $l$ is perpendicular to the plane $P$


*if lines $l_1$ and $l_2$ are perpendicular to $l_3$ then lines $l_1$ and $l_2$ are parallel


First statement is true because if the triple product is not zero then the lines can be not parallel and not intersect (not sure if the translation is correct but "skew lines")
Second statement is not correct because they can be not parallel and do not intersect at the same time
Third statement I think the normal vector will be perpendicular not parallel
Fourth statement I don't know how to approach
Fifth statement I believe is also correct but I don't know how to approach it as well
These are supposed to be simple self test questions in the book but there are no solutions to them.
Thank you!
 A: We obviously work in an affine Euclidean space of dimension 3, for example $(\Bbb R^3,\langle.,.\rangle)$.

*

*The first statement is indeed true. But your justification is not clear.
Suppose $l_1$ and $l_2$ intersect at $M$. Then $\overrightarrow{M_1M2}\in \text{span}(\vec{s_1},\vec{s_2})$ and $\det(\vec{s_1},\vec{s_2},\overrightarrow{M_1M2})=0$. So, if $\det(\vec{s_1},\vec{s_2},\overrightarrow{M_1M2})=2$, then $l_1$ and $l_2$ do not intersect.


*You're right : such lines are called skew lines.


*You're right.


*Let's call $l_1=M_1+\Bbb Rs_1$ and $l_2=M_2+\Bbb R s_2$ the two other lines and $l=M+\Bbb Rs$. $l$ perpendicular to $l_1$ and $l_2$ means $\langle s,s_1\rangle=0$ and $\langle s,s_2 \rangle=0$. Then $\forall \lambda, \mu \in \Bbb R, \langle s,\lambda s_1+\mu s_2\rangle=0$. In other words, the line $l$ is perpendicular to the plane $P$.


*You 're wrong. Let $i=(1,0,0),j=(0,1,0), k=(0,0,1), l=\Bbb Ri, l_1=\Bbb Rj,l_2=\Bbb Rk$
Then $l_1$ and $l_2$ are perpendicular to $ l$ but not parallel.
