Unique distance as a proof of a line and a plane parallelity

Question

Please consider a statement "a line is characterized by a unique distance from a plane." Is the unique distance a sufficient feature to prove that the line and plane are parallel?

Answer 1

The equation of the plane $P$ is given by $\vec n\cdot (\vec u-\vec u_0)=0$ where $\vec u_0$ lies in the plane, and $\vec n$ is a vector normal to $P$.

The parametric equation of the line $\ell$, passing through $\vec u_0$, is given by $\vec v(t)=\vec v_0+t\vec a$, where $\vec a=\vec v_0-\vec v_1$, the vector difference of any two points on $\ell.$

We have the following condition: $\ell \parallel P\Leftrightarrow \vec a\cdot \vec n=0$ and $P\cap\ell=\emptyset.$

I understand your question to be: suppose for each point $\vec u $ on the line, the perpendicular distance from $\ell$ to $P$ is some constant $d$. Then, $\ell \parallel P.$

Let's do the calculation. Letting $\vec v_0$ and $\vec v_1$ be arbitrary points on $\ell$, and $\vec u_0$ a fixed point lying in $P$. Then,

$d=\frac{(\vec v_0-\vec u_0)\cdot n}{\|\vec n\|}=\frac{(\vec v_1-\vec u_0)\cdot n}{\|\vec n\|}\Rightarrow (\vec v_0-\vec v_1)\cdot \vec n=0\Rightarrow \vec a\cdot \vec n=0.$

So the answer to your question is yes, as long as $P\cap \ell=\emptyset.$ But this is easy to check, by substitution of an arbitrary $(v_1,v_2,v_3)=\vec v\in \ell$ into the equation for $P$.