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

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?

• Welcome to MSE. Questions like "Here is the task. Solve it for me!" are poorly received on this site. Therefore try to improve your question with an edit. Improving could consist of providing some context concerning your task or by adding what you have tried so far and where did you struggle :) Oct 21 '18 at 11:17
• It's not clear to me what you mean here by "characterized". In the normal sense of the word, your statement is not true, even in the form "a line parallel to a plane is characterized by a unique distance from that plane." Can you put it another way? Oct 21 '18 at 11:17
• I am sorry to be unclear. The statement is a part of a patent claim. In a patent claim terminology „characterized“ is a special word denoting a characterizing portion of the patent claim. Oct 21 '18 at 11:32
• @TonyK Thank you very much for your reply. Does it change a matter if an axis and a plane parallelity are in a question? Oct 21 '18 at 13:59

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 $$P$$.