Showing that a line, given it does not lie in a plane, is parallel to the plane?

Question

The equation of the plane is given in its parametric form. Will it suffice to show that the direction vector of the line is a linear combination of two linearly independent vectors in the plane? This would mean that the direction vector of the line lies in the plane, so it would imply that it is parallel to the plane.

Answer 1

In parametric form, the plane is formed by points $\vec{p}(u, v)$ (and $u, v \in \mathbb{R}$), where $$\vec{p}(u, v) = \vec{p}_0 + u \, \vec{e}_u + v \, \vec{e}_v$$ and the vectors $\vec{e}_u$ and $\vec{e}_v$ are not parallel, i.e. $$\lVert \vec{e}_u \rVert \, \lVert \vec{e}_v \rVert \ne \lvert \vec{e}_u \cdot \vec{e}_v \rvert$$ or $$\vec{e}_u \times \vec{e}_v \ne 0$$

 
Let's say you have line $\vec{r}(t)$ ($t \in \mathbb{R}$), $$\vec{r}(t) = \vec{r}_0 + t \, \vec{r}_t$$

If the question is

Is it sufficient to show that $\vec{r}_t = \alpha \, \vec{e}_u + \beta \, \vec{e}_v$ with constants $\alpha, \beta \in \mathbb{R}$, when one wishes to prove that the line is parallel to the plane?

the answer is Yes, of course, because that is one way to define "parallel to a plane".

If the question is

How do I prove that the line $\vec{r}(t)$ is parallel to the plane $\vec{p}(u, v)$?

then the answer is something like two easiest methods are

• Showing that $$\vec{r}_t = \alpha \, \vec{e}_u + \beta \, \vec{e}_v$$ with constants $\alpha, \beta \in \mathbb{R}$, as already noted
   

• Showing that $$\vec{r}_t \cdot \left ( \vec{e}_u \times \vec{e}_v \right ) = 0$$which is just another way to express the definition of $\vec{r}_t$ being parallel to the plane $\vec{p}(u, v)$.

  Note that because this is a triple product, we can write the same equality as $$\vec{e}_u \cdot \left ( \vec{e}_v \times \vec{r}_t \right ) = 0$$ or as $$\vec{e}_v \cdot \left ( \vec{r}_t \times \vec{e}_u \right ) = 0$$

Personally, the latter method (using any one of the three triple product forms) is "easiest" for me, in particular because I don't need to introduce $\alpha$ and $\beta$ — but then again, I'm not a mathematician; I just use math as a tool for solving problems.

If we use $\vec{r}_t = ( x_t , y_t , z_t )$, $\vec{e}_u = ( x_u , y_u , z_u )$, and $\vec{e}_v = ( x_v , y_v , z_v )$, the triple product can be written as $$x_t ( y_u z_v - y_v z_u ) + y_t ( x_v z_u - x_u z_v ) + z_t (x_u y_v - x_v y_u) = 0$$