Tricky proof for a simple statement

Prove:

If a plane in $\unicode{x211D}^3$ contains two distinct points $P$ and $Q$, then the plane contains every point on the line containing $P$ and $Q$.

I tried a bunch of different ways of approaching this one (vector and cartesian equations of the plane, etc.), but didn't really get anywhere. It seems so obviously true that I'm having trouble formalizing it...

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It might help to translate the plane so it includes the origin, and then attack it with linear algebra. – John Stalfos Sep 17 '12 at 13:13
It depends on how your book defines a plane. – Thomas Andrews Sep 17 '12 at 14:25

A plane in $\mathbb{R}^3$ is given by $\{x \; | \; n\cdot x=n\cdot x_0\}$ for some point $x_0$ in the plane and normal vector $n$. Here '$\cdot$' is the standard scalar product.

Also, any point on the line containing $P$ and $Q$ can be written as $t \cdot P + (1-t) \cdot Q$ for a real number $t$.

What you need to show then is that, given $$n\cdot P= n\cdot Q= n\cdot x_0 ,$$ it follows $$n\cdot(tP + (1-t)Q) = n\cdot x_0.$$

But $$n\cdot(tP + (1-t)Q) = tn\cdot P + (1-t)n\cdot Q = t n\cdot x_0 + (1-t) n\cdot x_0 = n\cdot x_0.$$ QED.

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A plane $E$ in $\mathbb{R}^3$ is a set of the form $$\{x=p+\alpha u+\beta v:\ \alpha,\beta \in \mathbb{R}\}$$ where $p \in \mathbb{R}^3$ and $u, v \in \mathbb{R}^3\setminus\{0\}$ linearly independent.

Given $P, Q \in E$ there exist scalars $\alpha_1,\beta_1,\alpha_2,\beta_2$ such that $$P=p+\alpha_1u+\beta_1v,\ Q=p+\alpha_2u+\beta_2v.$$ Now any point on the line through $P$ and $Q$ is of the form $(1-t)P+tQ$ for some $t \in \mathbb{R}$.

So we have $$(1-t)P+tQ=(1-t)(p+\alpha_1u+\beta_1v)+t(p+\alpha_2u+\beta_2v)=p+\alpha u+\beta v \in E,$$ where $\alpha=(1-t)\alpha_1+t\beta_1$ and $\beta=(1-t)\alpha_2+t\beta_2$.

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