Let $P, Q, R \in \mathbb{R}^3$ be three points not on a line. Prove there is exactly one plane $E \subset \mathbb{R}^3$ containing $P,Q,R$, namely

Question

Let $P, Q, R \in \mathbb{R}^3$ be three points not on a line. Prove that there is exactly one plane $E \subset \mathbb{R}^3$ containing $P,Q,R$, namely

$$E = P + \mathbb{R} \cdot (P - Q) + \mathbb{R} \cdot (P - R)$$

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I know the equation of a plane is https://en.wikipedia.org/wiki/Euclidean_planes_in_three-dimensional_space

$$ax + by + cz + d = 0$$ or written otherwise $$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$

How am I supposed to prove that ? Any hint or help would be welcome, thank you !

Answer 1

Let $\mathcal G$ be an affine space, $G$ its associated linear space (over some field $K,$ not necessarily equal to $\Bbb R$), and $P,Q,R\in\mathcal G$ non-collinear, i.e. such that the two vectors $P-Q,P-R\in G$ are linearly independent. Then, $$\mathcal E:=P+K(P-Q)+K(P-R)\subset\mathcal G$$ is an affine plane (by definition) and contains $$P+0(P-Q)+0(P-R)=P,\quad P+(-1)(P-Q)+0(P-R)=Q,\quad P+0(P-Q)+(-1)(P-R)=R.$$ Conversely, let $\mathcal F\subset\mathcal G$ be an affine plane containing $P,Q,R.$ Then, the associated linear plane $F$ contains $P-Q,P-R$ hence is spanned by them, so that $\mathcal F=P+F=\mathcal E.$

Answer 2

In the lecture, they gave us this as "proof". I thought maybe someone will find it useful one day, so why not share it.

Claim: If $ P,Q,R \in \mathbb{R}^3$ are three points, which are not on one line, then there exists exactly one Plane $E \subset \mathbb{R}^3$, which contains $P,Q,R$, that is $E = P + \mathbb{R} \cdot (P-Q) + \mathbb{R} \cdot (P-R)$

Proof:

Part 1: $E$ contains $P,Q,R$ $$P = P + 0 \cdot (P-Q) + 0 \cdot (P-R)$$ $$Q = P + (-1) \cdot (P-Q) + 0 \cdot (P-R)$$ $$R = P + 0 \cdot (P-Q) + (-1) \cdot (P-R)$$

Part 2: $E$ is a Plane according to Parametrization Representation

We have to show that $(P-Q)$ and $(P-R)$ are linearly independent with (iii) from (Linear Independence) Show that for two vectors $v,w \in \mathbb{R}^n $, the conditions (i), (ii), (iii) are equivalent

Let $\lambda, \mu \in \mathbb{R}$ with $\lambda \cdot (P-Q) + \mu \cdot (P-R) = 0$ (*)

We have to show: $\lambda = \mu = 0$

Through contradiction: If $\mu \neq 0$ then $\lambda P - \lambda Q + \mu P - \mu R = 0$ according to (*) $$\implies (\lambda + \mu)P - \lambda Q - \mu R = 0$$ If $\mu \neq 0$ then $$\implies R = \frac{\lambda + \mu}{\mu}P - \frac{\lambda}{\mu}Q$$ that means $R$ is on a line with $P$ and $Q$ which contradicts the assumption.

So that means $\mu = 0$

Then it follows from (*) that $\lambda (P-Q) = 0$, that means $\lambda P = \lambda Q$

If $\lambda \neq 0$, then $P=Q$, that means $P,Q,R$ lie on a line, which contradicts the assumption.

So that means $\lambda = 0$

Part 3: $E$ is the only plane. Assume $E'$ is another plane, such that $E \neq E'$. Then $P,Q,R \in E \cap E'$.

$E \cap E'$ means Part 1 is a line as intersection of two planes, which contradicts the assumption.