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I am taking a first course in Vector calculus. I am attempting to write a proof for the following theorem.

The proof in the book splits this into two cases -

(i) either the $\vec {AB}$ is parallel to $\vec {CD}$, so $\vec {AB} = k \cdot\vec{CD}$

(ii) $\vec {AB}$ intersects $\vec {CD}$ at some point $P$, so $p=\frac{a+\lambda b}{1+\lambda}=\frac{c+\lambda' d}{1+\lambda'}$

Although I know the section formula, I couldn't come up with the above formulation on my own. The below is my proof. Could someone comment, on whether the proof is elegant, and also acceptable?

Theorem. Four points $A,B,C,D,$ no three of which are collinear, will lie in a plane when and only when there exist four non-zero numbers $\alpha,\beta,\gamma,\delta,$ such that

$$\alpha a+ \beta b + \gamma c + \delta d = 0 \\ \alpha + \beta + \gamma + \delta = 0 $$

Proof.

Idea: If the four points are coplanar, we should be able to construct a parallelogram of the diagonal on $\vec{CD}$, with sides as scalar multiples of $\vec {AB}$ and $\vec {CD}$.

Let

$\begin{aligned} \vec {CD} &= \lambda \vec{AB} + \lambda' \vec{BC}\\ (d - c) &= \lambda (b - a) + \lambda'(c - b)\\ d &= -\lambda a + (\lambda - \lambda')b + (1 + \lambda')c \end{aligned}$

We would like the coefficients of $a,b,c$ to be $\alpha,\beta,\gamma$ respective.

Set : $-\lambda = \frac{\alpha}{\alpha + \beta + \gamma}, 1+\lambda' = \frac{\gamma}{\alpha + \beta + \gamma}$. Solving for $\lambda,\lambda'$ we get :

$$\begin{aligned} \lambda' &= -\frac{(\alpha + \beta)}{\alpha + \beta + \gamma}\\ \lambda - \lambda' &= \frac{\beta}{\alpha + \beta + \gamma} \end{aligned}$$

So, we have :

$$\begin{aligned} d &= \frac{\alpha}{\alpha + \beta + \gamma}a + \frac{\beta}{\alpha + \beta + \gamma}b + \frac{\gamma}{\alpha + \beta + \gamma}c \\\\ (\alpha + \beta + \gamma)d &= \alpha a + \beta b + \gamma c \end{aligned}$$

Let $\delta = -(\alpha + \beta + \gamma)$. Thus,

$$ \alpha a + \beta b + \gamma c + \delta d = 0 $$

Further, these constants are non-zero, since $\lambda,\lambda'$ cannot be $0,\infty$.

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The proof in the book splits this into two cases

Don't quite follow this. There is just two cases only if the points are assumed to be coplanar already. Otherwise, it is entirely possible that $AB$ and $CD$ are skew i.e. neither parallel, nor intersecting.

The more direct way, in my opinion, would be to rewrite it in the equivalent form:

Theorem. Four points $A,B,C,D,$ no three of which are collinear, will lie in a plane when and only when there exist three non-zero numbers $\alpha,\beta,\gamma\,,$ with $\alpha+\beta+\gamma \ne 0\,$ such that $$\alpha a+ \beta b + \gamma c - (\alpha+\beta+\gamma) d = 0$$

This reduces to:

$$ d= \frac{\alpha a+ \beta b + \gamma c}{\alpha+\beta+\gamma} = a + \frac{\beta}{\alpha+\beta+\gamma}(b-a) + \frac{\gamma}{\alpha+\beta+\gamma}(c-a) \\[20px] \iff\quad\quad d-a = \lambda (b-a) + \mu(c-a) $$

The latter means $\vec{AD}$ is a linear combination of $\vec{AB}$ and $\,\vec{AC}\,$, which is the necessary and sufficient condition for the points to be coplanar.

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  • $\begingroup$ The proof in the book starts with those two possibilities. I have listed my proof however - beginning with the basic idea, the it should be possible to construct a parallelogram with diagonal as $\vec{CD}$ and sides $\lambda \vec{AB}$ and $\mu \vec{BC}$ - triangle law of addition. $\endgroup$ – Quasar Jul 15 '18 at 4:15
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    $\begingroup$ @Quasar I didn't say it didn't ;-) Just that I don't understand their reasoning. Neither of those possibilities covers the case where $AB,CD$ are the opposite edges of a tetrahedron, for example. $\endgroup$ – dxiv Jul 15 '18 at 4:19
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    $\begingroup$ Yes, I meant coplanar! Corrected the typo. $\endgroup$ – Quasar Jul 15 '18 at 5:06
  • $\begingroup$ Sorry, about being outright careless, while representing my attempt. $\endgroup$ – Quasar Jul 15 '18 at 5:13
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Let $x_0, x_1, x_2, x_3$ be the points. Then all four points are on the same plane if array whos rows are the coordinates of $x_1-x_0, x_2-x_0, x_3-x_0$ has a rank of two.

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