# Proof of test of collinearity and coplanarity

Statement : If there are 3 points with position vectors a, b and c. Then the points are collinear if and only if there exist scalars x,y,z, not all zero,such that x a + y b +z c = 0 where x+y+z =0.

Statement : If there are 4 points with position vectors a, b ,c and d. Then the points are copalanar if and only if there exist scalars x,y,z and w not all zero,such that x a + y b +z c +w d = 0 where x+y+z+w =0.

Can you give a proof for this using vectors?

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Points cannot be neither collinear nor coplanar, but vectors can. Also, what exactly did you mean by $a*, b*$ and $c*$? Same for $x*, y*, z*$. My guess is those are just typos or you didn't fully handle LaTeX. – Kaster Dec 15 '12 at 6:45
What?? You mean to say that 3 points cannot be collinear??? – Adwait Kumar Dec 15 '12 at 6:53
I just checked, points can be collinear if they lie on one line. I never heard about it though (not about points lying on one line but about calling them collinear) :) Never mind then. I'm posting the proof. – Kaster Dec 15 '12 at 7:08
It's perfectly usual to speak of three points being collinear and four points being coplanar. – joriki Dec 15 '12 at 7:22
Apparently not everywhere. – Kaster Dec 15 '12 at 7:38

1. Since $x+y+z = 0$ at least one of $x, y$ is not zero and : $$x\mathbf a + y\mathbf b + z\mathbf c = 0 \Leftrightarrow x(\mathbf a - \mathbf c) + y(\mathbf b - \mathbf c) = 0.$$ Two dependent vectors lie on a single line through the origin. This works the other way around as well.

2. Extend 1. to cover the coplanar case.

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Hint: Part 1

Assume $x+y \neq 0$

If a, b ,c are collinear iff

$${\bf c} = \frac{x{\bf a} + y{\bf b}}{x+y}$$

for some $x$ and $y$. Guess $z$ now!

Now observe that if $x+y = y+z = z+x = 0$ then all of them are zeroes.

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Since there is "iff" you should prove the proposition in both ways. – Kaster Dec 15 '12 at 7:40
I did not want to spell out everything in the hint. I am assuming he knows the formula for computing the position vector that divides a line joining two position vectors in the ratio $x$:$y$. – Isomorphism Dec 15 '12 at 7:49
Fair enough. Mentioned just in case. – Kaster Dec 15 '12 at 8:02

1) Let's denote points $A,B$ and $C$. Let's assume they are collinear, then $\overrightarrow{\pmb {AB}}$ and $\overrightarrow{\pmb {AC}}$ are collinear as well, so $\overrightarrow{\pmb {AB}} = k \overrightarrow{\pmb {AC}}$. On the other hand $\overrightarrow{\pmb {AB}} = \pmb b - \pmb a$ and $\overrightarrow{\pmb {AC}} = \pmb c - \pmb a$, so $\pmb b - \pmb a = k (\pmb c - \pmb a)$ which means $(k+1)\pmb a - \pmb b - k\pmb c = \pmb 0$, so one can take $x = k + 1, y = -1$ and $z = -k$.

Now let's assume that there are $x, y, z$ where $x + y +z = 0$ such that $x \pmb a + y\pmb b + z\pmb c = \pmb 0$. Since $x, y, z$ are not zeros all at once, at least one is non-zero. Let's assume it's $x \neq 0$. Due to the $x+y+z = 0$ at least one more number is non-zero. Let's assume it's $y \neq 0$

$x \pmb a + y \pmb b - (x + y) \pmb c = \pmb 0\\ (\pmb a - \pmb c) x + (\pmb b - \pmb c) y = \pmb 0$.

Last equation means that $\pmb a - \pmb c$ and $\pmb b - \pmb c$ are collinear. If one recalls that $\pmb b - \pmb a = \overrightarrow{\pmb {AB}}$ and $\pmb c - \pmb a = \overrightarrow{\pmb {AC}}$ then one can say that $A, B$ and $C$ are collinear as well.

2) Same thought can be used for coplanar case. With one more extra term.

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