B When are two vectors collinear and coplanar?

[PLAGUE]

TL;DR Summary

Why should, αa + βb + γc=0,and α + β + γ = 0,

mean A,B,C are collinear, and why should αa+βb+γc+δd=0, and α+β+γ+δ=0,

mean A,B,C,D are coplanar?

My book says, If the position vectors a, b, c of three points A,B,C and the scalars α, β, γ are such that

αa + βb + γc=0,and α + β + γ = 0,
then the three points A,B,C are collinear.

On the other hand,
If the position vectors a, b, c, d of the four points A,B,C,D (no three of which are collinear) and the non-zero scalars α,β,γ,δ are such that

αa+βb+γc+δd=0, and α+β+γ+δ=0,
then the four points A,B,C,D are coplanar.

But this book doesn't provide any reason why so. I tried a lot to prove these conditions, but failed. Why should, αa + βb + γc=0,and α + β + γ = 0,
mean A,B,C are collinear, and why should αa+βb+γc+δd=0, and α+β+γ+δ=0,
mean A,B,C,D are coplanar?

 

[WWGD]

Can you express both your questions as inner products <, ,>*<, , >=0?

 

[FactChecker]

Please try to make your problem statements complete. There must be more in the problem statement, otherwise $\alpha = \beta = \gamma = 0$ would always work no matter what A, B, and C were.

 

[mathwonk]

@PLAGUE: given three points A,B,C, the vector B-A is the arrow pointing from A to B. hence C lies on the line through the points A,B if and only if C can be written as C = A + t(B-A) where t is a real number. Do your equations permit that? (as fact-checker says, assume gamma ≠0.)

 

[Gavran]

\begin{align}
\alpha\vec A+\beta\vec B+\gamma\vec C=0&\nonumber\\
-\frac\alpha\beta\vec A+(-1)\vec B+(-\frac{\gamma}{\beta})\vec C=0&\nonumber\\
\frac{\beta+\gamma}\beta\vec A+(-1)\vec B+(-\frac{\gamma}{\beta})\vec C=0&\nonumber\\
\vec A-\vec B=(-\frac{\gamma}{\beta})(\vec A-\vec C)&\text{(condition of collinearity of three points}\nonumber\\
&\text{ with position vectors }\vec{A}\text{, }\vec{B}\text{ and }\vec{C}\text{)}\nonumber
\end{align}
\begin{align}
\alpha\vec A+\beta\vec B+\gamma\vec C+\delta\vec D=0&\nonumber\\
-\frac\alpha\gamma\vec A+(-\frac{\beta}{\gamma})\vec B+(-1)\vec C+(-\frac{\delta}{\gamma})\vec D=0&\nonumber\\
\frac{\beta+\gamma+\delta}\gamma\vec A+(-\frac{\beta}{\gamma})\vec B+(-1)\vec C+(-\frac{\delta}{\gamma})\vec D=0&\nonumber\\
\vec A-\vec C=(-\frac{\beta}{\gamma})(\vec A-\vec B)+(-\frac{\delta}{\gamma})(\vec A-\vec D)&\text{(condition of coplanarity of four points}\nonumber\\
&\text{ with position vectors }\vec{A}\text{, }\vec{B}\text{, }\vec{C}\text{ and }\vec{D}\text{)}\nonumber
\end{align}

 

[Curious Kev]

Vectors are collinear when they are parallel (which means they are coplanar) and share one end point.

 

[phinds]

Vectors are collinear when they are parallel (which means they are coplanar)

yes, you can construct a plane from two parallel vectors

and share one end point.

HUH ?

 

[fresh_42]

yes, you can construct a plane from two parallel vectors

HUH ?

AI?

 

[phinds]

AI?

I'm doubtful that even AI would put out something that far off base, but maybe ...

 

[fresh_42]

I'm doubtful that even AI would put out something that far off base, but maybe ...

If you start with parallel vectors and end up spanning a plane, then things like that may happen. But I admit I have read several noisy and necro-posts like this one by the OP.

 

[Gavran]

An additional explanation:
$$ \vec A-\vec B=(-\frac{\gamma}{\beta})(\vec A-\vec C) $$ Two vectors $\vec A-\vec B$ and $\vec A-\vec C$ are linearly dependent and that means the initial point and the terminal point of the vector $\vec A-\vec B$ are in the same line as the initial point and the terminal point of the vector $\vec A-\vec C$.
The initial point and the terminal point of the vector $\vec A-\vec B$ are two points with the position vectors $\vec B$ and $\vec A$ and the initial point and the terminal point of the vector $\vec A-\vec C$ are two points with the position vectors $\vec C$ and $\vec A$. So, three points with the position vectors $\vec A$, $\vec B$ and $\vec C$ are in the same line or they are collinear.
$$ \vec A-\vec C=(-\frac{\beta}{\gamma})(\vec A-\vec B)+(-\frac{\delta}{\gamma})(\vec A-\vec D) $$ Three vectors $\vec A-\vec C$, $\vec A-\vec B$ and $\vec A-\vec D$ are linearly dependent and that means their initial and terminal points are in the same plane.
The initial point and the terminal point of the vector $\vec A-\vec C$ are two points with the position vectors $\vec C$ and $\vec A$, the initial point and the terminal point of the vector $\vec A-\vec B$ are two points with the position vectors $\vec B$ and $\vec A$ and the initial point and the terminal point of the vector $\vec A-\vec D$ are two points with the position vectors $\vec D$ and $\vec A$. So, four points with the position vectors $\vec A$, $\vec B$, $\vec C$ and $\vec D$ are in the same plane or they are coplanar.