# Thread: "if and only if" proof of 3 vectors on a plane through the origin & linear dependence

1. ## "if and only if" proof of 3 vectors on a plane through the origin & linear dependence

Show that three vectors a, b, c lie on the same plane through the origin, if and only if there are three scalars α, β, γ not all zero such that αa + βb + γc = 0.

I believe in order to prove this, I need to show:
(1) If a, b, c lie on the same plane through the origin, then there are three scalars α, β, γ not all zero such that αa + βb + γc = 0 (i.e. the vectors are linearly dependent?).
(2) If there are three scalars α, β, γ not all zero such that αa + βb + γc = 0 (i.e. the vectors are linearly dependent?), then a, b, c lie on the same plane through the origin.

I've made an attempt at proving (1) by trying to prove the contrapositive "if there DOES NOT EXIST three scalars α, β, γ not all zero such that αa + βb + γc = 0 (i.e. the vectors are linearly independent?), then three vectors a, b, c DO NOT lie on the same plane through the origin" but I'm not sure if this is the right approach. I'm also unsure how to proceed from here.

2. ## Re: "if and only if" proof of 3 vectors on a plane through the origin & linear depend

I don't know if this constitutes a proof, but it makes sense that three nonzero vectors would have to be coplanar if there is a way to scale them so that the three vectors can form a triangle when added. If they go through the origin, then that means you could put two vectors starting at the origin and move the third (which you can do as long as you keep the same direction). Then if you picture the three vectors stretching out so that they form the triangle, and possibly reverse some directions so that if you start at the origin and go around the triangle you end up back where you started. This would demonstrate adding the vectors to get the zero vector, thus showing that if the vectors are coplanar and going through the origin, that \displaystyle \displaystyle \begin{align*} \alpha \mathbf{a} + \beta \mathbf{b} + \gamma \mathbf{c} = \mathbf{0} \end{align*}.

3. ## Re: "if and only if" proof of 3 vectors on a plane through the origin & linear depend Originally Posted by twilightmage13 Show that three vectors a, b, c lie on the same plane through the origin, if and only if there are three scalars α, β, γ not all zero such that αa + βb + γc = 0.

I believe in order to prove this, I need to show:
(1) If a, b, c lie on the same plane through the origin, then there are three scalars α, β, γ not all zero such that αa + βb + γc = 0 (i.e. the vectors are linearly dependent?).
(2) If there are three scalars α, β, γ not all zero such that αa + βb + γc = 0 (i.e. the vectors are linearly dependent?), then a, b, c lie on the same plane through the origin.

I've made an attempt at proving (1) by trying to prove the contrapositive "if there DOES NOT EXIST three scalars α, β, γ not all zero such that αa + βb + γc = 0 (i.e. the vectors are linearly independent?), then three vectors a, b, c DO NOT lie on the same plane through the origin" but I'm not sure if this is the right approach. I'm also unsure how to proceed from here.
I don't know what theorems, etc., you are allowed to use but I would use this:
A basis for an n dimensional vector space has three properties
a) the vectors are independent.
b) the vectors span the space.
c) there are n vectors in the set
And any two of those implies the third..

if three vectors are linearly independent, as you say, then they form a basis for $\displaystyle R^3$. Of course, that means they do NOT lie in a single plane (span a two dimensional set).

On the other hand, if three vectors do span a two dimensional set (lie in a single plane) then they do NOT span $\displaystyle R^3$ and cannot be independent.

4. ## Re: "if and only if" proof of 3 vectors on a plane through the origin & linear depend Originally Posted by twilightmage13 Show that three vectors a, b, c lie on the same plane through the origin, if and only if there are three scalars α, β, γ not all zero such that αa + βb + γc = 0.
By definition.

5. ## Re: "if and only if" proof of 3 vectors on a plane through the origin & linear depend

Given the definition of a plane through the origin as the set of all (x,y,z) which satisfy ax+by+cz=0

If e, f and g are on the plane:
ae1+be2+ce3=0
af1+bf2+cf3=0
ag1+bg2+cg3=0
Which is only true if the rows and columns are linearly dependent, ie, if
e+bf+cg=0

If e, f and g are linearly dependent, ie, if
a(e1,e2,e3) + b(f1,f2,f3) + c (g1+g2+g3) = 0
ae1+bf1+cg1=0
ae2+bf2+cf2=0
ae3+bf3+cg3=0
Which is only true if the rows and columns are linearly dependent, ie,
ae1+be2+ce3=0
af1+bf2+cf3=0
ag1+bg2+cg3=0, ie,
e, f and g are on the same plane:

There is also a geometric proof based on the geometric definition of vector addition, which is kind of awkward in words.

Or, you can take the OP as the definition of a plane, in which case the proof is by definition.

6. ## Re: "if and only if" proof of 3 vectors on a plane through the origin & linear depend Originally Posted by Hartlw There is also a geometric proof based on the geometric definition of vector addition, which is kind of awkward in words.
Which I tried to...

7. ## Re: "if and only if" proof of 3 vectors on a plane through the origin & linear depend

Something I missed. In order for second part of OP to be correct, a,b and c have to emanate from the origin. The statement αa+βb+γc=0 just requires a,b and c to be on a plane, not on a plane through the origin. With that understood, post 5 still applies and the geometric proof is as follows:

Assume Parallelogram Law of vector addition.

If a,b and c are on a plane through the origin, bring a,b, and c to the origin:
Through the tip of c draw a line parallel to b. It intersects a at αa.
Through the tip of c draw a line parallel to a. It intersects b at βb.
Then c= αa+ βb by Parallelogram Law of addition.

If c=αa+βb, a.b,c through the origin, then a,b and c are in a plane through the origin by parallelogram law of vector addition.

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### prove that for any three vectors a,b,c, the vectors b, c a, a b c are coplanar.

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