# Help with a proof

1. Feb 19, 2008

### uman

If we have two nonzero vectors in 3-space $\vec{V_1}=a_1\vec{i}+b_1\vec{j}+c_1\vec{k}$ and $\vec{V_2}=a_2\vec{i}+b_2\vec{j}+c_2\vec{k}$, define $\vec{V^'_1}=a^2_1\vec{i}+b^2_1\vec{j}+c^2_1\vec{k}$ and $\vec{V^'_2}=a^2_2\vec{i}+b^2_2\vec{j}+c^2_2\vec{k}$. How can we prove that if $\vec{V_1}-\vec{V^'_1}$ is parallel to $\vec{V_2}-\vec{V^'_2}$, then $\vec{V_1}$ is parallel to $\vec{V_2}$?

Any ideas? I've been thinking about this for a while and it's bugging me because I think it should be true but I can't figure out how to prove it.

Last edited: Feb 19, 2008
2. Feb 19, 2008

### uman

Also I don't know if this problem counts as "calculus" per se. It arised when doing a problem in Apostol's Calculus that I showed to be logically equivalent to the statement I posted above (at least I think so... let's hope I didn't make a mistake ) so I decided this forum was as good as any.

3. Feb 19, 2008

### Diffy

Well to start, if you know that two vectors are parallel, what does that mean? In your book do you have something that says "Two vectors are parallel if..."

Secondly what exactly is the vector $\vec{V_1}-\vec{V^'_1}$?

4. Feb 19, 2008

### uman

Two vectors $\vec{A}$ and $\vec{B}$ are parallel if there is a scalar $t$ such that $\vec{A}=t\vec{B}$. Also I wasn't aware there was more than one way to define subtraction of vectors but here is the one I am using: $\vec{V_1} - \vec{V^'_1}=(a1-a^2_1)\vec{i} + (b1-b^2_1)\vec{j} + (c1-c^2_1)\vec{k}$

5. Feb 19, 2008

### uman

Also I'm screwing up the latex somehow. There is supposed to be an arrow over $V^'_1$ and $V^'_2$. I don't know how to fix it :-/

6. Feb 19, 2008

### Diffy

Ok, so based on your definition, if $\vec{V_1}-\vec{V^'_1}$ is parallel to $\vec{V_2}-\vec{V^'_2}$ then there is some scalar $$t$$ such that $\vec{V_1}-\vec{V^'_1} = t (\vec{V_2}-\vec{V^'_2})$ ...

Can you take it from here? Think about the components of the vectors, can you get them to say that there is some scalar such that $\vec{V_1} = t\vec{V_2}$