Proving Parallelism of Vectors in 3-Space: Help with a Proof

[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.

 

[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.

 

[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}?

 

[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}

 

[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 :-/

 

[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}