can someone please help me with problem 1A-12 of the supplementary problems?

Question

can someone please help me with problem 1A-12 of  the supplementary problems?

anonymous · Answer

Which question are you talking about? I posted the two sections of PS1. If these are not the right problem sets, can you post which ones you are talking about? http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-a-vectors-determinants-and-planes/problem-set-1/MIT18_02SC_SupProb1.pdf or http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/1.-vectors-and-matrices/part-a-vectors-determinants-and-planes/problem-set-1/MIT18_02SC_pset1.pdf

anonymous · Answer

Actually from those problems sets, in the first file question 1A-12 has a * mark so the answer is not in the pages. 
The question is this:
Label the four vertices of a parallelogram in counterclockwise order as OPQR. Prove that the line segment from O to the midpoint of PQ intersects the diagonal PR in a point X that is 1/3 of the way from P to R.
(Let A = OP, and B = OR; express everything in terms of A and B.).

I have tried the problem with and without the hint and always end up with trivial equalities. Hope you can help me

joshdanziger23 · Answer

gazaso, here's how I looked at it:

Add a point $S$ at the mid point of $PQ$, and a point $T$ at the intersection of $OS$ and $PR$.  Let vector $a=OP$ and $b=OR$, so $PR=b-a$ and $OS=a+b/2$. Let scalar $u$ be the fractional distance along $OS$ where $T$ is sited, so $OT=uOS=u(a+b/2)$.  Let scalar $v$ be the fractional distance along $PR$ where $T$ is sited, so $PT=vPR=v(b-a)$. WE can get from $O$ to $S$ either direct (vector $u(a+b/2)$) or via $P$ (vector $a+v(b-a)$).  Equating these vectors gives $u(a+b/2)=a+v(b-a)$. If this is to apply for all $a$ and all $b$ then we can equate coefficents of $a$ and $b$ to get $u=1-v$ and $u/2=v$, which we can solve to get $v=1/3$ which is what we were asked to prove.

Does this help?

Josh

joshdanziger23 · Answer

I should just correct myself here: it is justified to equate coefficients of $a$ and $b$ provided  $a$ and $b$ are linearly independent vectors (not as previously stated because the reasoning has to apply to all $a$ and all$b$).  This means that for scalars $s$ and $t$, $sa+tb=0$ if and only if $s=t=0$.  To be linearly independent $a$ and $b$ must be neither zero nor colinear.