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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
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
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
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.