Do Lines EF and GH Intersect in Vector Geometry?

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The discussion revolves around proving the intersection of lines EF and GH in vector geometry, based on given position vectors of points A, B, and P. Participants express confusion regarding the calculations for midpoints E, F, G, and H, particularly the expressions for F and H, which some believe are incorrectly derived. Clarification is sought on the vector representations and the correct application of midpoint formulas. A participant suggests that their calculations indicate EF and GH are identical, implying they intersect everywhere. The conversation emphasizes the importance of accurately interpreting vector operations in the context of the problem.
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Homework Statement



This is taken from STEP II 2003 Q5.

The position vectors of the points A, B and P with respect to an origin O are ai , bj and li + mj + nk , respectively, where a, b, and n are all non-zero. The points E, F, G and H are the midpoints of OA, BP, OB and AP, respectively. Show that the lines EF and GH intersect.

Let D be the point with position vector dk, where d is non-zero, and let S be the point of intersection of EF and GH. The point T is such that the mid-point of DT is S. Find the position vector of T and hence find d in terms of n if T lies in the plane OAB.

Homework Equations



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The Attempt at a Solution



I can't seem to make any headway with this at all, some guidance would be appreciated. There's a solution given which I don't understand:

E = a/2, F = (p+b)/2, G = b/2, H = (p+a)/2 =>
EF = a/2+x[p+b-a]/2
GH= b/2+y[p+a-b]/2
for which there is a solution EF = GH = (p+a+b)/4 when x = y = 1/2
so we have s = (d+t)/2 = (p+a+b)/4 so the position vector of T is t = (p+a+b)/2-d
the plane OAB is the x-y plane i.e. z = 0 so the component of T in the k direction is 0, so considering the vectors in the k direction we have: 0 = n/2-d <=> d = n/2

I agree with E = a/2.

I don't understand how F = (p+b)/2. If OB + BP = OP, then BP = OP - OB, so F = ½BP = ½(p-b), not ½(p+b)?

I agree with G = b/2.

I don't understand how H = (p+a)/2, for the same reason that I don't agree with F = (p+b)/2. Can anyone help?

I must be doing something very wrong because I am getting that EF and GH are identical (so intersect everywhere).
 
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FeDeX_LaTeX said:

Homework Statement



This is taken from STEP II 2003 Q5.

The position vectors of the points A, B and P with respect to an origin O are ai , bj and li + mj + nk , respectively, where a, b, and n are all non-zero. The points E, F, G and H are the midpoints of OA, BP, OB and AP, respectively. Show that the lines EF and GH intersect.

Let D be the point with position vector dk, where d is non-zero, and let S be the point of intersection of EF and GH. The point T is such that the mid-point of DT is S. Find the position vector of T and hence find d in terms of n if T lies in the plane OAB.

Homework Equations



-

The Attempt at a Solution



I can't seem to make any headway with this at all, some guidance would be appreciated. There's a solution given which I don't understand:



I agree with E = a/2.
What does that mean? In the statement of the problem, above, you use bold face to mean vectors but the only "a" given is a number, not a vector.

I don't understand how F = (p+b)/2. If OB + BP = OP, then BP = OP - OB, so F = ½BP = ½(p-b), not ½(p+b)?

I agree with G = b/2.

I don't understand how H = (p+a)/2, for the same reason that I don't agree with F = (p+b)/2. Can anyone help?

I must be doing something very wrong because I am getting that EF and GH are identical (so intersect everywhere).
 
Apologies, I didn't make it clear -- a in non-bold text is just the constant a, whereas a or p indicate the vectors OA or OP.
 
Look at an example. If B= (1, 0, 0) and P= (0, 1, 0) the the point half way between them is the "average" ((1+0)/2, (0+1)/2, (0+0)/2)= (1/2, 1/2, 0). The vector form would be <1/2, 1/2, 0>= (1/2)<1, 0, 0>+ (1/2)<0, 1, 0>.

The difficulty with your reasoning is that "(1/2)BP" is NOT the vector from O to that midpoint. Yes, we can think of BP as a vector from B to P and (1/2)BP as a vector from B to the midpoint of BP. To represent (1/2)BP as a vector from O, we have to add B: <1, 0, 0>+ <-1/2, 1/2, 0>= <1/2, 1/2, 0>.
 
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