- #1
FeDeX_LaTeX
Gold Member
- 437
- 13
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:
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).
Last edited: