Vectors of Parallelograms and OctagonsDate: 07/28/98 at 05:49:41 From: Ivan Li Subject: Vectors I cannot get started with the following problem. Hope you can help. ABCDEFGH is a regular octagon and AB = p and BC = q. Express AH in terms of p and q and show that: AE + BH + CG + DF = 2(2 + sqrt2)(q - sqrt2*p) Thank you. Date: 07/28/98 at 17:05:04 From: Doctor Anthony Subject: Re: Vectors You must draw a fairly accurate figure so that you can draw in the various vectors and see how they are related. We start by writing in a few vectors; from these, others can be calculated: AC = p+q AD = q(1+sqrt(2)) CD = sqrt(2)q - p AE = AD + q - sqrt(2)p = q(2+sqrt(2)) - sqrt(2)p BH = -p(1+sqrt(2)) + q AH = -sqrt(2)p + q CG = -q - p + AH + CD (since HG = CD) = -q - p - sqrt(2)p + q + sqrt(2)q - p = -p(2 + sqrt(2)) + sqrt(2)q DF = BH = -p(1+sqrt(2)) + q So now we require: AE + BH + CG + DF = q(2+sqrt(2)) - sqrt(2)p - 2p(1+sqrt(2)) + 2q - p(2+sqrt(2)) + sqrt(2)q = 2q+sqrt(2)q - sqrt(2)p - 2p - 2sqrt(2)p + 2q - 2p - sqrt(2)p + sqrt(2)q = 4q - 4p + 2sqrt(2)q - 4sqrt(2)p = q(4+2sqrt(2)) - p(4 + 4sqrt(2)) = 2q(2+sqrt(2)) - 2sqrt(2)p(sqrt(2) + 2) = 2(2+sqrt(2))(q - sqrt(2)p) - Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 08/24/98 at 00:00:13 From: Ivan Li Subject: vectors I hope you can help me solve the following question. ABCDEFGH is a regular octagon. The vector AB = p, the vector BC = q. Express the vector AH in terms of p and q. Thank you. Date: 08/24/98 at 09:59:50 From: Doctor Anthony Subject: Re: Vectors If you start at A and follow a vector q from A you will be on a line CH. Let the point on CH be J. Then JH is parallel to -p and since triangle AJH is right-angled at A, the distance JH is p.sqrt(2). And so we have: AH = AJ + JH = q - p sqrt(2) - Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 08/24/98 at 23:05:13 From: Ivan Li Subject: Re: Vectors I hope you can solve the following question for me: The points P and R have position vectors 2a + b and a - 3b respectively, relative to an origin O. Given that OPQR is a parallelogram, express the vectors OQ and RP in terms of a and b. By evaluating two scalar products, show that if OPQR is a square, then |a|^2 = 2|b|^2 I got vector OQ = 3a - 2b and vector RP = a + 4b. Unfortunately, I cannot continue from here. Please help. Thank you. Date: 08/25/98 at 08:19:10 From: Doctor Anthony Subject: Re: Vectors I agree with you that: OQ = 3a - 2b RP = a + 4b We require to show that if OPQR is a square then |a|^2 = 2|b|^2. If OPQR is a square then |2a+b| = |a-3b|, so squaring both sides: 4a^2 + 4ab + b^2 = a^2 - 6ab + 9b^2 3a^2 + 10ab - 8b^2 = 0 (Equation 1) Also: |3a-2b| = |a+4b| 9a^2 -12ab + 4b^2 = a^2 + 8ab + 16b^2 8a^2 -20ab - 12b^2 = 0 4a^2 - 10ab - 6b^2 = 0 (Equation 2) Adding (1) and (2): 7a^2 - 14b^2 = 0 a^2 = 2b^2 |a|^2 = 2|b|^2 You could also have done this problem using the fact that the scalar product of OQ and RP is 0, since the diagonals of a square are at right angles. - Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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