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Vectors of Parallelograms and Octagons


Date: 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/   
    
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Linear Algebra
High School Triangles and Other Polygons

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