Expert: Paul Klarreich - 9/13/2007

Question
ABCDEF is a regular hexagon with sides of unit length.  Find the magnitude and direction of AB+AC+AD+AE+AF.

Answer
Questioner:   KN
Category:  Calculus
Private:  No
 
Subject:  Calculus
Question:  ABCDEF is a regular hexagon with sides of unit length.  Find the magnitude and direction of AB+AC+AD+AE+AF.
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Hi, KN,
(Is that what your mother calls you?)

Suppose we put one vertex, A, at the origin and the opposite vertex directly above it on the y-axis.  It looks something like this:

USE COURIER FONT TO VIEW THIS.  The diagram is pretty bad even in courier, and completely useless in a PS font.

    D
   / \
  /   \
 /     \
E/       \ C
|         |
|         |
|         |
|         |
F\       / B
 \     /
  \   /
   \ /
    A

Now I think the x-component of the vector sum will be zero by symmetry, so we only compute the vertical (y-) component.  For that, we need only the vertical components of AB, AC, AD.

In fact, once we get AB, the rest flows.

The interior angles are each 120 degrees, so if we look at:

    D
   /|\
  / | \
 /  |  \
E/---+---\ C
|   Y|    |
|    |    |
|    |    |
|   X|    |
F\---+---/ B
 \  |  /
  \ | /
   \ /
    A

Angle AXB = 90,  angle XAB = 60.  Also  AB = 1, so AX = cos 60 = sqrt(3)/2.
Also DY = sqrt(3/2) and XY = 1.  So our vertical components are:

AX =  sqrt(3)/2      = V.C. of AB
AY =  sqrt(3)/2 + 1  = V.C. of AC
AX =  sqrt(3)/2      = V.C. of AF
AY =  sqrt(3)/2 + 1  = V.C. of AE
AD = 2sqrt(3)/2 + 1  = V.C. of AD
----------------------------------
Total = 6 sqrt(3)/2 + 3

Total = 3 sqrt(3) + 3

OK, then -- since the horizontal component is zero, that number is the magnitude and the direction is obviously the same as the direction of AD, whatever it happens to be.  (Here we have made it vertical.)