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2015-04-15T22:46:03+05:30

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1)
       the distances are :
       AB = 20 km            CD = 4 km            CB = 8 km

finding:
   DB = CB - CD =  8 - 4 = 4 km
   AC = AB - CD = 20 - 8 = 12 km
 
   Displacement associated with direction also along with a magnitude.  Let us say that the displacements towards the right side direction are positive.  The displacements towards the left are negative.  Let the unit vector towards right be denoted by \hat{u}.

   Displacements:

\vec{AB}=20\ \hat{u}\ km,\ \ \ \vec{AC}=12\ \hat{u}\ km,\ \ \ \vec{CD}=4\ \hat{u}\ km,\ \ \ \vec{DB}=4\ \hat{u}\ km\\\\\vec{BA}=-20\ \hat{u}\ km,\ \ \ \vec{CA}=-12\ \hat{u}\ km,\ \ \ \vec{DC}=-4\ \hat{u}\ km,

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2)  distances :  AB = 15 km      AD = AB - DC - CB = 15 - 5- 8 = 2 km
     DC = 5 km             EB = DB - DE = 5 + 8 - 11 = 2 km
           CE = CB - EB = 8 - 2 = 6 km

     displacements :
\vec{AB}=15\ \hat{u}\ km,\ \ \ \vec{AD}=2\ \hat{u}\ km,\ \ \ \vec{DC}=5\ \hat{u}\ km,\ \ \ \vec{CE}=6\ \hat{u}\ km,\\\\\vec{BA}=-15\ \hat{u}\ km,\ \ \ \vec{EC}=-6\ \hat{u}\ km,\ \ \ \vec{BE}=-2\ \hat{u}\ km

you can write the other displacements from some more  points to others, in the same manner.


3 5 3
great answer sir
very good answer sir thanks,i hope you will help me in future
:)