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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 .

Displacements:

=====================

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 :

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

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 .

Displacements:

=====================

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 :

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