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(4) The technique of triangulation in surveying is to locate a position in R3 if the distance to 3 fixed points is known. This is similar to how global position systems (GPS) work. A GPS unit measures the time differences taken for a signal to travel from each of 4 satellites to a receiver on Earth. This is then converted to a difference in the distances from each satellite to the receiver, and this can then be used to calculate the distance to 4 satellites in known positions Let P (2,-1,4), P2 (3,4,-3), P (4,-2,6), P (6,4, 12) We wish to find a point P-(xy:) with r, 20 satisfying P is distance Δ from P. P is distance (Δ-12+ 9V3) from P2, P is distance A - 1 from Ps, and P i Pa s distance A-9 from
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Answer #1

P (x,y,z)

distance from P1 (2, -1,4)

\Delta2 = (x-2)2 + (y+1)2 + (z-4)2

P2(3,4,-3) : (\Delta-12+9\sqrt{3})2 = (x-3)2 + (y-4)2 + (z+3)2

P3(4,-2,6) ; (\Delta-1)2 = (x-4)2 + (y+2)2 + (z-6)2

P4(6,4,12) : (\Delta-9)2 = (x-6)2 + (y-4)2 + (z-12)2  

expanding the terms

\Delta2 = (x2 +y2 + z2) -4x +2y - 8z + 21

\Delta2 + (144 +81*3 -216\sqrt{3}) - (24-18\sqrt{3})\Delta = (x2 +y2 + z2) -6x -8y + 6z + 34

\Delta2 - 2\Delta +1 = (x2 +y2 + z2) -8x +4y - 12z

\Delta2 - 18\Delta +81 = (x2 +y2 + z2) - 12x - 18y - 24z

re-arrange the terms and put s = \Delta2 - (x2 +y2 + z2)

-4x +2y - 8z + 0 \Delta = s-21

-6x - 8y +6z + (24-18\sqrt{3}) \Delta = s +353 - 216\sqrt{3}

-8x +4y - 12z + 2 \Delta = s-55

-12x -8y - 24z + 18\Delta = s-115

We have 4 variables and 4 equations, solving them we get

x= -0.185s +37.22

y = 0.53s +16.39

z= 0.1s +2.07

\Delta = 0.3s +7.64

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