Question

7. A mining company discovered a valuable underground mineral deposit in the shape of an inverted pyramid (see figure below). The vertices of the inverted pyramid are points A, B, C and D. Point G is the location on the ground surface that the mining company will start drilling a borehole to intersect the top face ABC of the pyramid. Table of Point Coordinates Point (m) y (m) z (m) -50 0 425 250 70 410 C 100 -100 350 D -50 -50 G 150 -200 700 B k (vertical) (a) Determine point E on the plane holding points A, B, and C that is closest to point G. (b) Determine the vector GE and its length. (c) Determine the volume of the mineral deposit.

Answer 1

This is a very easy question.

Please check out my solution for better understanding.

Griven :- D C-50, -50,0) -50,0,425) K - BC-250,70,410) &(100,-700, 350) G (150,-200, 700) Solution - Equation of plane containing points A, B,Co x-x y-y, z-21 X₂X1 Y2-4, Z2-2, =0 - - -y, Zz-211 x+50 y Z-425 D -20070 – 15 20 150 -100 -75 | → (x+50)( 70 x(-75) – (-15) (-100)) + 9 (615) (150) – (-75) (-2001) __ +(2-425) (4200) (400) – 70x150) 20 → (2x+ 50) (-6750) +46-17250) + (2-425) (9500) 20 -6750 X - 172504 +9500 Z – 4375000 20

Point ê must lie on this plane e vector in direction of GE (its not GE vector) ? but in direction of GE vector = - 6750 î - 17250 ĝ + 9500 À direction ratio = 5-6750, -17250_, 9500 > _ (-135, -345, 1907 = < -27, -69,3> Equation of that line = 2-24 29-4, 2-2, za a bac a parameter 2-150 ~ 4+200-3-40022 -27 6938 xz - 272+150 y = -697-200 2: 382 + 700 Point E = (150-272, -697-200, 382 + 700) Since point & lie on plane containing the points A, B, C then it must satisfy the equation - 6750 x - 172504 +9500 Z-4375000 20 -27% – 699 +382 - 17500 30 >> -27 (150-272) -- 69(-699-200) +38(387+700) -17500 20 >>> 7292 +47612 +14442-4050 + 13800 +26600 - 17500 20 → 69342 + 18850 = 0 a = -2.72 E = (150-272 ,-691-200, 382+700) E: ( 223.44, -12.32 , 596.64)

b> GE= Ē - a > = (223.44 î – 12.321 +596-64ť ) -(150 –2009 + 700k) ->GE - 73.44 î. +187.689 – 103.36/ GET = V(73.413)2 + (187.68)2+(-103-302 →GE = 226.496 units IGE= 226.5 units] C AC = ²-3 = (100 î - 2007 +350k) -(6-50) ↑ +425k). - AC = 150 î - 100) - 754 BC = 2-B = (100Î – 1009 +350 ☆ ) - (250î + 70ſ +410€) » BC = - 150 1-30] - 60k De = ?-B= (1001 - 200] +350k) - (-501 - 507) >DE = 150 î – 50j +350 € Volume = ₃ [ AC BC Dc] = + 150 -100 -75 | -150 -30 -60 150 -50 350 = 1 x 150 x10x51 -10 -15 -1 -3 7 -5 -{X150x10x5 x| (-270 530 –120] * *x150x10x5*970) -12 701 Volume = 1212500 cubic unit
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