Three Dimentional Geometry

Answer :

centroid of ΔABC = centroid of ΔPQR as the triangles are equal

                   ∴ centroid of ΔABC = [{[9 + (– 7) + 8] / 3}, {[2 + 6 + (– 9)] / 3}, {[5 + 1 + 3] / 3}]

                                                = [(10 / 3), {(– 1) / 3}, 3].

Answer :

                   A(– 1, – 2, – 3), B(1, 2, 3), C(1, 2, 1)

                          centroid         = [{(– 1 + 1 + 1) / 3}, {(– 2 + 2 + 2) / 3}, {(– 3 + 3 + 1) / 3}]

                                      = [(1/3), (2/3), (1/3)]

                   Midpoint of BC = D = [{(1 + 1) / 2}, {(2 + 2) / 2}, {(3 + 1) / 2}] = (1, 2, 2)

                   ∴ ℓ(AD) = √(4 + 16 + 25) = √45 = 3√5 unit

Answer :

                   A(– 5, 7, 2), B(1, 3, 7). Let P & Q are points of trisection, Q divides

                   AB from A in ration 2 : 1

                         Q = [{[2(1) + 1(– 5)] / [2 + 1]}, {[2(3) + 1(7)] / [2 + 1]},{[2(7) + 1(2)] / [2 + 1]}]

                             = [– 1, (13 / 3), (16 / 3)]

                   P is midpoint of AQ

                   ∴ P = [{[– 1 – 5] / 2}, {[(13 / 3) + 7] / 2}, {[(16 / 3) + 2] / 2}]

                   ∴ P = [– 3, (17 / 3), (11 / 3)]

Answer :

Let position vector of A(a), B(o) and C(c).

                   Let P, Q divide AC from A in ratio of 1 : 2 and 2 : 1

                   ∴ P = [(2a + c) / 3], Q = [(a + 2c) / 3]

                   also   a ∙ c = AB ∙ BC = 0

                   and       BP² + BQ²         = (1/9)|2a + c|² + (1/9)|a + 2c|²

                                                = (1/9) [5|a|² + 5|c|²]

                                                = (5/9) [|a|² + |c|²]

                                                = (5/9) [AB² + BC²]

                   ∴ BP² + BQ² = (5/9) AC²

Answer :

                   in ΔABC, Let position vector of A, B, C is X, Y, Z. G is centroid of ΔABC.

                   ∴ position vector o = [(X + Y + Z) / 3]

                   ∴ X + Y + Z = 0

                   ⇒ 0 = GA + GB + GC

                   i.e. GA + GB = 0