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Achar a distância de P1 à P2, nos casos: 9º) P (1, 2, 3) e r: eixo Oz Eixo Oz (0, 0, 1) P x Oz = 𝑖Ԧ 𝑗Ԧ 𝑘ሬԦ 1 2 3 0 0 1 = 2 3 0 1 𝑖Ԧ - 1 3 0 1 𝑗Ԧ + 1 2 0 0 𝑘ሬԦ = (2 – 0) 𝑖Ԧ – (1 – 0) 𝑗Ԧ + (0 – 0) 𝑘ሬԦ = (-2, -1, 0) d = d (P, r) = |𝑃𝑥 . 𝑂𝑧| |𝑂𝑧| = |(−2,−1,0)| |(0,0,1)| = ඥ22+(−1)2+ 02 ඥ02+(0)2+ 12 = ξ5 ξ1 = ξ5 10º) P (1, 2, 3) e r: x = 1 e z = -1 A (1, 0, -1) 𝑣Ԧ = (0, 1, 0) 𝐴𝑃ሬሬሬሬሬԦ = P – A → (1, 2, 3) – (1, 0, -1) = (0, 2, 4) 𝑣Ԧ x 𝐴𝑃ሬሬሬሬሬԦ = 𝑖Ԧ 𝑗Ԧ 𝑘ሬԦ 0 1 0 0 1 4 = 1 0 1 4 𝑖Ԧ - 0 0 0 4 𝑗Ԧ + 0 1 0 1 𝑘ሬԦ = (4 – 0) 𝑖Ԧ – (0 – 0) 𝑗Ԧ + (0 – 0) 𝑘ሬԦ = (4, 0, 0) d = d (P, r) = | 𝑣Ԧ x 𝐴𝑃ሬሬሬሬԦ | |𝑣Ԧ| = ඥ42+(0)2+ 02 ඥ02+(1)2+ 02 = ξ16 ξ1 = 4 RESOLUÇÃO DO LIVRO VETORES E GEOMETRIA ANALÍTICA Achar a distância do ponto P ao plano , nos casos: 11º) P (2, -1, 2) e : 2x – 2y – z + 3 = 0 d (P, ) = |𝑎𝑥0+𝑏𝑦0+𝑐𝑧0+𝑑| ξ𝑎2+ 𝑏2+ 𝑐2 d (P, ) = |2 . 2−2 (−1)−1 . 2+3| ඥ22+ (−2)2+ 12 = |4+2 − 2+3| ξ9 = 7 ξ9 = 7 3 12º) P (3, -1, 4) e : x + y + z = 0 d (P, ) = |𝑎𝑥0+𝑏𝑦0+𝑐𝑧0+𝑑| ξ𝑎2+ 𝑏2+ 𝑐2 d (P, ) = |1 . 3+1 .(−1)+1 . 4| ඥ12+ (1)2+ 12 = |3−1+ 4| ξ3 = 6 ξ3 = 6 ξ3 . ξ3 ξ3 = 6ξ3 3 = 2ξ3 13º) P (1, 3, -6) e : 4x – y + z + 5 = 0 d (P, ) = |𝑎𝑥0+𝑏𝑦0+𝑐𝑧0+𝑑| ξ𝑎2+ 𝑏2+ 𝑐2 d (P, ) = |4 . 1−1 .(3)+1 . (−6)+5| ඥ42+ (−1)2+ 12 = |4−3−6+5| ξ18 = 0 ξ18 = 0 14º) P (0, 0, 0) e : 3x – 4y + 20 =0 d (P, ) = |𝑎𝑥0+𝑏𝑦0+𝑐𝑧0+𝑑| ξ𝑎2+ 𝑏2+ 𝑐2 d (P, ) = |3 . 0−4 .(0)+20| ඥ32+ (−4)2 = 20 ξ25 = 20 5 = 4 15º) P (1, 1, 1) e x = 2 + 2h + 3t : y = -1 + h + t z = 2 – h A (2, -1, 2) 𝐴𝑃ሬሬሬሬሬԦ = P – A → (1, 1, 1) – (2, -1, 2) = (-1, 2, -1) (𝑢ሬԦ, 𝑣Ԧ, 𝐴𝑃ሬሬሬሬሬԦ) = 𝑖Ԧ 𝑗Ԧ 𝑘ሬԦ 3 1 0 −1 2 −1 = 1 0 2 −1 . 2 - 3 0 −1 −1 . 1 + 3 1 −1 2 . (-1) = (-1 – 0) . 2 – (-3 + 0) . 1 + (6 + 1 ) . (-1) = -2 + 3 – 7 = -6 𝑣Ԧ x 𝑢ሬԦ = 𝑖Ԧ 𝑗Ԧ 𝑘ሬԦ 2 1 −1 3 1 0 = 1 −1 1 0 𝑖Ԧ - 2 −1 3 0 𝑗Ԧ + 2 1 3 1 𝑘ሬԦ = (0 + 1) 𝑖Ԧ – (0 + 3) 𝑗Ԧ + (2 – 3) 𝑘ሬԦ = (1, -3, -1) d (P, ) = |(𝑢ሬԦ,𝑣Ԧ,𝐴𝑃ሬሬሬሬԦ)| |𝑣Ԧ x 𝑢Ԧ | = −6 ඥ12+(−3) 2+(−1)2 = 6 ξ11 16º) Calcular a distância entre os planos parelelos: 1: x + y + z = 4 e 2: 2x + 2y + 2z = 5 𝑛ሬԦ1 = (1, 1, 1) e 𝑛ሬԦ2 = (2, 2, 2) Fazendo x = y = 0, temos que z = 4, então P (0, 0, 4) d (1, 2) = d (P1, 2) = |𝑎𝑥0+𝑏𝑦0+𝑐𝑧0+𝑑| ξ𝑎2+ 𝑏2+ 𝑐2 = d (P1, 2) = |2 .0 +2 .0 +2 .4−5| ξ22+ 22+ 22 = |8−5| ξ12 = 3 ξ12 = 3 2ξ3 = 3 2ξ3 . 2ξ3 2ξ3 = 6ξ3 4ξ9 = ξ3 2