Lista 7 de Geometria Analítica Professor Jaime Velasco

RESPOSTAS C(0,0,0) 3) L'índuce à d entre o ponto P(2,5,-1) e a reta r a que passa pelos pontos A(1,1,2) e B(2,-1,3).\n\n1) A Reta\n\nA = B - A = (2,-1,3)-(1,1,2) = (1,-2,1)\n\n(x,y,z) = (1,1,2) + t(1,0,-1) => eq. sd.\n\nx - 1 = t\n\ny - 1 = -t\n\nz - 2 = -t\n\n\nPara...\n\n- A distãncia entre P(2,5,-1) e r.\n\nP0 = (1,1,2)\n\nd(P0,r) = |(1/t) x(P-P0)| = |(1(4,1) x (1,6,-3))|\n\n|t|\n= (1,0,1)\n\nd(P0, r) = \\sqrt{(-6,-4,6)}\n= |(1,0,1)|\n\n\\sqrt{2}\n\nd(P0,n) = \\sqrt{36+16+36} = d(P0,n) = \\sqrt{82}/\\sqrt{2}\n\nd(P,r) = 2^{3/2}\n\n 4) Mostre que a reta r: \\displaystyle x+2 = \\frac{3y+1}{6} = \\frac{1-z}{3} \\in \\text{plano } D: 2x - 3y + 6z + 3 = 0\n\nd(P, n) = |\\frac{2x - 3y + 6z + 3}{\\sqrt{(2^{2}) + (-3)^{2} + (6^{2})}}|\n\n= |\\frac{12(0) - 3(1) + 6(9) + 3}{\\sqrt{49}}|\n\nd(P, n) = |10 + 3 + 6 + 1| \\frac{1}{\\sqrt{49}} = d (L,R) = 161 |\n\n\\frac{6}{7}\n\nx - 2 = \\frac{3y - 1}{6}\n\nx - 2 = \\frac{1 - z}{3}\n\n\\frac{6 - 6z = 3x - 6}\n\n-6 - 6z = 3x - 6 - 6\n\n-6z = 3x - 12\n\nz = 3x - 6 / -6\n\nt = \\frac{1}{3} x + 1\n\n 5) N.I. P.(1,-1,0) + t(k,2,2) + P.(2,-2,2) + 2P.(1,2,3)\n\nA1 = (1,-1,0) \\sqrt{(2,2,2)}\n\nP. = (-2,1,2)\\sqrt{(1,2,3)}\n\nd = d(n,m) = |(\\sqrt{j_1,j_2,A_1,A_2})| / |(j_1 x j_2)|\n\nd = d(n,m) = |10| / |10|\n\n= \\frac{1}{\\sqrt{2^{2}-(-4)^{2}+2^{2}}}\n\n= 2 - 18 + 4 + 12 - 4 / - 10\n\n= \\sqrt{8-18+8} + 2\n\n= d.[\\sqrt{10}]\n\n= d = d(n,m) = \\frac{5}{\\sqrt{6}}\n\n P: P(2,0,2) + t(9,4,1) P: P(2,2,0) + t(3,3,3) \n{\n x = x + 2 + 3t \n y = y + 3t \n z = z + 3t \n} \n J1 = (0,1,2) и (2,1,0) \nP = 1. \n J(2,1,0) - (9,0,2) -(1,1,1) \n (9,1,-2) \nline \nd(n,s) \n (0,2,0) \nd = |AxJ| / √ |J| \n √A3 \nd = d(n,s) |(-9,-9,0)| \nd = |(−9,9,0)| \n√(x,y,z) = 3√3| \n3−2 P: P(-4,0,-5) + t(3,4,-2) \nP: P(3,7,5) + λ(6,-4,-1) \n {\n x = -4 + 3t \n y = 4t \n z = -5 - 2t \n } \n {\n x = 3 + 6λ \n y = 7 - 4λ \n z = 5 - λ \n } \n A1 = (-4,-0,-5) \nA2 = (3,7,5) - A1 = \n(3,7,5) - (-4,0,-5) = \n(7,10) - A1A2 \n J1xJ2 \n{\n 3 4 2 \n 6 -4 6 \n 6 -4 7 \n} \n \n d(d(n,s)) = |(J1,J2,A1A2)| / |J1 x J2| \n {\n |507| \n |(-11-9-36)| \n} \n d · d(n,s) = 507 \n√(-12² + (-3)² + (-6)²) \n507 / √(144 + 81 + 1296) \n = 507/√1521 \nd = d(n,s) 507/13 \n39 / 13 \n3 = 13 P: P(0,1,2) + t(2,-2,-1) \n{\n x + y + z = 0 \n 2x - y + z = 0 \n} \n J(2,-1,1)//n \n J(3,0,3)//s \n d = ||(A)(B)(J x J)|| \n √(I x J) \n {I x J = (I) ||(B) / √(n) \n A = {(I)} \n {J} = {(2,-1,-1)} \n 5√\n d = ||(A·B)(J x J)|| = √S.c \n A = (JxJ) \n ≈ 5√3 \n 8 \n \n S(x,y) \n{\n 2 3 4 \n 0 1 −3 \n 2 −1 3 \n} \n A(3,7,3) / desconsiderando \n\n r x i+n3 \n {\n x + y + z = 0 \n y + z = 0 \n 1 / 2 \n 3x, y .\n} \n 6) Calcule a d entre os pontos P(2, 3, 1) e O(2, 2, 3) d(P, O) = |2x + y + z - 3| / \\sqrt{2^2 + 4^2 + 2^2} = |5.2 + 2.1 + 1.3 - 7| / \\sqrt{4 + 1} = |10 + 2| / \\sqrt{6} = 12 / \\sqrt{6}\n\n7) Encontre m tal que a (n, m) = \\sqrt{2}, onde \\{ \\begin{array}{l} x = -2 - t \\\\ y = 1 - t \\\\ z = 5 - t \\end{array} \\Rightarrow R_{-2, 1, 0}, \\{ n, 2, -2 \\}, S_{(1, m, 5)}, \\{ (-1, -4, -1) \\}, iS \\ = (1, m, 5) - (-2, 1, 0) R.S.\\ = (3, m - n, 5)\n\nd(n, m, n) - R.S. \\cdot (\\sqrt{u^2 + v^2}) = \\frac{13(m + 12)}{3\\sqrt{2}}\n\n\\(x + y) = \\sqrt{9 + 9} = 3\\sqrt{2}\nR.S. \\cdot (u \\times v) = 0 + 3m - 3 + 15\\n\\text{Ex. } (x, y) = 3\\text{m} + 12 = -6\\n3m + 12 = 6\\ 3m - c = 3m = -18\\n m = -2\\n m = -6\\\\n α: 2x - y + 3z + (5 + 3√7) = 0 ou α: 2x - y + 3z + (5 - 3√14) = 0 α: x + y + 3z + (1 - 3√11) = 0. (a) √(27/38) (b) 0 (c) 0 (d) 0 (e) 0 (f) 0 (g) √(165/29) (h) 0 (a) 0 (b) √(3/3) (c) 0 (d) 1 (e) 0 (f) 0 (g) 0 (h) 6/√11 x = -2 + t (1) y = 1 + 2t e z = 5 - t