Distâncias no Espaço - Geometria Analítica

Distâncias\n\n*Lembre: G.A. Plano vimos como calcular\nd (ponto, ponto)\nd (P,Q) = \\sqrt{(x_p - x_q)^2 + (y_p - y_q)^2}\n\nd (ponto, reta)\nr: ax + by + c = 0\nd(P,r) = lax_p + by_p + c |/\\sqrt{a^2 + b^2} Similarmente, estudaremos algumas distâncias entre objetos no espaço 3-dim.\n\nCalcularemos\nd (ponto, ponto)\nd (ponto, plano)\nd (plano, plano)\n\nd(1) d (ponto,ponto)\n\\{P = (x_p,y_p,z_p)\nQ = (x_q,y_q,z_q)\\}\n\nentão...\nd(P,Q) = \\sqrt{(x_p - x_q)^2 + (y_p - y_q)^2 + (z_p - z_q)^2}\n\nEx: Descreva (geometricamente e analiticamente) o conjunto dos pontos do espaço que equidistam de P:(3,4,-1) e Q:(0,0,2)\n\nQuero \\vec{u} = (x,y,z) tal que d (x)² = d (y)² ou d(x,y)² = d(z,z)²\n(x - 3)² + (y - 4)² + (z + 1)² = (x - 6)² + (y - 0)² + (z - 2)²\n-6x + 4y² - 8y + 16 + z² + 27 + 4 = x² + 12x + 36 + y² + z² - 47 + 4\n20 - 40 = -6x + 8y - 6z\n-14 = -6x + 8y - 6z \\rightarrow x = -3x + 2y + 3z\n\n(2) d(ponto,plano)\n\\{P = ponto\\}\n\\Pi = plano\n Precisamos de um vetor normal ao plano e um ponto que viva no plano\n\nd = (p, π) = || proj_\\vec{n} (\\vec{AP})|| = <\\vec{AP}, \\vec{n}> / ||\\vec{n}||\n\nEm coord.:\n\\{P = (x_p, y_p, z_p) \\} \\vec{n} \\text{ tem cood.} (a,b,c)\n\\Pi: ax + by + cz + d = 0\n\nd(p, π) = lax_p + by_p + cz_p + d / \\sqrt{a^2 + b^2 + c^2}\n\nEx: Calcular d(p, π) nos casos abaixo:\n\na) \\Pi = (1,3,4) e π:\\{ x = 1 + λ\ny = 1 + λ - μ\nz = 3λ\n\Veja que A: (1,1,0) \\in π (λ = μ = 0)\n\nT\\text{ambém } (1,1,0) e y = (0,-1,3)\n\\vec{u} \\times \\vec{v} = \\vec{n} = d + \\left( e_1 \\: e_2 \\: e_3 \\right) = e_1 \\: e_3 - e_2\\: e_3(-1 + 0) \n\\text{ }\\frac{1}{1\\:0}\n\n0 -1 3\n\\vec{n} = (3, -3, -1)\n\\vec{AP} = (0,2,4)\n\nd(p, π) = <\\vec{AP}, \\vec{n}> = | <3.0 +(-3).2 +(-1).4| * |0 – 6 – 4| = \\sqrt{10}\n|\\|\\vec{n}\\| = \\sqrt{(3^2 + (-3)^2 + (-1)^2)} b) P = (1,2,-1) e \\pi: 3x - 4y - 5z = 1 \\vec{n_\\pi} = (3,-1,-5)d(P,\\pi) = |3.1 + 4(-1).2 + 5(-1)| / \\sqrt{3^2 + (-4)^2 + (-5)^2}d(\\Phi,\\pi) = |3 - 8 + 5 - 1| / \\sqrt{9 + 16 + 25} = 1 / \\sqrt{50}Ex.: Obter os pontos de r: {x = 2 - y, 2y + z - 2 = 0} do plano \\pi: x - 2y - z = 1 *{(4-y - 2) \\Rightarrow (1,0) e (2y - 2 = 0) \\Rightarrow (0,2) |det(1 1 0; 0 2 1) = 2.0 - 0 2.0 + 1.0.0 = 0 |P/2}r: (0,2,0) 2° passo: usar d( plano, plano)\\sqrt{6} = |1(-1.+1) + 2(-1)(0) + (-1)(.2) + 1(1) + 3(-1)(6)|/(1^2 + 2^2 + 3^2)\\Rightarrow d(\\pi, \\pi)=|.7-1|=|7(-1)|/\\sqrt{6}\\Rightarrow \\lambda=1,1\\Rightarrow L: (x=2+\\lambda, y=5, z=2.16}\\Rightarrow (x,y,z) = (2 + \\lambda, y + 5, z + 10)\\Rightarrow (P1: (8,-7,16))(P2:(-3,5,-8))d(plano, plano): sejam \\pi e \\pi', d(\\pi,\\pi') = min{d(\\pi_i, \\pi_j), \\pi_i \\in \\pi, \\pi_j \\in \\pi'} note que se \\pi \\cap \\pi' \\neq \\emptyset então d(\\pi,\\pi') = 0 |L, paralelos No último caso,\\nEx: calcule d(\\pi,\\pi') nos seguintes casos: a)\\pi: (6x - 2y + z = 4 \\vec{n_\\pi} = (6,-2,1))\\pi': 3x - y + z = 0, b)\\vec{n_\\pi'} = (3,-1,1)|\\vec{n_\\pi} \\times \\vec{n_\\pi'} = det(6 - 2)\\Rightarrow (- 6,6,0);(- 2,-1)\\Rightarrow {(x = 2, y = 5\\Rightarrow \\Rightarrow(\\lambda=0)}\\Rightarrow \\vec{n_\\pi \\cdot \\vec{n_\\pi'}}=0 |\\Rightarrow d(\\pi, \\pi') = 4\\Rightarrow (x=9, z=3)\\Rightarrow d(\\pi,\\pi)=0 d(P,r) = min {d(P,Q) | Q ∈ r}\n\nObs: O ponto Q0 é tal que d(P,r) = d(P,Q0) é o \"pé da perpendicular\" à r passando por P.\n\nPara calcular d(P,r):\n1) Encontrar Q0\n2) Calcular d(P,Q0)\n\nEx: Calcule d(P,r) sendo P=(3,5,-1) e r paralela ao vetor u de coord.(1,2,-3) e que passa por A=(0,0,2)\nr: x=0 + λ\nPer: 3 = λ; 5 = 2.3.X\nY = 0 + 2λ\nZ = 2 - 3λ\n\nQueremos ver tal que <P0,u> = 0 <P0,u> = <λ-3,2λ-5,2-3λ> = (λ-3)·1 + (2λ-5)·2 + (2-3λ)·(-3) = 0\nλ - 3 + 4λ - 10 + 6 - 9λ = 0\n-3λ + 22 = 0 ⇒ λ=22/3\n\nx = λ ⇒ x = 11/7\ny = 2λ ⇒ y = 22/7\nz = 2 - 3λ ⇒ z0 = 2 - 3·(22/7) = 14 - 66/7 = -19/7\n\nd(P1,r) = d(P1,Q1) = √[(xP0-xA)² + (yP0-yA)² + (zP0-zA)²] \nd(P1,r) = √[(3-4)² + (5 - 2)² + (-1 - 2)²] \nd(P1,r) = √[(3-4)² + (5 - 2)² + (-1 - 2)²] d(r,π) = min {d(r,Qπ) | P ∈ r e Qπ ∈ π}\n\nObs: Se r ∩ π ≠ φ , então d(r,π) = 0\n\nSejam u~ um vetor diretor de r e n~π um vetor normal a π.\n\nSe <u~,n~π> = 0 : r ∩ π = 0 ⟹ d(r,π) = 0\n\nSe <u~,n~π> ≠ 0 : r // π : d(r,π) ≠ 0\n\nNos dois subcasos acima podemos calcular a distância d(r,π) = d(A,π) para qualquer A ∈ r. Ex: Calcular d(r, π) nos casos abaixo:\n\na) π: 2x + 3y - z = 0 e r: {4x + 6y = 2 \\ x + 4y + z = 0\n\n\\n_T = (0,3,0)\n\n\\langle \\mathbf{m}_{r}, \\mathbf{u}_{r}\\rangle = \\langle (2,6 + 3 \\cdot (4), 0) \\rangle\n\n\\text{det} \\begin{pmatrix}\n4 & 6 & 0 \n1 & 4 & 1 \\ \n\\end{pmatrix}\n\n= 2 \\cdot 4 - 6 \\cdot 1 (16 - 6)\\n\\n\\langle \\mathbf{n}_{π}\\rangle = \\mathbf{0}\n\n\\left\\{\\begin{array}{c}\nym + 6y = 2 \\ x = 1; y = -\\frac{1}{3} \\ \nx + 2y + z = 0 \\ z = -1 + 4y = -\\frac{2}{3} + -\\frac{1}{3}\n\\end{array}\\right.\n\\n\\mathbf{d}(p, \\pi) = 1 \\left[xa + by + cz + d\\right] \n\\sqrt{a^2 + b^2 + c^2}\n\nd(p, \\pi) = |2 + 3 + (-3) \cdots \\left(\\frac{1}{3} + (4)(-1)\\right)|\n\\backslash\\frac{0}{\\sqrt{2^2, 3^2, 0^2}}\n\nd(r, \\pi) = |2 - 1 - 1| = |1| = \\sqrt{\\frac{1}{13}} = 0\n\n\\mathbf{d}(p, \\pi) = |1 + 3 + 0| = 0\n\nb) π: x + y + z = 0 e r: {x = 3λ \\mathbf{u_r} = (3,1,0)\n\\n_T = (1,1,1)\n\\<<<< |(3,3,1,1,0) = 3 + 0\n\n\\langle \\mathbf{n}_{r}, \\mathbf{u}_{r}\\rangle = 2\n-->n(r, \\pi) \\neq \\emptyset = 0\\n\n\nc) π: x + y - 2 = 6 e r: {x = 6 + λ \\ y = λ \\ z = 1 + 2λ\n\\n_T = (4,3,-1)\n\\langle \\mathbf{n}_{π}, \\mathbf{u}_{r}\\rangle = (1,4 - 2) => \\langle (1,3), u\\rangle = 0