Sol - Lista 2 - Geometria Analítica

SOLUCIONÁRIO DA LISTA 2\nLISTA # 1\n1. \\mathbf{u}=(5,4,1),\\mathbf{v}=(1,2,2),\\mathbf{w}=(2,0,-3)\n(a) \\mathbf{u}\\cdot(\\mathbf{v}\\times\\mathbf{w})\n= \\begin{bmatrix}\n- 2 & 1 & -1 \\\\\n 2 & 0 & 2 \\\\\n 0 & -4 & -1\n\\end{bmatrix}\n=\\left(-(6+10)+(3+4)(-1)+(0-4)(1)\\right)\n=-18+7-4=-29\n(b) \\mathbf{u}\\cdot(\\mathbf{w}\\times\\mathbf{v})\n= \\begin{bmatrix}\n 2 & -3 & 0\\\\\n -2 & 3 & 0 \\\\\n 0 & 2 & 0\n\\end{bmatrix}\n=\\left(-2(-3)-6(-1)+(0+1)(2)\\right)\n=8-0-21=-29\n(c) \\mathbf{v}\\cdot(\\mathbf{u}\\times\\mathbf{w})\n= \\begin{bmatrix}\n 1 & 2 & 2 \\\\\n(0-3)(2-9)+(2-9)(2)+(2-0)2\n\\end{bmatrix}\n= -3-22+4=-29\n\n985 Um momento para fazer, mas para resolver uma questão\nSiga lembrar do produto scalar e, \\mathbf{u}=(\\mathbf{u}\\cdot(\\mathbf{v}\\times\\mathbf{w})\\,\n\\mathbf{v}=(\\mathbf{u}\\cdot(\\mathbf{w}\\times\\mathbf{u})\n\n2. \\mathbf{u}\\cdot(\\mathbf{v}\\times\\mathbf{w})=0 \\text{ colineares}\n(a) \\mathbf{u}=\\mathbf{v}-(\\mathbf{w}\\cdot \\mathbf{u})\n=\\begin{bmatrix}-1 & -1 & 0 \\\\ 2 & 2 & 4 \\\\ 3 & -1 & 0\n\\end{bmatrix}\n=\\left(-(8-3)\\right)=(41+12)\\cdot(-1)+(3+4)2=0\n=\\text{NÃO COLINEARES}\n(b) \\mathbf{u}=\\mathbf{w}\\times\\mathbf{v}\n=\\begin{bmatrix}2\\-3&0\\\\2&4\\\\1 & 3\\\\\n\\end{bmatrix}\n=\\left(2(2)+2(-1)\\right)\\times(\\mathbf{u})=8-8+0=0\n\\text{COLINEARES}\n LISTA 2 # 2\n\\mathbf{u}=(1, 0)\\\n\\mathbf{v}=(2, 0, 1)\n\\mathbf{w}=(1, 0, 0)\n\\mathbf{u}=1(-1)\\\n\\mathbf{w}=8(0,0,0)-2(-2,-2)=(3,8)-3\\\\-0\\\n8(HE,1,0)+(-4)+(-2)\\\n=\\left(4(4,0)-(2\\times\\mathbf{w}\\cdots \\\n=8 + \\mathbf{v}=(1,-0,0)-\\mathbf{w}=(-1,0,0)-[(1,0,0)-0,0]=0\\ \n\\left[-1(1,2)-(4)\\cdots (2,0,-2)]-(-1,0,0) - 3(-6)-(1,-0,0)-(-1-2)\\in\\ 0\\\nVol(\\mathbf{u=3}+ - 0,0,\\mathbf{-1,-8,-2})\\\\ \n VOL = - 5.00(6.8)\\\n\\n\\ \nVol =\\left(\\mathbf{u}\\cdot(\\mathbf{v}\\times\\mathbf{w})\\right)\\\\ \n=|\\mathbf{u}=((\\mathbf{x}\\times \\mathbf{u})=\\mathbf{y})\\cdots -(1)\\\n1(1)=2\\2.\\[\\cout = \\mathbf{u}\\cdot\\mathbf{v} . (\mathbf{d})$\\\n-(-1) - 2\\hspace{bmatrix}\n0 = x^{(1,0)}(4-2)\n\\] LISTA 2 # 3\n4. \\mathbf{L}_1(5,-5,6)\\mathbf{P_2(4,-1,12)}\\\nF: \\mathbf{x}=3+ 4(y)+ 7 - 2\\\\\n-1 2 -2\\\\ \n-1 +(2)-(-2)\\ =\\text{PROTOXIS})\n\\\n-\\mathbf{d}_{2}\\\nF=\\(-5, -1)( (\\mathbf{d^{2}})\\cdots \n-\\mathbf{F}\\to B )\\ \n3+\\pm(2)\\mathbf{A}{(-1)(1,2,-2)}=-(0)\\\n0 =\\mathbf{P}^{2}(0)=0\\\\\n pF = 0 = Art __ \n6(8) = [distributivas] \nP(1,-7,1)\\ \\ PROTOXY\n=\\mathbf{p.part= -\mathbf{u}\\dot(\\mathbf{x},\\lambda=1)(\\ (4,-1,2)} \n Lista 2 # 5\n(c) x = -4t\n\\{ x = -2 + t \n y = -4 - t \n z = -4 + 2t \n\\}\nE\n\\begin{array}{cc}\n(0,0,0) & C(0,8,0) \n(2,0,4) & D(2,3,0) \nB(1,0,0) & E(2,0,0) \n\\end{array}\n\n- Resolviendo por A\n\\{(x,y,z) = (2,0,4) + t(2,0,0)\\} \n\\{ x = 2 + 2t\\}\n\\{ y = 0\\}\n\\{ z = -4 + 0\\}\n\\}\n- Resolviendo por B\n\\{(x,y,z) = (0,2,0) + t(0,2,0)\\}\n\\{ x = 2t\\}\n\\{ y = 0\\}\n\\{ z = 0\\}\\} (a) B ⊂ C\n \\vec{EB} = (0,0,0) - (0,0,4) = (0,0,-4)\n * Parametrizando por B\n (x,y,z) = (0,0,4) + t(0,0,-4)\n \\begin{cases}\n x = 0 \\\\\n y = 0 \\\\\n z = 4 - 4t\n \\end{cases}\n * Parametrizando por C\n (x,y,z) = (0,5,-4) + t(0,0,4)\n \\begin{cases}\n x = 0 \\\\\n y = 5 \\\\\n z = -4 + 4t\n \\end{cases}\n (d) B ⊂ C\n \\vec{EB} = (0,0,4) - (0,0,0) = (0,0,4) \n \\vec{EC} = (0,5,-4)\n * Parametrizando por B\n (x,y,z) = (0,0,4) + t(0,5,-4)\n \\begin{cases}\n x = 0 \\\\\n y = 5t \\\\\n z = 4 - 4t\n \\end{cases}\n * Parametrizando por C\n (x,y,z) = (0,5,-4) + t(0,0,4)\n \\begin{cases}\n x = 0 \\\\\n y = 5 \\\\ \\\\\n z = -4 + 4t\n \\end{cases}\n\n(e) B ⊂ E\n \\vec{B} = (2,0,0) - (2,0,4) = (0,0,-4)\n \\vec{E} = (0,2,0) - (0,2,0)\n * Parametrizando por B\n (x,y,z) = (2,0,0) + t(0,0,-4)\n \\begin{cases}\n x=2 \\\\\n y=0 \\\\\n z = 0 - 4t\n \\end{cases}\n * Parametrizando por E\n (x,y,z) = (0,2,0) + t(0,0,0)\n \\begin{cases}\n x=0 \\\\\n y=2 \\\\\n z=0\n \\end{cases} 11. O ángulo entre dos vectores v1 y v2 viene dado a través de la siguiente fórmula:\n \\vec{v_1} = (1,-1,2)\n \\vec{v_2} = (-2,2,1)\n Sabiendo que \\cos(\\theta) = \\frac{\\vec{v_1} \\cdot \\vec{v_2}}{||\\vec{v_1}|| ||\\vec{v_2}||}\n \\vec{v_1} \\cdot \\vec{v_2} = (1)(-2) + (-1)(2) + (2)(1) \n = -2 - 2 + 2 = -2\n ||\\vec{v_1}|| = \\sqrt{(1)^{2} + (-1)^{2} + (2)^{2}} = \\sqrt{1 + 1 + 4} = \\sqrt{6}\n ||\\vec{v_2}|| = \\sqrt{(-2)^{2} + (2)^{2} + (1)^{2}} = \\sqrt{4 + 4 + 1} = 3\n \\cos(\\theta) = \\frac{-2}{\\sqrt{6}(3)}\n \\cos(\\theta) = \\frac{-2}{3\\sqrt{6}}\n => \\theta = \\arccos(\\frac{-2}{3\\sqrt{6}})\n\n12. (x,y) = (4, 8)\n de la fórmula que: y = b + mx \n x- y = 1 \\\\\n \\frac{x+y}{n} = \\frac{x + \\frac{y}{2} + 2y}{...}\n = m - \\sqrt{2}\n (a)(x,y,z) = (4,0,-2) + t(2,4,5)\n = (x - 4)/2 = (y + 2)/4 = (z - 5)/5\n\n x = 8 - 2t\n y = 2t - 10\n z = 5t + 3\n\n(b)\n (x,y,z) = (2,1,-2) + s(1,-2,3)\n = (x - 2)/1 = (y - 1)/-2 = (z + 2)/3\n\n x = 2 + s\n y = -2s + 1\n z = 3s - 2\n\n OBS: O mesmo resultado foi obtido utilizando o ponto B. (c)\n u = (2,-1,3) - (-1,2,3)\n = (3,-3,0)\n (x,y,z) = (1,-2,3) + t(3,0,0)\n\n (x+3)/5 = (y+4)/-2 = (z-5)/0\n\n y = -x - 1\n x = 2\n\n y = -x + 1\n OBS: O mesmo resultado foi obtido utilizando o ponto B. 14. v = P1 - P2 = (1,2,3) - (-1,0,3)\n = (2,2,0)\n (x1,y1,z1) = (1,-1,2) + t(-2,2,4) PONTO T1\n\n (x+x1)/2 = (y+y1)/2 = (z+z1)/4\n\n x + x1 = 2\n y -1 = 3\n\n x = 2 - y\n y = -y + 1\n z = z - 2\n\n OBS: O mesmo resultado foi obtido utilizando o ponto P2. y0 = 3x - 4 - x + 4\n\n1: y = 2x - 3\n\n y = 2x - 2\n\n-m 0 - 6\n\nF3 = (1, 4, -2)\n\nA3 = (0, 4, 0)\n\n- Si f r = u t o c u p a n t o\n\n (c y1 x X1 1 A2 A3) = m 2 0\n l 1 o. (AB - O) m - (1 - e 2m)(1 - 3) + (7 - 6 - 2m)(0) = 0\n1bm - 14 m0\n\nsem - y2 = 8\nm = 2/2 L:1ST2 # A7 L:1ST2 # A7\nr: (x - m - t)\ny = 2x - 2t\nz = 2t\n\na1 = x1 - 1 1 4 + 2 - 3 y\n\nx2 = 3 - m + 2m\n\nm and re evaluating\n x = m - e y - 1 + 1 y e \n 2x - 20 - z2 =/ge/ 9 8\nx = 2 - 4/ sin m\n\n-10/3\nm = -1/2\n-2 \n y = 4 + m 3/ 2m = 4m + 2m m = 4 20. P1(0,0,3) + t(1,2,0)\n\nT = (0.0,3)-(1.0,0) - (0,-4,0,-2)\n\nd = |F-A| = | 1 - 4 + 0 + 0 \n0 | \n \n= 12. \n\nF - K. j = \n\n(-1 1 - 2 6\n1 0 0 \n- 2 F\n= d67.0.07.8116 \n|\nd = 1 ( 1/4 - 22)\n= 6/4 plus 14/ more treat \n