Resolução de Alguns Exercícios de Distâncias boldrini

Exercícios Distância\n2-) A(1,1,4) B(-6,6,4)\nd(A,B)=d(P,B) P(X,Y,Z)\n((X-1)²+(Y-1)²+(Z-4)²)-((6-X)²+(6-Y)²+(4-Z)²)\nX²-2X+1+Y²-2Y+1+Z²-16-(6-X)²-(6-Y)²-(4-Z)²\nX²-2X+1+Y²-2Y+1+Z²-16-(-6)²-(6)²-(Z-4)²\nX²-2X+1+Y²-2Y+1+Z²-36+12X+12Y-8Z+16-8Z+Z²\n-2X-2Y-24=0 <=> 0=0\n-12X+10Y=70 => 10Y=70-12X => 7X=0 => X=0\nY=7.0+Y=7Y=2t=>P(7,P)\n\n3-) A(2,1,3)\nR: {X=1,2+t/ Y=2+t/ A=(1,2,2)}\nY=2+t\nZ=2t-t.A=(0,2,1)\n\nD= d=(√(X)×AP)\n|1| = \n| 1 K K |\n| 1 2 2 |\n| 0 2 1 0 |\n\n|√(X)×AP|=√(16+4+16)=√36=6\n|√(1)|=√1+4+1; |1=3|\n\nD= 6/3=> d=√2 b) P(1,2,3)\nd(P,B)=d(0,F)\n0B:F)\n2x2-3y=(0,-3,2)\n(2x-2,0)\n2\n|1|=1\n|1 K 1 |\n| 1 2 |\n| 2 3 1 2 |\n3y=0 d=3 d=√3\n\n√(y)=0\n\nC(2-2i=3);\n\nD=√5 x) A(-3,i(4,5), B(-4,-1,0) C(-X,3,5)\nd=|102|xBA| \n| (0,4,5) = | (1,2,4) | = 6i+5g-yXk+0 (6,5,-x1)\n\n√(k)101 16i s1y\nIBc(BA):1√(8g+25b+6=√77 d=|B C|=√16+25^(-√41)\n\n(√(172 -√44)=√3152=>√–√3/6) b) r(A(1,2,1) B(4,-1,2)) 5 (C(0,4,-2) D(4,1,1)) \nAB = (2,-1,-1) \nA(2) = (1,4,-3) \n= -2 1 2 -1 \n 1 0 3 \n 4 -1 1 \n= 3 -1 + 6 3 = (5) \n-6 3 3 -1 \n \\ |\\ \nA B x C D = -2 \\ \n | 3 | 1 \n | 2 3 | \n = | 3 ; | + 5 y ; h x = (-3,1) \n 9 25 + 1 ; \\sqrt{35} \n d= 5 \sqrt{35} - 5 \sqrt{35} = 5 \\sqrt{35} \n 35 135 135 35 7 \n(c) (x = 3 y = t + 5 { x = 1 \n y = 2 t + y \\rightarrow \n \n 2 \n √2(3,2,0) \n ⟶ x \n ( \n (√3,√2,√1) = 3 2 0 8 3 2 \n 1 × 4 \n 2 2 0 2 2 2 \n