Enunciado<\/b><\/span><\/p>\nNum referencial o.n. tridimensional, est\u00e3o representados o ponto $A\\,(2,-3,1)$ e o ponto $B\\,(3,2,6)$.<\/p>\n
\n- Determine a intersec\u00e7\u00e3o da reta AB com o plano xOy.<\/li>\n
- Determine o lugar geom\u00e9trico dos pontos equidistantes de A e de B.<\/li>\n<\/ol>\n
Resolu\u00e7\u00e3o >><\/a><\/span><\/div><\/div>\n\n\n
Resolu\u00e7\u00e3o<\/b><\/span><\/p>\n\n- Como $A\\,(2,-3,1)$ e $B\\,(3,2,6)$, ent\u00e3o o vetor $\\overrightarrow{AB}=(1,5,5)$ \u00e9 director da reta AB, podendo esta ser definida por \\[\\frac{x-2}{1}=\\frac{y+3}{5}=\\frac{z-1}{5}\\]
\nComo o plano xOy pode ser definido pela condi\u00e7\u00e3o $z=0$, temos: \\[\\begin{array}{*{35}{l}}
\n\\left\\{ \\begin{array}{*{35}{l}}
\n\\frac{x-2}{1}=\\frac{y+3}{5}=\\frac{z-1}{5}\u00a0 \\\\
\nz=0\u00a0 \\\\
\n\\end{array} \\right. & \\Leftrightarrow\u00a0 & \\left\\{ \\begin{array}{*{35}{l}}
\n\\frac{x-2}{1}=\\frac{y+3}{5}=-\\frac{1}{5}\u00a0 \\\\
\nz=0\u00a0 \\\\
\n\\end{array} \\right. & \\Leftrightarrow\u00a0 & \\left\\{ \\begin{array}{*{35}{l}}
\nx=\\frac{9}{5}\u00a0 \\\\
\ny=-4\u00a0 \\\\
\nz=0\u00a0 \\\\
\n\\end{array} \\right.\u00a0 \\\\
\n\\end{array}\\]
\nLogo, o ponto de intersec\u00e7\u00e3o da reta AB com o plano xOy tem coordenadas $(\\frac{9}{5},-4,0)$.
\n\u00ad<\/li>\n