Find the equation of the plane through the points (1,1,1),(-4,0,2) and (3,-1,0)
"A(1;1;1);B(-4;0;2);;C(3;-1;0);\\\\\\begin{vmatrix}\n x-x_1 & x_2-x_1 &x_3-x_1 \\\\\n y-y_1 & y_2-y_1 &y_3-y_1\\\\z-z_1 &z_2-z_1 &z_3-z_1\n\\end{vmatrix}=0\\rightarrow\\\\\\begin{vmatrix}\n x-1 & -5 &2 \\\\\n y-y_1 & -1&-2\\\\z-z_1 &1 &-1\n\\end{vmatrix}=0\\rightarrow\\\\(x-1)\\times(-1)^2\\times\\begin{vmatrix}\n -1 & -2 \\\\\n 1 & -1\n\\end{vmatrix}+\\\\+(y-1)\\times(-1)^3\\times\\begin{bmatrix}\n -5 & 2 \\\\\n 1 & -1\n\\end{bmatrix}+\\\\+(z-1)\\times(-1)^4\\times\\begin{vmatrix}\n -5 & 2 \\\\\n -1 & -2\n\\end{vmatrix}=0\\rightarrow\\\\3(x-1)-3(y-1)+12(z-1)=0\\rightarrow\\\\3x-3y+12z-12=0\\\\\nAnswer:3x-3y+12z-12=0"
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