Question

Write an equation whose graph consists of the set of points \(P(x, y, z)\) that are twice as far from \(A(0,-1,1)\) as from \(B(1,2,0)\)

Short answer

The equation of the set of points \(P(x, y, z)\) that are twice as far from \(A(0, -1, 1)\) as they are from \(B(1, 2, 0)\) is \(3x^2 - 8x + 3y^2 + 4y + 3z^2 - 2z = 0\).

Step-by-step solution

Define the Points

Define \(P(x, y, z)\) as the point of interest, \(A(0, -1, 1)\) as point A, and \(B(1, 2, 0)\) as point B.

Formulate the Equation

According to the problem statement, the distance from P to A is twice the distance from P to B. Considering the Euclidean distance formula in 3D (i.e., for any two points \((x_1, y_1, z_1)\) and \((x_2, y_2, z_2)\), their distance d is given by \(\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2 - z_1)^2}\), we can formulate an equation according to the problem statement: \((x - 0)^2 + (y + 1)^2 + (z - 1)^2 = 2^2((x-1)^2+(y-2)^2+(z-0)^2)\)

Simplify the Equation

Simplify further to remove the square root and the brackets, leaving the expression in terms of squared distances: \(x^2 + (y + 1)^2 + (z - 1)^2 = 4 * ((x - 1)^2 + (y - 2)^2 + z^2)\)

Expand and Rearrange the Equation

Expand out the terms and consolidate like terms to obtain the final form of the quadratic surface equation: \(3x^2 - 8x + 3y^2 + 4y + 3z^2 - 2z = 0\)