Find the locus of points P(x, y, z) that are equidistant from the two parallel planes x + 2y - 2z = 1 and x + 2y - 2z = 7.
Answer and explanation
Correct answer: C. x + 2y - 2z = 4
- Step 1: The two planes have the same normal vector n = (1, 2, -2).
- Step 2: The distance from P to the first plane is |x + 2y - 2z - 1| / 3.
- Step 3: The distance to the second plane is |x + 2y - 2z - 7| / 3.
- Step 4: Equating distances: x + 2y - 2z - 1 = -(x + 2y - 2z - 7), so 2(x + 2y - 2z) = 8, which gives x + 2y - 2*z = 4.
Answer
x + 2y - 2z = 4.
Why not the others: x + 2y - 2z = 3 is offset towards the first plane; x + 2y - 2z = 8 lies outside both planes; 2x + 4y - 4*z = 5 simplifies to constant 2.5, which is not midway.
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