The equation of the planes parallel to the plane x - 2y + 2z - 3 = 0 which are at unit distance from the point (1, 2, 3) is ax + by + cz + d = 0. If (b - d) = k(c - a), then the positive value of k is :

Solution

Maths sir · 4 steps

A parallel plane keeps the same normal, so only the constant changes. Find it from the distance condition.

1. Parallel plane. Any plane parallel to x - 2y + 2z - 3 = 0 has the same normal (1, -2, 2). So it can be written as x - 2y + 2z + d = 0, with a = 1, b = -2, c = 2.

   (|1 - 2(2) + 2(3) + d|)/√(1² + (-2)² + 2²) = 1

2. Solve for d. The numerator is |3 + d| and the denominator is √9 = 3. So |3 + d| = 3, which gives d = 0 or d = -6. There are two planes, one on each side of the point.
3. Find k. Here c - a = 2 - 1 = 1. For d = 0, b - d = -2, so k = -2. For d = -6, b - d = -2 + 6 = 4, so k = 4. Scaling the plane equation scales both sides equally, so k does not change.
4. So. The positive value is k = 4.

The common slip: 2. Forgetting to divide by √(1 + 4 + 4) = 3 gives |3 + d| = 1, so d = -4 and k = 2.

The idea to keep. For a plane parallel to a given one, copy the normal and keep only the constant unknown. A fixed distance from a point then gives two planes, one on each side.

Answer from the NTA final answer key. Solution by PrepWiser.