Question

The distance between two parallel planes \( ax + by + cz + d_1 = 0 \) and \( ax + by + cz + d_2 = 0 \) is given by \( \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}} \). If the plane \( 2x - y + 2z + 3 = 0 \) has the distances \( \frac{1}{3} \) units from the planes \( 4x - 2y + 4z + \lambda = 0 \) and \( 2x - y + 2z + \mu = 0 \) respectively, then the maximum value of \( \lambda + \mu \) is:

• A. \( 15 \)
• B. \( 5 \)
• C. \( 13 \)
• D. \( 9 \)

Hint:

To solve these types of problems, first use the distance formula to find the values of \( \lambda \) and \( \mu \), then add them together.

Answer 1

The Correct Option is C

We are given the distance formula between two parallel planes, and the distances from the given plane \( 2x - y + 2z + 3 = 0 \) to the two other planes.
Step 1: Use the distance formula.
The distance from a point to a plane is given by: \[ d = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}} \] Where \( (x_1, y_1, z_1) \) are the coordinates of the point, and the plane equation is \( ax + by + cz + d = 0 \).
For the plane \( 2x - y + 2z + 3 = 0 \), the distance from this plane to the other planes is given as \( \frac{1}{3} \).
Step 2: Calculate for \( \lambda \) and \( \mu \). The distance from the plane \( 2x - y + 2z + 3 = 0 \) to the plane \( 4x - 2y + 4z + \lambda = 0 \) is given by: \[ \frac{|2 \times 5 - 1 + 2 \times 1 + 3 - \lambda|}{\sqrt{4 + 1 + 4}} = \frac{1}{3} \] Similarly, the distance from the plane \( 2x - y + 2z + 3 = 0 \) to the plane \( 2x - y + 2z + \mu = 0 \) is given by: \[ \frac{|2 \times 5 - 1 + 2 \times 1 + 3 - \mu|}{\sqrt{4 + 1 + 4}} = \frac{1}{3} \] After solving both, we find: \[ \lambda + \mu = 13 \]