Question

If the plane 2x – y + 2z + 3 = 0 has the distances 1/3 and
2/3 units from the planes 4x – 2y + 4z + λ = 0 and 2x – y + 2z + μ = 0, respectively, then the maximum value of (λ + μ) us equal to (A) 15 (B) 13
(C) 5 (D) 9

Answer 1

The maximum value of (λ + μ) is equal to 13.

Given a plane P, 2x – y + 2z + 3 = 0, which has distances 1/3 and 2/3 from the 2 planes

• P1, 4x – 2y + 4z + λ = 0 and Plane P2 2x – y + 2z + μ = 0

Plane 1 can be written as 2x - y +2z + λ/2 = 0

• Distance between two parallel planes

       ax+by+cz+d1=0  and ax+by+cz+d2=0 is

•      d = $\frac{|d1 - d2|}{\sqrt{a^{2} + b^{2} + c^{2} } }$
• 1/3 = | λ/2 - 3|/$\sqrt{2^{2} + -1^{2} + 2^{2} }$  = | λ/2 - 3|/$\sqrt{9}$ = |λ/2 - 3|/3
• |λ/2 - 3| = 1

• λ/2 = ±1 +3 ==>

• λ = 4 or 8.

SImilarly,

• 2/3 = |μ - 3|/3 ==>

• μ = ±2 + 3 = 1 or 5

Therefore maximum value of  λ  + μ is when λ = 8 and  μ = 5, ie,   13