If the plane $2 x-y+2 z+3=0$ has the distances $\frac{1}{3}$ and $\frac{2}{3}$ units from the planes $4 x-2 y +4 z+\lambda=0$ and $2 x-y+2 z+\mu=0$, respectively, then the maximum value of $\lambda+\mu$ us equal to

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Three Dimensional Geometry

If the plane

2 x - y + 2 z + 3 = 0

has the distances

\frac{1}{3}

and

\frac{2}{3}

units from the planes

4 x - 2 y + 4 z + λ = 0

and

2 x - y + 2 z + μ = 0

, respectively, then the maximum value of

λ + μ

us equal to

(A)15
(B)13✓ Correct
(C)5
(D)9

Answer: (B)

Solution

The plane

2 x - y + 2 z + 3 = 0

has normal of length

3

. Doubling it to

4 x - 2 y + 4 z

, the first comparison plane is parallel, and its distance

\frac{∣ λ - 6∣}{16 + 4 + 16} = \frac{∣ λ - 6∣}{6} = \frac{1}{3}

gives

∣ λ - 6∣ = 2

, so

λ = 8

4

. For

2 x - y + 2 z + μ = 0

the distance

\frac{∣ μ - 3∣}{3} = \frac{2}{3}

gives

∣ μ - 3∣ = 2

, so

μ = 5

1

. The largest sum is

λ + μ = 8 + 5 = 13

JEE Main · 2019 · 10 Apr 2019 · Shift 2 · Single correct option

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