JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Three Dimensional Geometry

The equation of the planes parallel to the plane

x - 2 y + 2 z - 3 = 0

which are at unit distance from the point

(1, 2, 3)

a x + b y + cz + d = 0

. If

(b - d) = K (c - a)

, then the positive value of

K

is ____

Answer: 4

Solution

A plane parallel to

x - 2 y + 2 z - 3 = 0

shares its normal, so it has the form

x - 2 y + 2 z + d = 0

. Its distance from

(1, 2, 3)

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

∣3 + d ∣ = 3

and

d = 0

d = - 6

. Here

a = 1

b = - 2

and

c = 2

, and the relation

b - d = K (c - a)

gives

K = \frac{- 2 - 0}{2 - 1} = - 2

for

d = 0

and

K = \frac{- 2 + 6}{2 - 1} = 4

for

d = - 6

. The positive value of

K

4

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Integer answer

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