JEE Advanced · 2015 · 24 May 2015 · Paper 1 · Three Dimensional Geometry

R^{3}

, consider the planes

P_{1} : y = 0

and

P_{2} : x + z = 1

. Let

P_{3}

be a plane, different from

P_{1}

and

P_{2}

, which passes through the intersection of

P_{1}

and

P_{2}

. If the distance of the point

(0, 1, 0)

from

P_{3}

is 1 and the distance of a point

(α, β, γ)

from

P_{3}

is 2, then which of the following relations is (are) true?

Answer: (B), (D)

Solution

The line of intersection of

P_{1} : y = 0

and

P_{2} : x + z = 1

{(1 - z, 0, z) : z \in R}

.

A plane through this line has the form

P_{3} : y = 0 + λ (x + z - 1)

for some

λ

, i.e.,

λ x + y + λ z - λ = 0

.

Distance from

(0, 1, 0)

P_{3}

\frac{∣ λ ( 0 ) + 1 + λ ( 0 ) - λ ∣}{λ ^{2} + 1 + λ ^{2}} = \frac{∣1 - λ ∣}{2 λ ^{2} + 1} = 1

∣1 - λ ∣^{2} = 2 λ^{2} + 1

1 - 2 λ + λ^{2} = 2 λ^{2} + 1

- λ^{2} - 2 λ = 0

λ (λ + 2) = 0

λ = 0

(gives

P_{1}

, excluded) or

λ = - 2

.

With

λ = - 2

P_{3} : - 2 x + y - 2 z + 2 = 0

, or

2 x - y + 2 z - 2 = 0

.

Distance from

(α, β, γ)

P_{3}

is 2:

\frac{∣2 α - β + 2 γ - 2∣}{4 + 1 + 4} = \frac{∣2 α - β + 2 γ - 2∣}{3} = 2

∣2 α - β + 2 γ - 2∣ = 6

This gives:

2 α - β + 2 γ - 2 = 6 or 2 α - β + 2 γ - 2 = - 6

2 α - β + 2 γ - 8 = 0 or 2 α - β + 2 γ + 4 = 0

These correspond to options (B) and (D).

JEE Advanced · 2015 · 24 May 2015 · Paper 1 · One or more correct options