JEE Advanced · 2014 · 25 May 2014 · Paper 1 · Three Dimensional Geometry

From a point

P (λ, λ, λ)

, perpendiculars

P Q

and

P R

are drawn respectively on the lines

y = x, z = 1

and

y = - x, z = - 1

. If

P

is such that

∠ QP R

is a right angle, then the possible value(s) of

λ

is(are)

Answer: (C)

Solution

The line

y = x, z = 1

can be written as

\frac{x - 0}{1} = \frac{y - 0}{1} = \frac{z - 1}{0}

. The foot of perpendicular from

P (λ, λ, λ)

to this line is

Q (λ, λ, 1)

. Similarly, the line

y = - x, z = - 1

\frac{x - 0}{1} = \frac{y - 0}{- 1} = \frac{z + 1}{0}

, and the foot of perpendicular from

P

R (0, 0, - 1)

. For

∠ QP R = 9 0^{\circ}

, we need

P Q \cdot P R = 0

. We have

P Q = (0, 0, 1 - λ)

and

P R = (- λ, - λ, - 1 - λ)

. Thus

0 + 0 + (1 - λ) (- 1 - λ) = 0

, giving

(λ - 1) (λ + 1) = 0

, so

λ = \pm 1

. At

λ = 1

, point

P (1, 1, 1)

lies on the first line, so

λ = - 1

is the answer.

JEE Advanced · 2014 · 25 May 2014 · Paper 1 · One or more correct options