JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Three Dimensional Geometry

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Let Q and R be the feet of perpendiculars from the point P(a, a, a) on the lines x=y, z=1 and x=-y, z=-1 respectively. If ∠QPR is a right angle, then

12 a^{2}

is equal to ____

Answer: 12

Solution

Line 1 (

x = y, z = 1

): points

(r, r, 1)

. Line 2 (

x = - y, z = - 1

): points

(k, - k, - 1)

Q = (r, r, 1)

is the foot of the perpendicular from

P (a, a, a)

P Q \cdot (1, 1, 0) = 0 ⟹ (a - r) + (a - r) = 0 ⟹ r = a

, so

Q = (a, a, 1)

R = (k, - k, - 1)

P R \cdot (1, - 1, 0) = 0 ⟹ (a - k) - (a + k) = 0 ⟹ k = 0

, so

R = (0, 0, - 1)

P Q = (0, 0, 1 - a)

P R = (- a, - a, - (1 + a))

. For

∠ QP R = 9 0^{\circ}

P Q \cdot P R = 0 + 0 + (1 - a) (- (1 + a)) = - (1 - a^{2}) = a^{2} - 1 = 0 ⟹ a^{2} = 1

12 a^{2} = 12

JEE Main · 2024 · 31 Jan 2024 · Shift 1 · Integer answer

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