Let P be the foot of the perpendicular from the point (1,2,2) on the line L:1x−1=−1y+1=2z−2. Let the line r=(−i^+j^−2k^)+λ(i^−j^+k^),λ∈R, intersect the line L at Q . Then 2(PQ)2 is equal to:
(A)27✓ Correct
(B)25
(C)29
(D)19
Answer: (A)
Solution
Foot of perpendicular P from A(1,2,2) on L:1x−1=−1y+1=2z−2=μ.
General point on L: (μ+1,−μ−1,2μ+2). Then AP=(μ,−μ−3,2μ), and perpendicularity with direction (1,−1,2) gives
μ+(μ+3)+4μ=0⇒6μ=−3⇒μ=−21
so P=(21,−21,1).
For Q, equate the general point on L with the general point on r=(−1,1,−2)+λ(1,−1,1), i.e. (λ−1,1−λ,λ−2):
μ+1=λ−1,−μ−1=1−λ,2μ+2=λ−2
The first two give λ=μ+2; substituting into the third: 2μ+2=μ⇒μ=−2,λ=0, so Q=(−1,1,−2).
PQ2=(23)2+(23)2+32=49+49+9=227
2(PQ)2=27
JEE Main · 2025 · 29 Jan 2025 · Shift 2 · Single correct option
