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JEE Main 202522 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2025 · 22 January · Evening Shift · Mathematics · Three Dimensional Geometry

Let the distance between two parallel lines be 5 units and a point...

Let the distance between two parallel lines be 5 units and a point PP lie between the lines at a unit distance from one of them. An equilateral triangle PQRP Q R is formed such that QQ lies on one of the parallel lines, while RR lies on the other. Then (QR)2(Q R)^2 is equal to _______ -.Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one

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Step-by-step solution

Let the distance between two parallel lines be 5 units and a point P lie between the lines at a unit distance from one
PR=cosec⁡θ,PQ=4sec⁡(30+θ)P R=\operatorname{cosec} \theta, P Q=4 \sec (30+\theta) For equilateral d=PR=PQ⇒cos⁡(θ+30∘)=4sin⁡θ⇒32cos⁡θ−12sin⁡θ=4sin⁡θ⇒tan⁡θ=133QR2=d2=cosec⁡2θ=28\begin{aligned} & \mathrm{d}=\mathrm{PR}=\mathrm{PQ} \\ & \Rightarrow \cos \left(\theta+30^{\circ}\right)=4 \sin \theta \\ & \Rightarrow \frac{\sqrt{3}}{2} \cos \theta-\frac{1}{2} \sin \theta=4 \sin \theta \\ & \Rightarrow \tan \theta=\frac{1}{3 \sqrt{3}} \\ & \mathrm{QR}^2=\mathrm{d}^2=\operatorname{cosec}^2 \theta=28 \end{aligned}
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