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JEE Main 20246 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
6 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2024 · 6 April · Evening Shift · Mathematics · Three Dimensional Geometry

Let P(α, β, γ) be the image of the point Q(3,-3,1) in the line...

Let P(α,β,γ)\mathrm{P}(\alpha, \beta, \gamma) be the image of the point Q(3,−3,1)\mathrm{Q}(3,-3,1) in the line x−01=y−31=z−1−1\frac{x-0}{1}=\frac{y-3}{1}=\frac{z-1}{-1} and R\mathrm{R} be the point (2,5,−1)(2,5,-1). If the area of the triangle PQRP Q R is λ\lambda and λ2=14K\lambda^2=14 K, then KK is equal to :

Options

  1. A36
  2. B81
  3. C72
  4. D18

Correct answer

B. 81

Step-by-step solution

RQ=1+64+4=69RQ→=ℓ^−8j^+2k^RS→=ℓ^+j^−k^cos⁡θ=RQ→⋅RS→∣RQ→∣∣RS→∣=∣1−8−2693∣=9323cos⁡θ=323=RSRQ=RS69RS=33sin⁡θ=1423=QS69\begin{aligned} & \mathrm{RQ}=\sqrt{1+64+4}=\sqrt{69} \\ & \overrightarrow{\mathrm{RQ}}=\hat{\ell}-8 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{RS}}=\hat{\ell}+\hat{\mathrm{j}}-\hat{\mathrm{k}} \\ & \cos \theta=\frac{\overrightarrow{\mathrm{RQ}} \cdot \overrightarrow{\mathrm{RS}}}{|\overrightarrow{\mathrm{RQ}}||\overrightarrow{\mathrm{RS}}|}=\left|\frac{1-8-2}{\sqrt{69} \sqrt{3}}\right|=\frac{9}{3 \sqrt{23}} \\ & \cos \theta=\frac{3}{\sqrt{23}}=\frac{\mathrm{RS}}{\mathrm{RQ}}=\frac{\mathrm{RS}}{\sqrt{69}} \\ & \mathrm{RS}=3 \sqrt{3} \\ & \sin \theta=\frac{\sqrt{14}}{\sqrt{23}}=\frac{\mathrm{QS}}{\sqrt{69}}\end{aligned}QS=42 area =12⋅2QS⋅RS=42⋅33λ=914λ2=81.14=14kk=81\begin{aligned} & \mathrm{QS}=\sqrt{42} \\ & \text{ area }=\frac{1}{2} \cdot 2 \mathrm{QS} \cdot \mathrm{RS}=\sqrt{42} \cdot 3 \sqrt{3} \\ & \lambda=9 \sqrt{14} \\ & \lambda^2=81.14=14 \mathrm{k} \\ & \mathrm{k}=81\end{aligned}$
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