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JEE Main 202524 JanuaryEvening Shift
Exam
JEE Main
Year
2025
Date
24 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2025 · 24 January · Evening Shift · Mathematics · Three Dimensional Geometry

Let P be the image of the point Q(7,-2,5) in the line L:...

Let P be the image of the point Q(7,−2,5)\mathrm{Q}(7,-2,5) in the line L:x−12=y+13=z4\mathrm{L}: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4} and R(5,p,q)\mathrm{R}(5, \mathrm{p}, \mathrm{q}) be a point on LL. Then the square of the area of △PQR\triangle P Q R is ________.

Correct answer

0

Step-by-step solution

 Let R(2λ+1,3λ−1,4λ)2λ+1=5λ=2R(5,5,8) let T(2λ+1,3λ−1,4λ)QT→=(2λ−6)i^+(3λ+1)j^+(4λ−5)k^b→=2i^+3j^+4k^QT→⋅b→=04λ−12+9λ+3+16λ−20=0\begin{aligned} & \text{ Let } \mathrm{R}(2 \lambda+1,3 \lambda-1,4 \lambda) \\ & 2 \lambda+1=5 \\ & \lambda=2 \\ & \mathrm{R}(5,5,8) \\ & \text{ let } \mathrm{T}(2 \lambda+1,3 \lambda-1,4 \lambda) \\ & \overrightarrow{\mathrm{QT}}=(2 \lambda-6) \hat{\mathrm{i}}+(3 \lambda+1) \hat{\mathrm{j}}+(4 \lambda-5) \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{QT}} \cdot \overrightarrow{\mathrm{b}}=0 \\ & 4 \lambda-12+9 \lambda+3+16 \lambda-20=0\end{aligned}λ=1T(3,2,4)QT=33RT=29( area of  ΔPQR)2=(1229⋅233)2=957\begin{aligned} & \lambda=1 \\ & \mathrm{ T}(3,2,4) \\ & \mathrm{QT}=\sqrt{33} \quad \mathrm{RT}=\sqrt{29} \\ & (\text{ area\text{ of } } \Delta \mathrm{PQR})^2=\left(\frac{1}{2} \sqrt{29} \cdot 2 \sqrt{33}\right)^2 \\ & =957\end{aligned}$
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