Verified Solution • 2025–26 Session moderate

QJEEM25Maecf7: JEE Main 2025 Mathematics - Class 12 JEEM25Maecf7 Solution: Let P be the foot of the perpendicular from the point on the…

Let P be the foot of the perpendicular from the point Q(10,−3,−1)\mathrm{Q}(10,-3,-1) on the line x−37=y−2−1=z+1−2\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2} . Then the area of the right angled triangle PQRP Q R , where RR is the point (3,−2,1)(3,-2,1) , is

Multiple Choice Options

  • A 30\sqrt{30}
  • B 9159 \sqrt{15}
  • C 3303 \sqrt{30}
  • D 8158 \sqrt{15}

Step-by-Step Verified Solution

JEE Main 2025 (Online) 23rd January Morning Shift Mathematics - 3D Geometry Question 8 English Explanation

amp;amp;x−37=y−2−1=z+1−2=λamp;⇒7λ+3,−λ+2,−2λ−1amp; dr’s of QP ⇒amp;7λ−7,−λ+5,−2λamp; Now amp;amp;(7λ−7)⋅7−(−λ+5)+(2λ)⋅2=0amp;54λ−54=0⇒λ=1amp;∴P=(10,1,−3)amp;PQ→=−4j^+2k^amp;PR→=−7i^−3j^+4k^\begin{aligned}&\begin{aligned}& \frac{x-3}{7}=\frac{y-2}{-1}=\frac{z+1}{-2}=\lambda \\& \Rightarrow 7 \lambda+3,-\lambda+2,-2 \lambda-1 \\& \text { dr's of QP } \Rightarrow \\& 7 \lambda-7,-\lambda+5,-2 \lambda\end{aligned}\\&\text { Now }\\&\begin{aligned}& (7 \lambda-7) \cdot 7-(-\lambda+5)+(2 \lambda) \cdot 2=0 \\& 54 \lambda-54=0 \Rightarrow \lambda=1 \\& \therefore \mathrm{P}=(10,1,-3) \\& \overrightarrow{\mathrm{PQ}}=-4 \hat{\mathrm{j}}+2 \hat{\mathrm{k}} \\& \overrightarrow{\mathrm{PR}}=-7 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}\end{aligned}\end{aligned}

Area =|12|ijk0−42−7−34||=330

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