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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2025 · 23 January · Morning Shift · Mathematics · Three Dimensional Geometry

Let P be the foot of the perpendicular from the point Q(10,-3,-1)...

Let PP be the foot of the perpendicular from the point Q(10,−3,−1)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

Options

  1. A9159 \sqrt{15}
  2. B30\sqrt{30}
  3. C8158 \sqrt{15}
  4. D3303 \sqrt{30}

Correct answer

D. 3303 \sqrt{30}

Step-by-step solution

x−37=y−2−1=z+1−2=λ⇒7λ+3,−λ+2,−2λ−1dr’s of  QP ⇒7λ−7,−λ+5,−2λ\begin{aligned} & \frac{\mathrm{x}-3}{7}=\frac{\mathrm{y}-2}{-1}=\frac{\mathrm{z}+1}{-2}=\lambda \\ & \Rightarrow 7 \lambda+3,-\lambda+2,-2 \lambda-1 \\ & \text{dr's\text{ of } QP } \Rightarrow 7 \lambda-7,-\lambda+5,-2 \lambda \end{aligned} Now (7λ−7)⋅7−(−λ+5)+(2λ)⋅2=054λ−54=0⇒λ=1∴P=(10,1,−3)PQ→=−4j^+2k^PR→=−7i^−3j^+4k^Area =12∣ijk0−42−7−34∣=330\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}} \\ & \text{Area } =\frac{1}{2}\left|\begin{array}{ccc} \mathrm{i} & \mathrm{j} & \mathrm{k} \\ 0 & -4 & 2 \\ -7 & -3 & 4 \end{array}\right|=3 \sqrt{30} \end{aligned}
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