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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

The perpendicular distance, of the line...

The perpendicular distance, of the line x−12=y+2−1=z+32\frac{x-1}{2}=\frac{y+2}{-1}=\frac{z+3}{2} from the point P(2,−10,1)\mathrm{P}(2,-10,1), is :

Options

  1. A6
  2. B525 \sqrt{2}
  3. C434 \sqrt{3}
  4. D353 \sqrt{5}

Correct answer

D. 353 \sqrt{5}

Step-by-step solution

x−12=y+2−1=z+32=λ (let )(2λ+1,−λ−2,2λ−3)∵PA→⋅n→=0⇒(2λ−1)2+(−λ+8)(−1)+(2λ−4)2=0⇒4λ−2+λ−8+4λ−8=0⇒9λ−18=0⇒λ=2∴A(5,−4,1)∴AP=32+62+02=45=35\begin{aligned} & \left.\frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+2}{-1}=\frac{\mathrm{z}+3}{2}=\lambda \text{ (let }\right) \\ & (2 \lambda+1,-\lambda-2,2 \lambda-3) \\ & \because \overrightarrow{\mathrm{PA}} \cdot \overrightarrow{\mathrm{n}}=0 \\ & \Rightarrow(2 \lambda-1) 2+(-\lambda+8)(-1)+(2 \lambda-4) 2=0 \\ & \Rightarrow 4 \lambda-2+\lambda-8+4 \lambda-8=0 \\ & \Rightarrow 9 \lambda-18=0 \Rightarrow \lambda=2 \\ & \therefore \mathrm{ A}(5,-4,1) \\ & \therefore \mathrm{AP}=\sqrt{3^2+6^2+0^2}=\sqrt{45}=3 \sqrt{5}\end{aligned}
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