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JEE Main 20257 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
7 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2025 · 7 April · Evening Shift · Mathematics · Three Dimensional Geometry

Consider the lines L_1: x-1=y-2=z and L_2: x-2= y=z -1 . Let the...

Consider the lines L1:x−1=y−2=z\mathrm{L}_1: \mathrm{x}-1=\mathrm{y}-2=\mathrm{z} and L2:x−2=\mathrm{L}_2: \mathrm{x}-2=\mathrm{y}=\mathrm{z}−1-1. Let the feet of the perpendiculars from the point P(5,1,−3)\mathrm{P}(5,1,-3) on the lines L1\mathrm{L}_1 and L2\mathrm{L}_2 be QQ and RR respectively. If the area of the triangle PQR is A , then 4 A24 \mathrm{~A}^2 is equal to :

Options

  1. A139
  2. B147
  3. C151
  4. D143

Correct answer

B. 147

Step-by-step solution

L1:x−11=y−21=z−02 Let Q(λ+1,λ+2,λ)PQ→=(λ−4,λ−1,λ+3)PQ→⋅m→=0\begin{aligned} & \mathrm{L}_1: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-0}{2} \\ & \text{ Let } \mathrm{Q}(\lambda+1, \lambda+2, \lambda) \\ & \overrightarrow{\mathrm{PQ}}=(\lambda-4, \lambda-1, \lambda+3) \\ & \overrightarrow{\mathrm{PQ}} \cdot \overrightarrow{\mathrm{m}}=0\end{aligned}⇒λ−4+λ+1,λ+3=0⇒3λ=0λ=0\begin{aligned} \Rightarrow & \lambda-4+\lambda+1, \lambda+3=0 \\ \Rightarrow & 3 \lambda=0 \\ & \lambda=0\end{aligned} L2:x−21=y−01=z−12 Let R(μ+2,μ,μ+1)PR→=(μ−3,μ−1,μ+4)PR→⋅n→=0μ−3+μ−1+μ+4=0≠μ=0\begin{aligned} & L_2: \frac{x-2}{1}=\frac{y-0}{1}=\frac{z-1}{2} \\ & \text{ Let } \mathrm{R}(\mu+2, \mu, \mu+1) \overrightarrow{\mathrm{PR}}=(\mu-3, \mu-1, \mu+4) \\ & \overrightarrow{\mathrm{PR}} \cdot \overrightarrow{\mathrm{n}}=0 \\ & \mu-3+\mu-1+\mu+4=0 \\ & \neq \mu=0\end{aligned}  Area of  △PQR(A)=12∣PQ→×PR→∣A=12∣(−4i^+j^+3k^)×(−3i^+j^+4k^)∣A=12∣7(i^+j^+k^)∣∣i^j^k^−413−3−14∣=7i^+7j^+7k^4A2=49×3=147\begin{aligned} & \text{ Area\text{ of } } \triangle \mathrm{PQR}(\mathrm{A})=\frac{1}{2}|\overrightarrow{\mathrm{PQ}} \times \overrightarrow{\mathrm{PR}}| \\ & \mathrm{A}=\frac{1}{2}|(-4 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}) \times(-3 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}})| \\ & \mathrm{A}=\frac{1}{2}|^{7}(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})| \\ & \left|\begin{array}{crr}\hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -4 & 1 & 3 \\ -3 & -1 & 4\end{array}\right| \\ & =7 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}+7 \hat{\mathrm{k}} \\ & 4 \mathrm{ A}^2=49 \times 3=147\end{aligned}
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