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JEE Main 202522 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
22 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Three Dimensional Geometry

JEE Main 2025 · 22 January · Morning Shift · Mathematics · Three Dimensional Geometry

Let L_1: (x-1)/(3)=(y-1)/(-1)=(z+1)/(0) and L_2:...

Let L1:x−13=y−1−1=z+10\mathrm{L}_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0} and L2:x−22=y0=z+4α,α∈R\mathrm{L}_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in \mathbf{R}, be two lines, which intersect at the point BB. If PP is the foot of perpendicular from the point A(1,1,−1)A(1,1,-1) on L2L_2, then the value of 26α( PB)226 \alpha(\mathrm{~PB})^2 is _________

Correct answer

0

Step-by-step solution

Point BB(3λ+1,−λ+1,−1)≡(2μ+2,0,αμ−4)3λ+1=2μ+2−λ+1=0−1=αμ−4λ=1,μ=1,α=3B(4,0,−1)\begin{aligned} & (3 \lambda+1,-\lambda+1,-1) \equiv(2 \mu+2,0, \alpha \mu-4) \\ & 3 \lambda+1=2 \mu+2 \\ & -\lambda+1=0 \\ & -1=\alpha \mu-4 \\ & \lambda=1, \mu=1, \alpha=3 \\ & B(4,0,-1) \end{aligned} Let Point ’ P ’ is (2δ+2,0,3δ−4) Dr’s of  AP \text{ Let Point ' } P \text{ ' is }(2 \delta+2,0,3 \delta-4) \text{ Dr's\text{ of } AP }< 2 \delta+1,-1,3 \delta-3> A P \perp L_2 \Rightarrow \delta=\frac{7}{13} P\left(\frac{40}{13}, 0, \frac{-31}{13}\right) \therefore 26 \alpha(P B)^2=26 \times 3 \times\left(\frac{144}{169}+\frac{324}{169}\right) =216
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