Vectors & 3D GeometrymediumPYQ · JEE Main · 4 Apr 2026 · Shift 1 (Morning)Free

Line Intersecting Two Lines: 225α² = 1014 | JEE Main 2026

JEE Maths question with a full step-by-step solution.

Question
A line with direction ratios 1,−1,21,-1,2 intersects the lines x2=y3=z+13\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z+1}{3} and x+1−1=y−21=z4\dfrac{x+1}{-1}=\dfrac{y-2}{1}=\dfrac{z}{4} at the points PP and QQ respectively. If the length of the line segment PQPQ is α\alpha, then 225α2225\alpha^2 is equal to
A10241024
B10141014correct
C11041104
D12041204
Solution
Step 1: Let P=(2λ,3λ,3λ−1)P=(2\lambda,3\lambda,3\lambda-1) on the first line and Q=(−μ−1,μ+2,4μ)Q=(-\mu-1,\mu+2,4\mu) on the second. Step 2: Direction ratios of PQPQ are (2λ+μ+1, 3λ−μ−2, 3λ−4μ−1)(2\lambda+\mu+1,\ 3\lambda-\mu-2,\ 3\lambda-4\mu-1), proportional to (1,−1,2)(1,-1,2):
2λ+μ+11=3λ−μ−2−1=3λ−4μ−12.\frac{2\lambda+\mu+1}{1}=\frac{3\lambda-\mu-2}{-1}=\frac{3\lambda-4\mu-1}{2}.
 Step 3: Solving gives λ=15\lambda=\dfrac15, μ=−815\mu=-\dfrac{8}{15}, so
P(25,35,−25),Q(−715,2215,−3215).P\left(\frac25,\frac35,-\frac25\right),\qquad Q\left(-\frac{7}{15},\frac{22}{15},-\frac{32}{15}\right).
 Step 4:
PQ2=(1315)2+(1315)2+(2615)2=169+169+676225=1014225.PQ^2=\left(\frac{13}{15}\right)^2+\left(\frac{13}{15}\right)^2+\left(\frac{26}{15}\right)^2=\frac{169+169+676}{225}=\frac{1014}{225}.
 So α2=1014225\alpha^2=\dfrac{1014}{225} and 225α2=1014225\alpha^2=1014. Correct answer: (2)
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