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MHT CET202525 Apr 2025Evening ShiftMathematicsThree Dimensional GeometryActual

If the lines x=a y-1=z-2 and x=3 y-2=b z-2(a b 0) are coplanar, then

Options

  1. Aa=1, ~b = 1 2
  2. Ba=2, ~b =2
  3. Ca= 1 2 , ~b = 1 2
  4. Db =1, a R - 0

Correct answer

D. b =1, a R - 0

Step-by-step solution

The lines are coplanar if the scalar triple product of the vectors P₁P₂ , d₁ , and d₂ vanishes. Line 1 passes through P₁(0, 1/a, 2) with direction vector d₁ = (1, 1/a, 1) , and line 2 passes through P₂(0, 2/3, 2/b) with direction vector d₂ = (1, 1/3, 1/b) . The condition becomes: vmatrix 0 & 2a-3 3a & 2-2b b 1 & 1/a & 1 1 & 1/3 & 1/b vmatrix = 0 Expanding along the first row yields: - 2a-3 3a ( 1 b -1 ) + 2(1-b) b ( 1 3 - 1 a ) = 0 Multiplying through by 3ab gives: -(2a-3)(1-b) + 2(1-b)(a-3) = 0 Factoring out (1-b)

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