MHT CET202523 Apr 2025Morning ShiftMathematicsThree Dimensional GeometryActual
The equation of the line passing through the point of intersection of x-1 2 = y-2 3 = z-3 4 and x-4 5 = y-1 2 =z and also through the point (2,1,-2) is
Options
- Ar =(- i - j - k )+ ( i +2 j + k )
- Br =(- i - j + k )+ (2 i +2 j + k )
- Cx+1 3 = y+1 2 = z+1 -1
- Dx-1 3 = y-1 2 = z+1 1
Correct answer
C. x+1 3 = y+1 2 = z+1 -1
Step-by-step solution
Find the intersection point of lines L₁ and L₂ . A general point on L₁ is P₁(2k₁+1, 3k₁+2, 4k₁+3) . A general point on L₂ is P₂(5k₂+4, 2k₂+1, k₂) . Equating coordinates for intersection: 2k₁ - 5k₂ = 3 3k₁ - 2k₂ = -1 4k₁ + 3 = k₂ Substitute k₂ = 4k₁ + 3 into the second equation: 3k₁ - 2(4k₁ + 3) = -1 3k₁ - 8k₁ - 6 = -1 -5k₁ = 5 k₁ = -1 Then k₂ = 4(-1) + 3 = -1 The intersection point is P_ int (-1, -1, -1) Find the line through P_ int and point A(2, 1, -2) . The direction vector is A - P_ int = 3 i + 2 j - k The line