MHT CET202519 Apr 2025Morning ShiftMathematicsThree Dimensional GeometryActual
If the points A (2-x, 2,2), B (2,2-y, 2), C (2,2,2-z) and D (1,1,1) are coplanar, then the locus of point P (x, y, z) is
Options
- A1 x + 1 y + 1 z =1
- B1 x + 1 y + 1 z =0
- C1 1+x + 1 1+y + 1 1+z =1
- D1 x + 1 2 y + 1 3 z =0
Correct answer
A. 1 x + 1 y + 1 z =1
Step-by-step solution
Given points A (2-x, 2,2) , B (2,2-y, 2) , C (2,2,2-z) , and D (1,1,1) , the vectors from point D are: DA = (1-x, 1, 1) DB = (1, 1-y, 1) DC = (1, 1, 1-z) The scalar triple product must vanish for coplanarity: vmatrix 1-x & 1 & 1 1 & 1-y & 1 1 & 1 & 1-z vmatrix = 0 Expanding the determinant yields: (1-x)[(1-y)(1-z) - 1] - 1[1 (1-z)-1 1] + 1[1 1-1 (1-y)] =(1-x)(-y-z+yz) + z + y Simplifying gives xy + xz + yz - xyz = 0 . Dividing by xyz (assuming nonzero parameters) produces: 1 x + 1 y + 1 z = 1 The locus corresponds