MHT CET202523 Apr 2025Morning ShiftMathematicsThree Dimensional GeometryActual
The distance of the point P (3,4,4) from the point of intersection of the line joining the points Q (3,-4,-5), R (2,-3,1) and the plane 2 x+y+z=7 is
Options
- A7 units
- B9 units
- C11 units
- D6 units
Correct answer
A. 7 units
Step-by-step solution
Find the distance from point P (3,4,4) to the intersection of line QR with plane 2x+y+z=7 . A point on line QR through Q (3,-4,-5) and R (2,-3,1) is parametrized as (3-t, -4+t, -5+6t) . Substitute into the plane equation: 2(3-t) + (-4+t) + (-5+6t) = 7 6 - 2t - 4 + t - 5 + 6t = 7 -3 + 5t = 7 5t = 10 t = 2 The intersection point is S (1,-2,7) when t=2 . The distance from P (3,4,4) to S (1,-2,7) is (1-3)^2 + (-2-4)^2 + (7-4)^2 = 4 + 36 + 9 = 49 = 7 Final answer: 7 units