JEE MainMathematicsThree Dimensional Geometry
Let the line L be the intersection of the planes x - 2y = 1 and x + z = 4 . Let Q be the image of the point P(2, -1, -1) in the line L . A line L₂ passing through Q is perpendicular to the plane x + 2y + 2z = 10 and intersects the xy -plane at the point S . The value of 2QS is equal to
Options
- A9
- B24
- C3
- D18
Correct answer
A. 9
Step-by-step solution
The direction vector of the line L is given by the cross product of the normal vectors of the two planes: d = ( i - 2 j ) ( i + k ) = -2 i - j + 2 k To find a point on L , we can set y = 0 . From x - 2y = 1 , we get x = 1 . From x + z = 4 , we get z = 3 . Thus, (1, 0, 3) lies on L . The symmetric equation of L is x-1 2 = y 1 = z-3 -2 = t . Let M(2t+1, t, -2t+3) be the foot of the perpendicular from P(2, -1, -1) to L . The direction ratios of PM are (2t-1, t+1, -2t+4) . Since PM is perpendicular to L , their dot pro