JEE MainMathematicsThree Dimensional Geometry
Let P' be the image of the point P(2, -1, 1) in the plane x + 2y + 2z = 11 . A line L passes through P' and has direction ratios proportional to (2, 2, 1) . Let Q be a point on the line L such that it is at a distance of 9 units from P' and has a negative x -coordinate. The perpendicular distance of the point Q from the plane 2x - y + 2z = 0 is :
Options
- A1 3
- B9
- C7
- D1
Correct answer
D. 1
Step-by-step solution
First, we find the image P' of the point P(2, -1, 1) in the plane x + 2y + 2z - 11 = 0 . The coordinates (x, y, z) of the image P' are given by: x - 2 1 = y + 1 2 = z - 1 2 = -2 (1(2) + 2(-1) + 2(1) - 11) 1^2 + 2^2 + 2^2 x - 2 1 = y + 1 2 = z - 1 2 = -2 (2 - 2 + 2 - 11) 9 x - 2 1 = y + 1 2 = z - 1 2 = -2 ( -9 9 ) = 2 Solving for x, y, z : x = 2 + 2 = 4 y = -1 + 4 = 3 z = 1 + 4 = 5 Thus, P' (4, 3, 5) . The line L passes through P'(4, 3, 5) and has direction ratios (2, 2, 1) . The magnitude of the direction ratios is