JEE MainMathematicsThree Dimensional Geometry
Let the lines L ₁: x-1 2 = y-2 1 = z-3 2 and L ₂: x-2 3 = y-2 2 = z- 1 , R intersect at the point R . Let P and Q be the points lying on lines L ₁ and L ₂ , respectively, such that PR = 3 and ( PQ )² = 13 7 . If the x -coordinate of the point P is strictly greater than 5 , and S is the sum of the coordinates of the point Q , then the value of 7S is equal to
Options
- A82
- B154
- C142
- D112
Correct answer
C. 142
Step-by-step solution
Any point on L ₁ is given by (2 +1, +2, 2 +3) and on L ₂ is given by (3 +2, 2 +2, + ) . For intersection point R , we equate the coordinates: 2 +1 = 3 +2 2 - 3 = 1 +2 = 2 +2 = 2 Solving these gives = 1 and = 2 . Thus, R is (5, 4, 7) . Let P be (2 _P+1, _P+2, 2 _P+3) . PR ^2 = (2 _P-4)^2 + ( _P-2)^2 + (2 _P-4)^2 = 9( _P-2)^2 Given PR = 3 PR ^2 = 9 9( _P-2)^2 = 9 ( _P-2)^2 = 1 _P = 3 or 1 Since the x -coordinate of P is strictly greater than 5 , 2 _P+1 > 5 _P > 2 . Thus, _P = 3 . So, P is (7, 5, 9) . Let Q be a point