JEE MainMathematicsThree Dimensional Geometry
Let the lines L ₁: r = (2 i + j + k ) + ( i + 2 j - k ) and L ₂: r = (3 i + 3 j ) + (2 i - j + k ) intersect at the point R . Let P and Q be points lying on lines L ₁ and L ₂ , respectively, such that PR = 2 6 and PQ = 2 14 . If the z -coordinate of P is negative and Q has integer coordinates, then 7d² , where d is the perpendicular distance from the origin to the line passing through P and Q , is equal to
Options
- A10
- B546
- C378
- D346
Correct answer
D. 346
Step-by-step solution
Equating the position vectors of L ₁ and L ₂ : 2+ = 3+2 , 1+2 = 3- , and 1- = Solving these gives = 1 and = 0 . Thus, the point of intersection is R (3, 3, 0) . Let P be (2+ _P, 1+2 _P, 1- _P) . PR ^2 = ( _P-1)^2 + (2 _P-2)^2 + (- _P+1)^2 = 6( _P-1)^2 Given PR = 2 6 PR ^2 = 24 6( _P-1)^2 = 24 ( _P-1)^2 = 4 _P = 3 or -1 For P to have a negative z -coordinate, 1- _P 1 . Thus, _P = 3 . So, P is (5, 7, -2) . Let Q be (3+2 _Q, 3- _Q, _Q) . PQ ^2 = (2 _Q-2)^2 + (- _Q-4)^2 + ( _Q+2)^2 = 6 _Q^2 + 4 _Q + 24 Given PQ = 2 14