JEE MainMathematicsThree Dimensional Geometry
Let a variable point P move on the line L ₃: x 1 = y -1 = z+4 1 . Let Q and R be the feet of the perpendiculars drawn from P to the lines L ₁: x-1 1 = y 1 = z -1 and L ₂: x 1 = y-1 1 = z 1 respectively. If the minimum value of PQ ^2 + PR ^2 is M , then the value of 3 M is equal to :
Options
- A68
- B156
- C52
- D17
Correct answer
C. 52
Step-by-step solution
Let the coordinates of the variable point P on L ₃ be (t, -t, t-4) . For line L ₁ , a point on it is A (1,0,0) and its direction vector is v ₁ = i + j - k . The square of the perpendicular distance PQ ^2 is given by | AP v ₁|^2 | v ₁|^2 . AP = (t-1) i - t j + (t-4) k . AP v ₁ = vmatrix i & j & k t-1 & -t & t-4 1 & 1 & -1 vmatrix = 4 i + (2t-5) j + (2t-1) k . PQ ^2 = 16 + (2t-5)^2 + (2t-1)^2 3 = 8t^2 - 24t + 42 3 . For line L ₂ , a point on it is B (0,1,0) and its direction vector is v ₂ = i + j + k . The square of