JEE Main202226 Jul 2022Morning ShiftMathematicsThree Dimensional GeometryActual
Let Q and R be two points on the line x + 1 2 = y + 2 3 = z - 1 2 at a distance 26 from the point P 4 , 2 , 7 . Then the square of the area of the triangle P Q R is ________.
Correct answer
0
Step-by-step solution
Given, L : x + 1 2 = y + 2 3 = 2 - 1 2 Plotting the diagram of given value's in question we have, Let T be any point on line whose coordinates are T 2 t - 1 , 3 t - 2 , 2 t + 1 Now P T ⊥ Q R by diagram, So, 2 2 t - 5 + 3 3 t - 4 + 2 2 t - 6 = 0 ⇒ 17 t = 34 ⇒ t = 2 , so T 3 , 4 , 5 Now the value of P T = 1 + 4 + 4 = 3 And Q T = 26 - 9 = 17 by using pythagorus theorem ∴ Area of ∆ P Q R = 1 2 × 2 17 × 3 = 3 17 ∴ Square of ar ∆ P Q R = 153 .