JEE Main202126 Aug 2021Evening ShiftMathematicsThree Dimensional GeometryActual
Let Q be the foot of the perpendicular from the point P ( 7 , - 2 , 13 ) on the plane containing the lines x + 1 6 = y - 1 7 = z - 3 8 and x - 1 3 = y - 2 5 = z - 3 7 Then ( P Q ) 2 , is equal to ______.
Correct answer
0
Step-by-step solution
Equation of the plane A ( x + 1 ) + B ( y - 1 ) + C ( z - 3 ) = 0 where 6   A + 7   B + 8 C = 0 and 3   A + 5   B + 7 C = 0 By cross multiplication method, we get A 1 = B - 2 = C 1 ⇒ 1 ( x + 1 ) - 2 ( y - 1 ) + 1 ( z - 3 ) = 0 ⇒ x - 2 y + z = 0 Let Q ( α , β , γ ) α - 7 1 = β + 2 - 2 = γ - 13 1 = - ( 7 + 4 + 13 ) 1 + 4 + 1 = - 4 Q ( α , β , γ ) = Q ( 3 , 6 , 9 ) P Q = 7 - 3 2 + - 2 - 6 2 + 13 - 9 2 ( P Q ) 2 = 16 + 64 + 16 ( P Q ) 2 = 96