JEE Main202523 Jan 2025Morning ShiftMathematicsThree Dimensional GeometryActual
Let P be the foot of the perpendicular from the point Q(10,-3,-1) on the line x-3 7 = y-2 -1 = z+1 -2 . Then the area of the right angled triangle P Q R , where R is the point (3,-2,1) , is
Options
- A9 15
- B30
- C8 15
- D3 30
Correct answer
D. 3 30
Step-by-step solution
aligned & x -3 7 = y -2 -1 = z +1 -2 = & 7 +3,- +2,-2 -1 & dr's of QP 7 -7,- +5,-2 aligned Now aligned & (7 -7) 7-(- +5)+(2 ) 2=0 & 54 -54=0 =1 & P =(10,1,-3) & PQ =-4 j +2 k & PR =-7 i -3 j +4 k & Area = 1 2 | array ccc i & j & k 0 & -4 & 2 -7 & -3 & 4 array |=3 30 aligned