JEE Main20257 Apr 2025Evening ShiftMathematicsThree Dimensional GeometryActual
If the equation of the line passing through the point (0,- 1 2 , 0 ) and perpendicular to the lines r = ( i +a j +b k ) and r =( i - j -6 k )+ (-b i + a j +5 k ) is x -1 -2 = y +4 ~d = z - c -4 , then a + b + c + d is equal to :
Options
- A10
- B14
- C13
- D12
Correct answer
B. 14
Step-by-step solution
Line is ^ r to 2 line line will be parallel to (i+a j +b k ) (-b i +a j +5 k ) Parallel vector along the required line is i (5 a - ab )- j ( ~b ^2+5 )+ k ( a + ab ) Dr's of required line (5 a-a b),- (b^2+5 ),(a+a b) Also Dr's of required line -2, ~d ,-4 Also point (0, -1 2 , 0 ) will lie on x -1 -2 = y +4 ~d = z - c -4 0-1 -2 = -1 2 +4 d = 0-c -4 d=7, c=2 aligned & From (1) 5 a - ab -2 = - b ^2-5 7 = a + ab -4 & 5 a - ab -2 = a + ab -4 ; - b ^2-5 7 = a + ab -4 aligned array c|c -20 a+4 a b=-2 a-2 a b & 4 b^2+20=70+