JEE Advanced2010MathematicsThree Dimensional GeometryActual
Equation of the plane containing the straight line x 2 = y 3 = z 4 and perpendicular to the plane containing the straight lines x 3 = y 4 = z 2 and x 4 = y 2 = z 3 is
Options
- Ax+2 y-2 z=0
- B3 x+2 y-2 z=0
- Cx-2 y+z=0
- D5 x+2 y-4 z=0
Correct answer
C. x-2 y+z=0
Step-by-step solution
The DR's of normal to the plane containing x 3 = y 4 = z 2 and x 4 = y 2 = z 3 . n ₁= | array ccc i & j & k 3 & 4 & 2 4 & 2 & 3 array |=(8 i - j -10 k ) Also, equation of plane containing x 2 = y 3 = z 4 and DR's of normal to be n ₂=a i +b j +c k array ll & a x+b y+c z=0 where, n ₁ n ₂=0 & 8 a-b-10 c=0 and & n ₂ (2 i +3 j +4 k ) & 2 a+3 b+4 c=0 array From Eqs. (ii) and (iii), we get aligned & a -1 -10 = b 8 = c -1 a -4+30 & = b -20-32 = c 24+2 a 26 & = b -52 = c 26 a 1 & = b -2 = c 1 aligned From Eqs. (i) and (iv),