AP EAMCET202018 Sep 2020Morning ShiftMathematicsThree Dimensional GeometryActual
Equation of the line passing through the intersection of the plane (x+2 y+3 z=4 ) and the line ( x-1 2 = y+1 1 = z-1 -1 ) and parallel to the vector ((2 i -3 j ) ( i +2 j - k ) ) is
Options
- A( x-5 3 = y-1 2 = z+1 -7 )
- B( x-5 -3 = y-1 -2 = z-1 7 )
- C( x-5 -3 = y-1 -2 = z+1 -7 )
- D( x-5 -3 = y-1 2 = z+1 7 )
Correct answer
C. ( x-5 -3 = y-1 -2 = z+1 -7 )
Step-by-step solution
The general point (p ) on the line ( x-1 2 = y+1 1 = z-1 -1 =r (Let) (i) ) is (P(2 r+1, r-1,1-r) ). Let the point (P ) is the intersection of line (i) and the plane (x+2 y+3 z=4 ), so (2 r+1+2 r-2+3-3 r=4 r=2 ) So, point (P(5, 1,-1) ) ( aligned & Now, (2 i -3 j ) ( i +2 j - k )= | array ccc i & j & k 2 & -3 & 0 1 & 2 & -1 array | & = i (3)- j (-2)+ k (4+3)=3 i +2 j +7 k aligned ) ( ) Equation of required line is ( x-5 3 = y-1 2 = z+1 7 ) or ( x-5 -3 = y-1 -2 = z+1 -7 )