AP EAMCET201922 Apr 2019Morning ShiftMathematicsThree Dimensional GeometryActual
The distance of the plane (3 x+4 y+5 z+19=0 ) from the point ((1,-1,1) ) measured along a line parallel to the line with direction ratios (2,3,1 ) is
Options
- A( 23 5 2 )
- B( 71 5 2 )
- C( 14 )
- D( 23 )
Correct answer
C. ( 14 )
Step-by-step solution
According to given information Equation of lines is passing through ((1,-1,1) ) and having DC's is ((2,3,1) ) ( x-1 2 = y+1 3 = z-1 1 =r ) Here, ((2 r+1,3 r-1, r+1) ) lie on plane. ( ) These points satisfy the equation of plane. ( aligned & 3(2 r+1)+4(3 r-1)+5(r+1)+19=0 & 6 r+3+12 r-4+5 r+5+19=0 & 23 r+23=0 r=-1 aligned ) So, point is ((-2+1,-3-1,-1+1) ) i.e. ((-1,-4,0) ) Now, required distance (= (-2)^2+(-3)^2+1^2 ) (= 4+9+1 = 14 )