MHT CET20243 May 2024Morning ShiftMathematicsThree Dimensional GeometryActual
The image of the line x-1 3 = y-3 1 = z-4 -5 in the plane 2 x-y+z+3=0 is the line
Options
- Ax+3 -3 = y-5 -1 = z+2 5
- Bx-3 3 = y+5 1 = z-2 -5
- Cx-3 -3 = y+5 -1 = z-2 5
- Dx+3 3 = y-5 1 = z-2 -5
Correct answer
D. x+3 3 = y-5 1 = z-2 -5
Step-by-step solution
Given line x-1 3 = y-3 1 = z-4 -5 passes through the point (1,3,4) Let the required line passes through the point (p, q, r) Now, according to the given condition, we get Distance between point (1,3,4) and the given plane = Distance between point (p, q, r) and the given plane. | 2(1)-(3)+(4)+3 (2)^2+(-1)^2+(1)^2 |= | 2( p )-( q )+( r )+3 (2)^2+(-1)^2+(1)^2 | |2 p - q + r +3|=6 Note that line given in option (D) passes through the point (-3,5,2) and this point satisfies the condition given in equation (i). Option (D)