COMEDK2014MathematicsThree Dimensional Geometry
Determine the plane through the intersection of the planes x+2 y+3 z-4=0 and 2 x+y-z+5=0 and perpendicular to the plane 5 x+3 y+6 z+8=0
Options
- A-51 x-15 y-50 z-173=0
- B51 x+15 y-50 z+173=0
- C51 x-15 y+50 z-173=0
- D51 x+50 y+15 z+173=0
Correct answer
B. 51 x+15 y-50 z+173=0
Step-by-step solution
Equation of plane through the intersection of planes x+2 y+3 z-4=0 and 2 x+y-z+5=0 is (x+2 y+3 z-4)+k(2 x+y-z+5)=0 or (1+2 k) x+(2+k) y+(3-k) z+(5 k-4)=0 ...(i) D.R.' s of normal of plane (i) are = Given, 5 x+3 y+6 z+8=0 ...(ii) D.R.'s of plane (ii) are . Since Eq. (i) is Perpendicular to the plane (ii), aligned & 5(1+2 k)+(2+k) 3+6(3-k)=0 & 5+10 k+6+3 k+18-6 k=0 & 7 k+29=0 k= -29 7 aligned Required equation of plane is aligned &(x-2 y+3 z-4)+ (- 29 7 )(2 x+y-z+5)=0 & 7 x+14 y+21 z-28-58 x-29 y+29 z-145=0 & -51 x-1