A) \[2x+17y+9z=0\]
B) \[28x-17y+9z=0\]
C) \[2x+17y-9z=0\]
D) None of these
Correct Answer: B
Solution :
The equation of plane passing through the intersection line of the planes \[2x-y=0\] and \[3z-y=0\] is \[(2x-y)+\lambda (3z-y)=0\] \[\Rightarrow \] \[2x-(1+\lambda )y+3\lambda z=0\] ... (i) According to the question, plane (i) will be perpendicular to \[4x+5y-3z=8\] ... (ii) \[\therefore \] \[2\cdot 4-(1+\lambda )5+3\lambda (-3)=0\] \[\Rightarrow \] \[8-5-5\lambda -9\lambda =0\] \[\Rightarrow \] \[14\lambda =3\Rightarrow \lambda =\frac{3}{14}\] Thus, required plane is \[(2x-y)+\frac{3}{14}(3z-y)=0\] \[\Rightarrow \] \[28x-17y+9z=0\]You need to login to perform this action.
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