A) \[28x-17y+9z=0\]
B) \[28x+17y+9z=0\]
C) \[28x-17y-9z=0\]
D) \[7x-3y+z=0\]
Correct Answer: A
Solution :
Equation of plane containing the line of intersection of planes is \[(2x-y)+\lambda (y-3z)=0\] ... (i) Also, plane (i) is perpendicular to \[4x+5y-3z-8=0\] \[\therefore \] \[4(2)+5(\lambda -1)-3(-3\lambda )=0\] \[\Rightarrow \] \[14\lambda =-3\] \[\Rightarrow \] \[\lambda =\frac{-3}{14}\] Putting the value of 'k in Eq. (i), we get \[28x-17y+9z=0\] which is the required plane.You need to login to perform this action.
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