A) \[\frac{x-2}{3}=\frac{y}{-3}=\frac{z+4}{5}\]
B) \[\frac{x-3}{2}=\frac{y}{-3}=\frac{z-4}{5}\]
C) \[\frac{x-3}{2}=\frac{-y}{3}=\frac{z+4}{5}\]
D) \[\frac{x+3}{2}=\frac{y}{3}=\frac{z-4}{5}\]
E) \[\frac{x-2}{3}=\frac{y}{3}=\frac{z+4}{5}\]
Correct Answer: C
Solution :
Comparing given plane\[2x-3y+5z-7=0\] with the plane\[lx+my+nz+d=0\] \[\Rightarrow \] \[l=2,m=-3,z=5\] Since, the required line is perpendicular to the given plane. Therefore, these are the direction cosine of the line. \[\therefore \]Equation of line is \[\frac{x-3}{2}=\frac{y-0}{-3}=\frac{z+4}{5}\] \[\Rightarrow \] \[\frac{x-3}{2}=-\frac{y}{3}=\frac{z+4}{5}\]
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