Question

The equation of the line passing through the point $ (3,0,-4) $ and perpendicular to the plane $ 2x-3y+5z-7=0 $ is

• 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} $

Answer 1

Answer: (C)

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} $