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
$\frac{x-3}{2}=\frac{y-0}{-3}=\frac{z+4}{5}$
$\Rightarrow$ $\frac{x-3}{2}=-\frac{y}{3}=\frac{z+4}{5}$