Question

Equation of the line passing through $i+j-3k$ and perpendicular to the plane $2x-4y+3x+5=0$ is

• A. $\frac{x-1}{2}=\frac{1-y}{-4}=\frac{z-3}{3}$
• B. $\frac{x-1}{2}=\frac{1-y}{4}=\frac{z+3}{3}$
• C. $\frac{x-2}{1}=\frac{y+4}{1}=\frac{z-3}{3}$
• D. $\frac{x-1}{-2}=\frac{1-y}{-4}=\frac{z-3}{3}$
• E. $\frac{x-1}{2}=\frac{1-y}{4}=\frac{z-3}{-3}$

Answer 1

The correct option is B $\frac{x-1}{2}=\frac{1-y}{4}=\frac{z+3}{3}$

We know that, the direction cosines of any line which is perpendicular to any plane, i.e., normal to the plane, is proportional to the direction cosines of the plane.
So, the equation of lines passing through the point $(1,1,-3)$ and perpendicular to the plane $2x-4y+3z+5=0$ is
$\frac{x-1}{2}=\frac{1-y}{-4}=\frac{z-(-3)}{3}$
$\Rightarrow \frac{x-1}{2}=\frac{1-y}{4}=\frac{z+3}{3}$