Question

The equation of the line passing through the points $(1,–2,3)$ and parallel to the planes $x-y+2z-5=0$ and $3x+y+z-6=0$ is

• A. $\frac{x-1}{-3}=\frac{y+2}{5}=\frac{z-3}{4}$
• B. $\frac{x-1}{1}=\frac{y+2}{-1}=\frac{z-3}{2}$
• C. $\frac{x-1}{3}=\frac{y+2}{5}=\frac{z-3}{-4}$
• D. $\frac{x-1}{1}=\frac{y+2}{1}=\frac{z-3}{2}$

Answer 1

The correct option is A $\frac{x-1}{-3}=\frac{y+2}{5}=\frac{z-3}{4}$

Let $a,b,c$ be the direction ratios of the required line.
As it is parallel to both the given planes, it will be perpendicular to their normals.
$\Rightarrow a-b+2c=0$ and $3a+b+c=0$
$\Rightarrow \frac{a}{-3}=\frac{b}{5}=\frac{c}{4}$
Hence, equation of the required line is
$\frac{x-1}{-3}=\frac{y+2}{5}=\frac{z-3}{4}$