Question

The image of the line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ in the plane $2x-y+z+3=0$ is the line:

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

Answer 1

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

The direction ratios of given line $\frac{x-1}{3}=\frac{y-3}{1}=\frac{z-4}{-5}$ is $(3,1,-5)$ and direction ratios of given plane $2x-y+z+3=0$ is $(2,-1,1)$
Check, line is parallel or perpendicular to plane
$\therefore 3\times 2+1\times -1+-5\times 1=0$
Hence, line is parallel to plane
Consider any point $M(x,y,z)$ on the plane


The equation of line which is perpendicular to line $l_1$ and parallel to the normal vector plane is
$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\lambda$
$\therefore x=2\lambda +1,y=-\lambda +3,z=\lambda +4$
point $(x,y,z)$ pass through the plane
$\therefore 4\lambda +2+\lambda -3+\lambda +4+3=0$
$\Rightarrow \lambda =-1$
So, $x=-1,y=4$ and $z=3$
Point $M$ is the mid point of $A$ and $B$
So cordinates of point B are $(-3,5,2)$
Hence, The equation of line $l_2$ is
$\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$