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

Answer: (A)

$\frac{a-1}{2}=\frac{b-3}{-1}=\frac{c-4}{1}=\lambda$
$\Rightarrow a=2\lambda +1$
$b=3-\lambda$
$c=4+\lambda$
$P\equiv \left(\lambda +1,3-\frac{\lambda }{2},4+\frac{\lambda }{2}\right)$
$2(\lambda +1)-\left(3-\frac{\lambda }{2}\right)+\left(4+\frac{\lambda }{2}\right)+3=0$
$2\lambda +2-3+\frac{\lambda }{2}+4+\frac{\lambda }{2}+3=0$
$3\lambda +6=0$
$\Rightarrow \lambda =-2$
$a=-3,b=5,c=2$
So the equation of the required line is
$\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$