Question

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