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)

$\frac{x+3}{3}=\frac{y-5}{1}=\frac{z-2}{-5}$

Answer 2

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