Question

On which of the following lines lies the point of intersection of the line, $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}$ and the plane, $x+y+z=2$?

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

Answer 1

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

Let $\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}=\lambda$
$\therefore x=2\lambda +4,y=2\lambda +5,z=\lambda +3$ lies on the plane $x+y+z=2$
so, $2\lambda +4+2\lambda +5+\lambda +3=2$
$\Rightarrow \lambda =-2$
Hence, $(x,y,z)\equiv (0,1,1)$
which is satisfying $\frac{x-1}{1}=\frac{y-3}{2}=\frac{z+4}{-5}$