Question

A line $L$ meets the plane $P$ at point $B$. Let $A(2,1,6)$ be a point on the line $L$ such that the image of $A$ about $P$ is $A^{\prime }$. If the equations of the line $L$ and the plane $P$ are $L:\frac{x-2}{3}=\frac{y-1}{4}=\frac{z-6}{5}$ and $P:x+y-2z=3$ respectively, then which of the following is/are correct?

• A. Coordinates of $A^{\prime }$ is $(6,5,-2)$
• B. Coordinates of $B$ is $(-10,-15,-14)$
• C. Image of $L$ about the plane $P$ is $\frac{x-6}{3}=\frac{y-5}{5}=\frac{z+2}{3}$
• D. Image of $L$ about the plane $P$ is $\frac{x+10}{2}=\frac{y+15}{5}=\frac{z+14}{5}$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Coordinates of $A^{\prime }$ is $(6,5,-2)$
B Coordinates of $B$ is $(-10,-15,-14)$
C Image of $L$ about the plane $P$ is $\frac{x-6}{3}=\frac{y-5}{5}=\frac{z+2}{3}$

Any point on the line $L$ is $(3k+2,4k+1,5k+6),k\in R$
$\because$ this point lies on the plane,
$\therefore 3k+2+4k+1-10k-12=3\Rightarrow k=-4$
Therefore, coordinates of $B$ is $(-10,-15,-14)$

Let the coordinates of $A^{\prime }$ is $(a,b,c)$
$\Rightarrow \frac{a-2}{1}=\frac{b-1}{1}=\frac{c-6}{-2}=\frac{-2(-12)}{6}\Rightarrow (a,b,c)=(6,5,-2)$

Points $A^{\prime }$ and $B$ both lie on the image of $L$ about the plane $P.$
$\therefore$ Image of $L$ is
$\frac{x-6}{6+10}=\frac{y-5}{5+15}=\frac{z+2}{-2+14}\Rightarrow \frac{x-6}{16}=\frac{y-5}{20}=\frac{z+2}{12}$

Required image of the line $L$ is
$\frac{x-6}{4}=\frac{y-5}{5}=\frac{z+2}{3}$