A ray of light comes along line L and strikes the plane mirror kept along the plane $P = 0$ at point B. $A (2, 1, λ)$ is a point on the line L whose image about $P = 0$ is $A^{'}$ . It is given that line L is $\frac{x - 2}{3} = \frac{y - 1}{4} = \frac{z - 6}{5}$ and plane $P = 0$ is $x + y - 2 z = 3$ , then which of the following is/are TRUE?

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The coordinates of $A^{'}$ are (6,5,2)

The coordinates of $A^{'}$ are (6,5,-2)

Equation of the reflected ray is $\frac{x + 10}{4} = \frac{y - 15}{4} = \frac{z + 2}{3}$

Equation of the reflected ray is $\frac{x + 10}{4} = \frac{y + 15}{5} = \frac{z + 14}{3}$

answer is A, C.

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Detailed Solution

Any point on line L is (3k+2, 4k+1, 5k+6) $∴ k = 0$ $∴ A' (6, 5, - 2)$ A is (2,1,6)
Image $A' (x_{2}, y_{2}, z_{2}); \frac{x_{2} - 2}{3} = \frac{y_{2} - 1}{4} = \frac{z_{2} - 6}{5} = - 2 (\frac{2 + 1 - 12 - 3}{6}) = 4$
Putting (3k+2, 4k+1, 5k+6) in equation of plane, we get B
$3 k + 2 + 4 k + 1 - 10 k - 12 = 3 \Rightarrow k = - 4$ $∴$ B is (-10,-15,-14)
$∴$ Equation of reflected ray BA’ is $\frac{x + 10}{4} = \frac{y + 15}{5} = \frac{z + 14}{3}$