3D Geometry
Three Dimensional Geometry
star_batch_jee_advanced_2025
Grade 12
Question:
A ray of light is coming along the line $\frac{x-2}{3} = \frac{y-1}{4} = \frac{z-6}{5}$ and strikes the plane mirror kept along the plane through the points $(2, 1, 0)$, $(5, 0, 1)$ and $(4, 1, 1)$. Then the equation of reflected ray is:
$\frac{x+10}{4} = \frac{y+15}{5} = \frac{z+14}{3}$
$\frac{x-10}{4} = \frac{y-15}{5} = \frac{z-14}{3}$
$\frac{x-15}{-4} = \frac{y-14}{-5} = \frac{z-10}{3}$
None of these
Step-by-Step Solution
Key Concept: Use the determinant formula for the plane equation, then apply the reflection formula using the perpendicular distance to the plane.
To find the equation of the plane through $(2, 1, 0)$, $(5, 0, 1)$, and $(4, 1, 1)$, we use the determinant form $\begin{vmatrix} x-2 & y-1 & z-0 \\ 5-2 & 0-1 & 1-0 \\ 4-2 & 1-1 & 1-0 \end{vmatrix} = 0$. Expanding gives $(x-2)(-1) - (y-1)(3) + z(2) = 0$, which simplifies to $x + y - 2z = 3$. The reflection of point $(2, 1, 6)$ with respect to this plane is found using the reflection formula: the reflected point is $(6, 5, -2)$. The reflected ray passes through this point and point $Q = (-10, -15, -14)$, giving the reflected line equation $\frac{x+10}{4} = \frac{y+15}{5} = \frac{z+14}{3}$.
Correct Answer: 1