<p>Find the image of point <i>A</i>(2, 1, 6) about the mirror plane <i>x</i> + <i>y</i> - 2<i>z</i> = 3.</p>
Step-by-Step Solution
Key Concept: The image point lies on the normal to the plane through the original point. The midpoint must satisfy the plane equation.
Solution: Let Q ( x _2, y _2, z _2) be the image of A (2, 1, 6). The midpoint of AQ lies on the plane and AQ is perpendicular to the plane with normal n = (1, 1, -2). Using the formula: $\frac{x_2 - 2}{1} = \frac{y_2 - 1}{1} = \frac{z_2 - 6}{-2} = \frac{-2(2 + 1 - 12 - 3)}{1^2 + 1^2 + (-2)^2} = \frac{-2(-12)}{6} = 4$ Therefore: x _2 = 2 + 4 = 6, y _2 = 1 + 4 = 5, z _2 = 6 - 8 = -2 ∴ Q ≡ (6, 5, -2)
Correct Answer: B