CBSE 2025, Set 1

Question

Let the polished side of the mirror be along the line $\frac{x}{1} = \frac{1-y}{-2} = \frac{2z-4}{6}$. A point $P(1, 6, 3)$, some distance away from the mirror, has its image formed behind the mirror. Find the coordinates of the image point and the distance between the point $P$ and its image.

Edvaya · Accepted Answer

The correct answer is Subjective. First, rewrite the equation of the line representing the mirror in standard symmetric form ($\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$): $\frac{x-0}{1} = \frac{-(y-1)}{-2} = \frac{2(z-2)}{6}$ $\frac{x-0}{1} = \frac{y-1}{2} = \frac{z-2}{3} = \lambda$ (let this be equal to a scalar $\lambda$) Any general point $Q$ on this line has coordinates: $Q(\lambda, 2\lambda + 1, 3\lambda + 2)$ Let $Q$ be the foot of the perpendicular drawn from the point $P(1, 6, 3)$ to the line. The direction ratios of the line segment $PQ$ are: $(\lambda - 1, 2\lambda + 1 - 6, 3\lambda + 2 - 3) = (\lambda - 1, 2\lambda - 5, 3\lambda - 1)$ The direction ratios of the given line are $(1, 2, 3)$. Since $PQ$ is perpendicular to the line, the sum of the products of their direction ratios is zero: $1(\lambda - 1) + 2(2\lambda - 5) + 3(3\lambda - 1) = 0$ $\lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 0$ $14\lambda - 14 = 0 \implies \lambda = 1$ Substitute $\lambda = 1$ back into the coordinates of $Q$ to get the foot of the perpendicular: $Q(1, 2(1)+1, 3(1)+2) = Q(1, 3, 5)$ **Coordinates of the Image:** Let the image of point $P$ be $P'(x', y', z')$. In a plane mirror, the foot of the perpendicular $Q$ acts as the midpoint of the line segment joining $P$ and $P'$. Using the midpoint formula: $Q(1, 3, 5) = (\frac{x'+1}{2}, \frac{y'+6}{2}, \frac{z'+3}{2})$ Equating the respective coordinates: $\frac{x'+1}{2} = 1 \implies x'+1 = 2 \implies x' = 1$ $\frac{y'+6}{2} = 3 \implies y'+6 = 6 \implies y' = 0$ $\frac{z'+3}{2} = 5 \implies z'+3 = 10 \implies z' = 7$ Thus, the coordinates of the image point are $(1, 0, 7)$. **Distance between $P$ and its image:** Using the 3D distance formula between $P(1, 6, 3)$ and $P'(1, 0, 7)$: Distance $= \sqrt{(1-1)^2 + (0-6)^2 + (7-3)^2}$ $= \sqrt{0 + (-6)^2 + (4)^2}$ $= \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$ units.

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