How to Find the Image of a Point in the Plane

The image of a point with respect to a plane in three-dimensional space is the reflection of that point across the plane, such that the line connecting the original point and its image is perpendicular to the plane and bisected by the plane. This procedure is a standard result in coordinate geometry and is fundamental for solving problems involving symmetries and reflections across coordinate planes.

Algebraic Computation of the Image of a Point in a Plane

Consider a point $P(x_1, y_1, z_1)$ and a plane given by the equation $Ax + By + Cz + D = 0$. Let the image of $P$ in the plane be $P'(x', y', z')$. The reflection $P'$ must satisfy two properties: the line segment $PP'$ is perpendicular to the plane, and the midpoint of $PP'$ lies on the plane.

Let the direction ratios of the normal to the plane be $(A, B, C)$. Since $PP'$ is perpendicular to the plane, the vector $\overrightarrow{PP'} = (x' - x_1,\, y' - y_1,\, z' - z_1)$ is in the same or opposite direction as the normal. Hence, there exists a real scalar $\lambda$ such that:

\begin{align*} x' &= x_1 + A\lambda \\ y' &= y_1 + B\lambda \\ z' &= z_1 + C\lambda \end{align*}

The midpoint $M$ of $PP'$ is given by:

\begin{align*} M &= \left( \frac{x_1 + x'}{2},\, \frac{y_1 + y'}{2},\, \frac{z_1 + z'}{2} \right) \end{align*}

Since $M$ lies on the plane, its coordinates satisfy the equation of the plane:

A \left( \frac{x_1 + x'}{2} \right) + B \left( \frac{y_1 + y'}{2} \right) + C \left( \frac{z_1 + z'}{2} \right) + D = 0

Substituting the values of $x'$, $y'$, $z'$ in terms of $\lambda$ gives:

A \left( \frac{x_1 + x_1 + A\lambda}{2} \right) + B \left( \frac{y_1 + y_1 + B\lambda}{2} \right) + C \left( \frac{z_1 + z_1 + C\lambda}{2} \right) + D = 0

Expanding this expression leads to:

\frac{A(2x_1 + A\lambda) + B(2y_1 + B\lambda) + C(2z_1 + C\lambda)}{2} + D = 0

The numerator expands as follows:

2A x_1 + A^2\lambda + 2B y_1 + B^2\lambda + 2C z_1 + C^2\lambda = 2(A x_1 + B y_1 + C z_1) + (A^2 + B^2 + C^2)\lambda

Therefore, the previous equation simplifies to:

\frac{2(A x_1 + B y_1 + C z_1) + (A^2 + B^2 + C^2)\lambda}{2} + D = 0

Multiply both sides by $2$:

2(A x_1 + B y_1 + C z_1) + (A^2 + B^2 + C^2)\lambda + 2D = 0

Isolate $\lambda$:

(A^2 + B^2 + C^2)\lambda = -2(A x_1 + B y_1 + C z_1 + D)

\implies \lambda = \frac{ -2(A x_1 + B y_1 + C z_1 + D) }{ A^2 + B^2 + C^2 }

Therefore, the coordinates of the image $P'(x', y', z')$ are given by:

\begin{align*} x' &= x_1 + A\lambda \\ y' &= y_1 + B\lambda \\ z' &= z_1 + C\lambda \end{align*}

Substitute the value of $\lambda$ to get the explicit formulas:

\begin{aligned} x' &= x_1 - \frac{2A(A x_1 + B y_1 + C z_1 + D)}{A^2 + B^2 + C^2} \\ y' &= y_1 - \frac{2B(A x_1 + B y_1 + C z_1 + D)}{A^2 + B^2 + C^2} \\ z' &= z_1 - \frac{2C(A x_1 + B y_1 + C z_1 + D)}{A^2 + B^2 + C^2} \end{aligned}

Result: The image of $P(x_1, y_1, z_1)$ in the plane $Ax + By + Cz + D = 0$ is $P'\bigg(x',\, y',\, z'\bigg)$ given by the above formulas.

Determination of Images in Standard Coordinate Planes

The equations of the standard coordinate planes are $xy$-plane: $z = 0$, $yz$-plane: $x = 0$, $zx$-plane: $y = 0$. The process for determining the image simplifies significantly for these cases.

For the $xy$-plane ($z = 0$), coefficients: $A = 0$, $B = 0$, $C = 1$, $D = 0$. The formula above becomes:

\begin{aligned} x' &= x_1 \\ y' &= y_1 \\ z' &= z_1 - \frac{2z_1}{1} = -z_1 \end{aligned}

Result: The image of $P(x_1, y_1, z_1)$ in the $xy$-plane is $(x_1,\, y_1,\, -z_1)$.

For the $yz$-plane ($x = 0$), $A = 1$, $B = 0$, $C = 0$, $D = 0$:

\begin{aligned} x' &= x_1 - \frac{2x_1}{1} = -x_1 \\ y' &= y_1 \\ z' &= z_1 \end{aligned}

Result: The image in the $yz$-plane is $(-x_1,\, y_1,\, z_1)$.

For the $zx$-plane ($y = 0$), $A = 0$, $B = 1$, $C = 0$, $D = 0$:

\begin{aligned} x' &= x_1 \\ y' &= y_1 - \frac{2y_1}{1} = -y_1 \\ z' &= z_1 \end{aligned}

Result: The image in the $zx$-plane is $(x_1,\, -y_1,\, z_1)$.

For more detailed treatments of coordinate geometry concepts, refer to Coordinate Geometry.

Worked Examples: Image of a Point in the Specified Plane

Given: The point $P(5, 4, -3)$ and the $xy$-plane ($z = 0$).

Substitute: $x_1 = 5$, $y_1 = 4$, $z_1 = -3$.

Using the formula $(x_1, y_1, -z_1)$, compute:

\begin{align*} x' &= 5 \\ y' &= 4 \\ z' &= -(-3) = 3 \end{align*}

Result: The image of $(5, 4, -3)$ in the $xy$-plane is $(5, 4, 3)$.

Given: The point $Q(-2, 0, 0)$ and the $xy$-plane ($z = 0$).

Substitute: $x_1 = -2$, $y_1 = 0$, $z_1 = 0$.

\begin{align*} x' &= -2 \\ y' &= 0 \\ z' &= -0 = 0 \end{align*}

Result: The image of $(-2, 0, 0)$ in the $xy$-plane is $(-2, 0, 0)$.

Given: The point $R(-3, 4, 7)$ and the $yz$-plane ($x = 0$).

Substitute: $x_1 = -3$, $y_1 = 4$, $z_1 = 7$.

\begin{align*} x' &= -(-3) = 3 \\ y' &= 4 \\ z' &= 7 \end{align*}

Result: The image of $(-3, 4, 7)$ in the $yz$-plane is $(3, 4, 7)$.

Given: The point $S(-7, 2, -1)$ and the $zx$-plane ($y = 0$).

Substitute: $x_1 = -7$, $y_1 = 2$, $z_1 = -1$.

\begin{align*} x' &= -7 \\ y' &= -2 \\ z' &= -1 \end{align*}

Result: The image of $(-7, 2, -1)$ in the $zx$-plane is $(-7, -2, -1)$.

Given: The point $T(-4, 0, 1)$ and the $zx$-plane ($y = 0$).

Substitute: $x_1 = -4$, $y_1 = 0$, $z_1 = 1$.

\begin{align*} x' &= -4 \\ y' &= -0 = 0 \\ z' &= 1 \end{align*}

Result: The image of $(-4, 0, 1)$ in the $zx$-plane is $(-4, 0, 1)$.

To extend your understanding of image transformations, reflections, and related functions, see Functions and Their Types.