Reflection of a point across a plane in 3D

2026-10-22 huynhvietkhanh Comments: 0

Find a reflected point from its orthogonal projection and the midpoint relation; check the geometry and coordinate-plane cases.

High school3D Coordinate Geometry

Content updated: October 22, 2026

To reflect $M$ across a plane, first find the orthogonal projection $H$ of $M$ onto that plane. The reflected point $M'$ is then determined by requiring $H$ to be the midpoint of $MM'$.

Learning goals: find the coordinates of a reflection across a plane; check the midpoint and perpendicular-direction conditions; handle a point already in the plane.

Prerequisites: normal vectors, orthogonal projection onto a plane, and the midpoint coordinate formula.

On this page: Idea · Method · Examples · Video · Practice

1. From projection to reflection

Suppose $M\notin(P)$. Let $H$ be the orthogonal projection of $M$ onto $(P)$. On the line $MH$, choose $M'$ on the side opposite to $M$ relative to $H$, with $HM'=HM$. Then $M'$ is the reflection of $M$ across $(P)$.

The diagram shows M outside (P): H is the midpoint of MM′, and MM′ is perpendicular to (P).

Both conditions are needed: $H$ is the midpoint of $MM'$, and the line $MM'$ is perpendicular to $(P)$. Choosing any point on the other side of the plane does not determine the reflection.

If $M\in(P)$, its reflection is $M$ itself. Then $H=M=M'$; $MM'$ does not define a line, since the two points coincide.

2. Method and coordinate formula

Let $M(x_0;y_0;z_0)$ and the plane $(P):ax+by+cz+d=0$ be given, with $a^2+b^2+c^2>0$. A normal vector to $(P)$ is $\vec n=(a;b;c)$.

Step 1. Find the projection $H$ of $M$ onto $(P)$. Following the lesson on projecting a point onto a plane (English lesson), set

$$t=\frac{ax_0+by_0+cz_0+d}{a^2+b^2+c^2}.$$

Then

$$H(x_0-at;\ y_0-bt;\ z_0-ct).$$

Step 2. Use the condition that $H$ is the midpoint of $MM'$. By the midpoint formula,

$$\begin{cases}x_{M'}=2x_H-x_0,\\y_{M'}=2y_H-y_0,\\z_{M'}=2z_H-z_0.\end{cases}$$

Substituting the coordinates of $H$ yields the direct formula:

$$\begin{cases}x_{M'}=x_0-2at,\\y_{M'}=y_0-2bt,\\z_{M'}=z_0-2ct.\end{cases}$$

In compact coordinate notation, $M'=2H-M=M-2t\vec n$.

Since $a^2+b^2+c^2>0$, the value $t$ is uniquely determined. If $M\in(P)$, then $t=0$, so $M'=M$. If $M\notin(P)$, then $t\ne0$ and $\overrightarrow{MM'}=-2t\vec n$, so $MM'\perp(P)$.

Step 3. Check the answer. Find the midpoint $H$ of $MM'$ and check $H\in(P)$. When $M'\ne M$, also check that $\overrightarrow{MM'}$ is parallel to $\vec n$. These two conditions verify the required reflection.

3. Worked examples

Example 1. Continuing the projection example

Find the point $M'$ obtained by reflecting $M(2;1;3)$ across $(P):x-y+z-1=0$.

Solution. A normal vector is $\vec n=(1;-1;1)$. Then

$$t=\frac{2-1+3-1}{1^2+(-1)^2+1^2}=1.$$

Hence $H(1;2;2)$. Since $H$ is the midpoint of $MM'$,

$$\begin{aligned}x_{M'}&=2\cdot1-2=0,\\y_{M'}&=2\cdot2-1=3,\\z_{M'}&=2\cdot2-3=1.\end{aligned}$$

Thus $M'(0;3;1)$.

Check: the midpoint of $M(2;1;3)$ and $M'(0;3;1)$ is $H(1;2;2)$, with $1-2+2-1=0$. Also, $\overrightarrow{MM'}=(-2;2;-2)=-2\vec n$, so $MM'\perp(P)$.

Example 2. Fractional coordinates

Find the reflection of $M(1;2;0)$ across $(P):2x-y+2z+3=0$.

Solution. With $\vec n=(2;-1;2)$, we calculate

$$t=\frac{2\cdot1-2+2\cdot0+3}{2^2+(-1)^2+2^2}=\frac13.$$

The projection is $H\left(\frac13;\frac73;-\frac23\right)$. By the reflection formula,

$$\begin{aligned}x_{M'}&=1-2\cdot2\cdot\frac13=-\frac13,\\y_{M'}&=2-2(-1)\cdot\frac13=\frac83,\\z_{M'}&=0-2\cdot2\cdot\frac13=-\frac43.\end{aligned}$$

Thus $M'\left(-\frac13;\frac83;-\frac43\right)$.

Check: the midpoint of $MM'$ is $H$; substitution of $H$ into the equation of $(P)$ yields $\frac23-\frac73-\frac43+3=0$. Also, $\overrightarrow{MM'}=-\frac23(2;-1;2)$, so the line $MM'$ is perpendicular to $(P)$.

Example 3. Reflection across a coordinate plane

Let $M(-2;3;5)$. The projection of $M$ onto $(Oxy)$ is $H(-2;3;0)$. Since $H$ is the midpoint of $MM'$, the reflection across $(Oxy)$ is $M'(-2;3;-5)$.

For a general point $M(x_0;y_0;z_0)$:

Across $(Oxy)$: $M'(x_0;y_0;-z_0)$.
Across $(Oyz)$: $M'(-x_0;y_0;z_0)$.
Across $(Oxz)$: $M'(x_0;-y_0;z_0)$.

Reflection across a coordinate plane retains the two coordinates in the plane and changes the sign of the remaining coordinate. Projection sets that coordinate to $0$; reflection reverses its sign. If $M$ lies in the plane, both transformations fix $M$.

4. Video lesson

The Vietnamese-language video below explains Example 1 by finding the projection and applying the midpoint formula. Its line parameter has the opposite sign from $t$ in this lesson; both conventions yield $H(1;2;2)$ and $M'(0;3;1)$.

Watch on YouTube.

5. Practice

Find the reflection $M'$ of $M$ across the given plane. Check the midpoint and perpendicular direction yourself before opening the answers.

$M(1;0;2)$, $(P):x+y+z=0$.
$M(1;-1;2)$, $(P):2x+2y+z-1=0$.
$M(-4;2;-3)$, plane $(Oyz)$.
$M(1;2;-1)$, $(P):x-y+2z+3=0$.

Show answers and hints

$t=1$ and $H(0;-1;1)$, so $M'(-1;-2;0)$. The midpoint of $MM'$ is $H$, and $\overrightarrow{MM'}=-2(1;1;1)$.
$t=\frac19$ and $H\left(\frac79;-\frac{11}9;\frac{17}9\right)$, so $M'\left(\frac59;-\frac{13}9;\frac{16}9\right)$. Check the midpoint and $\overrightarrow{MM'}=-\frac29(2;2;1)$.
$(Oyz):x=0$, so $H(0;2;-3)$ and $M'(4;2;-3)$.
Since $1-2+2(-1)+3=0$, we have $M\in(P)$. Thus $M'=M=(1;2;-1)$.

6. Common mistakes

Stopping at the projection $H$ when the problem asks for the reflection $M'$.
Writing $M'=M-2H$. The midpoint formula instead yields $M'=2H-M$ componentwise.
Omitting the factor $2$ from the direct reflection formula.
Changing all coordinate signs when reflecting across a coordinate plane.
Checking only that the two points are equally far from the plane while omitting the midpoint and perpendicular-direction conditions.

Key idea: find the projection $H$, then use $H$ as the midpoint to find $M'$. Reflection fixes every point of the plane.

7. Prerequisite, next lesson, and additional resources

Prerequisite: Orthogonal projection of a point onto a plane (English lesson).
Next lesson: Orthogonal projection of a point onto a line (English lesson).
Additional practice material on Scribd (external link).