Orthogonal projection of a point onto a plane in 3D

Content updated: October 22, 2026

To project a point orthogonally onto a plane, follow the normal direction from the point and find the intersection with the plane. This lesson derives the coordinate formula, works through examples, and explains how to check the result.

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

1. How is the projection defined?

If $M$ does not lie in the plane $(P)$, the orthogonal projection of $M$ onto $(P)$ is the point $H$ satisfying both $H\in(P)$ and the condition that the line $MH$ is perpendicular to $(P)$. Since the normal vector $\vec n$ is perpendicular to $(P)$, the line through $M$ in the direction $\vec n$ meets $(P)$ at the required point.

If $M\in(P)$, its projection is $M$ itself. Then $H=M$; we do not call $MH$ a line, since its two defining points coincide.

2. Method and coordinate formula

In three-dimensional Cartesian space $Oxyz$, let $M(x_0;y_0;z_0)$ be a point and let the plane be

where $a^2+b^2+c^2>0$. Then $\vec n=(a;b;c)$ is a normal vector to $(P)$.

Step 1. Write the line $\Delta$ through $M$ perpendicular to $(P)$:

The minus signs reflect our choice of parameter. With this choice, every point on $\Delta$ has coordinates $(x_0-at;y_0-bt;z_0-ct)$.

Step 2. Find the intersection $H$ of $\Delta$ with $(P)$. Substitute the coordinates above into the plane equation:

After simplifying,

Since $a^2+b^2+c^2>0$, the equation determines the unique value

If $M\in(P)$, the numerator is $0$, so $t=0$ and $H=M$. If $M\notin(P)$, then $t\ne0$, and $\overrightarrow{MH}=-t\vec n$ is a vector different from $\vec0$ and parallel to the normal vector. Thus $MH\perp(P)$.

Step 3. Check the answer. Substitute the coordinates of $H$ into the equation of $(P)$. If $H\ne M$, also check that $\overrightarrow{MH}$ is parallel to $\vec n$. Checking $H\in(P)$ alone does not establish that $H$ is the projection.

3. Worked examples

Example 1. Integer coordinates

Find the orthogonal projection of $M(2;1;3)$ onto the plane $(P):x-y+z-1=0$.

Solution. A normal vector to $(P)$ is $\vec n=(1;-1;1)$. The line through $M$ perpendicular to $(P)$ has equations

Substituting into the equation of $(P)$,

or $3-3t=0$, so $t=1$. Hence $H(1;2;2)$.

Check: $1-2+2-1=0$, so $H\in(P)$. Also, $\overrightarrow{MH}=(-1;1;-1)=-\vec n$, so $MH\perp(P)$. Therefore $H(1;2;2)$ is the required projection.

Example 2. Negative coefficients and fractional coordinates

Find the orthogonal projection of $M(1;2;0)$ onto the plane $(P):2x-y+2z+3=0$.

Solution. Here $a=2$, $b=-1$, $c=2$, and $d=3$. Thus

Calculate each coordinate of $H$:

Thus $H\left(\frac13;\frac73;-\frac23\right)$. In the second coordinate, subtracting a negative number amounts to addition.

Check: $\frac23-\frac73-\frac43+3=0$ and $\overrightarrow{MH}=\left(-\frac23;\frac13;-\frac23\right)=-\frac13(2;-1;2)$. Both conditions defining the projection hold.

Example 3. Projection onto a coordinate plane

Let $M(-2;3;5)$. The plane $(Oxy)$ has equation $z=0$ and normal vector $(0;0;1)$. Thus the projection of $M$ onto $(Oxy)$ is $H(-2;3;0)$.

More generally, for $M(x_0;y_0;z_0)$:

• Projection onto $(Oxy)$: $H(x_0;y_0;0)$.
• Projection onto $(Oyz)$: $H(0;y_0;z_0)$.
• Projection onto $(Oxz)$: $H(x_0;0;z_0)$.

Setting one coordinate equal to $0$ works for the corresponding coordinate plane, not for an arbitrary plane.

4. Video lesson

The Vietnamese-language video below explains Example 1. Its parameter runs in the direction opposite to the choice made in this lesson; the resulting line and projection are the same.

Watch on YouTube.

5. Practice

Find $H$ yourself and check the result before opening the answers.

1. Find the projection of $M(1;0;2)$ onto $(P):x+y+z=0$.
2. Find the projection of $M(1;-1;2)$ onto $(P):2x+2y+z-1=0$.
3. Find the projection of $M(-4;2;-3)$ onto the plane $(Oyz)$.
4. Find the projection of $M(1;2;-1)$ onto $(P):x-y+2z+3=0$.

6. Common mistakes

• Misreading the sign of the constant term $d$: first write the plane equation in the form $ax+by+cz+d=0$ before substituting values.
• Using the wrong denominator: the denominator of $t$ is $a^2+b^2+c^2$, not its square root.
• Taking an absolute value in the numerator: retain the sign of $t$ to obtain the correct coordinates of $H$.
• Mixing parameter conventions: this lesson uses $H(x_0-at;y_0-bt;z_0-ct)$ with the stated formula for $t$.
• Checking that the point lies in the plane while omitting the perpendicular-direction condition when $H\ne M$.

Key idea: follow the normal direction, use $H\in(P)$ to determine the parameter, and check the result. If $M$ already lies in $(P)$, its projection is $M$ itself.

7. Further study and resources

• Review: Three-dimensional Cartesian coordinates (English lesson).
• Next lesson: Reflection of a point across a plane (English lesson).
• Additional practice material on Scribd (external link).