Distance from a Point to a Plane

Why Distance to a Plane Uses a Normal

A plane extends without bound in two directions, so measuring from a point to an arbitrary point on the plane would not give a unique distance. The distance from a point to a plane means the shortest distance to any point on that plane. The shortest segment meets the plane perpendicularly, and the plane’s normal vector points in that perpendicular direction.

This geometric fact turns the problem into a projection. The point’s displacement from the plane can be measured along a unit normal, while a non-unit normal must first be accounted for by dividing by its length. The same calculation also gives a signed distance, which records which side of the plane contains the point relative to a chosen normal orientation.

Distance Formula

Let a plane contain a point with position vector \(\mathbf{r}_0\), and let \(\mathbf{n}\ne\mathbf{0}\) be a normal vector to the plane. Let \(P\) be a point with position vector \(\mathbf{p}\). The plane consists of points \(\mathbf{x}\) satisfying \(\mathbf{n}\cdot(\mathbf{x}-\mathbf{r}_0)=0\).

Definition: For the chosen orientation \(\mathbf{n}\), the signed distance from \(P\) to the plane is \[ \frac{\mathbf{n}\cdot(\mathbf{p}-\mathbf{r}_0)}{\|\mathbf{n}\|}. \] The distance from \(P\) to the plane, without a sign, is the absolute value: \[ \frac{\left|\mathbf{n}\cdot(\mathbf{p}-\mathbf{r}_0)\right|}{\|\mathbf{n}\|}. \]

The norm \(\|\mathbf{n}\|\) is the length of the normal. Dividing by it is essential unless the chosen normal is already a unit vector. Replacing \(\mathbf{n}\) with \(-\mathbf{n}\) reverses the signed distance but leaves the ordinary distance unchanged.

If the plane is written in general form \(ax+by+cz+d=0\), then \(\mathbf{n}=\langle a,b,c\rangle\) is a normal vector. For \(P=(x_0,y_0,z_0)\), the formula becomes:

\text{signed distance} = \frac{ax_0+by_0+cz_0+d}{\sqrt{a^2+b^2+c^2}}, \qquad \text{distance} = \frac{|ax_0+by_0+cz_0+d|}{\sqrt{a^2+b^2+c^2}}.

The denominator is nonzero because a plane’s normal vector cannot be the zero vector. The numerator is zero exactly when \(P\) lies on the plane, so the formula gives distance zero in that case.

Why the Formula Works

Choose any point \(Q\) on the plane, with position vector \(\mathbf{q}\). The displacement \(\mathbf{p}-\mathbf{q}\) can be decomposed into a component parallel to the plane and a component parallel to its normal. The parallel component does not contribute to the shortest distance. Taking the dot product with the normal removes that parallel component, because every direction in the plane is perpendicular to \(\mathbf{n}\).

Derivation: Since \(Q\) lies on the plane, \(\mathbf{n}\cdot(\mathbf{q}-\mathbf{r}_0)=0\). Therefore \[ \mathbf{n}\cdot(\mathbf{p}-\mathbf{r}_0) = \mathbf{n}\cdot\big((\mathbf{p}-\mathbf{q})+(\mathbf{q}-\mathbf{r}_0)\big) = \mathbf{n}\cdot(\mathbf{p}-\mathbf{q}). \] The dot-product inequality gives \[ |\mathbf{n}\cdot(\mathbf{p}-\mathbf{q})| \le \|\mathbf{n}\|\,\|\mathbf{p}-\mathbf{q}\|. \] Thus every point \(Q\) on the plane satisfies \[ \|\mathbf{p}-\mathbf{q}\| \ge \frac{|\mathbf{n}\cdot(\mathbf{p}-\mathbf{r}_0)|}{\|\mathbf{n}\|}. \] Equality occurs when \(\mathbf{p}-\mathbf{q}\) is parallel to \(\mathbf{n}\), which is precisely the perpendicular segment from \(P\) to the plane. Hence the lower bound is attained and equals the shortest distance.

This argument also explains the sign: the dot product is positive when \(\mathbf{p}-\mathbf{r}_0\) has a component in the direction of \(\mathbf{n}\), negative when the component points against \(\mathbf{n}\), and zero when \(P\) is on the plane.

Using the Point-Normal Form

When a plane is given by a point and a normal, use that normal directly. First subtract the known plane point from the point whose distance is sought. Take the dot product with the normal, divide by the normal’s length, and use absolute value only if the question asks for ordinary distance.

Worked Example: Signed Distance from a Point-Normal Equation

Find the signed distance and ordinary distance from \(P=(4,1,2)\) to the plane through \(Q=(1,-1,0)\) with normal \(\mathbf{n}=\langle 2,1,2\rangle\).

The displacement from the plane point to \(P\) is \[ \overrightarrow{QP}=\langle 4-1,1-(-1),2-0\rangle=\langle 3,2,2\rangle. \] Its dot product with the normal is \[ \mathbf{n}\cdot\overrightarrow{QP} =2(3)+1(2)+2(2)=6+2+4=12. \] The normal length is \[ \|\mathbf{n}\|=\sqrt{2^2+1^2+2^2}=\sqrt{9}=3. \] Therefore the signed distance is \(12/3=4\), and the ordinary distance is \(|4|=4\).

To check, the plane equation is \(2(x-1)+(y+1)+2z=0\), or \(2x+y+2z-1=0\). At \(P\), its left side is \(8+1+4-1=12\). Dividing by the normal length \(3\) gives \(4\), as before. The positive sign means \(P\) lies on the side indicated by the chosen normal.

Using the General Equation

For a plane in general form, the coefficients of \(x,y,z\) supply the normal. The constant term belongs in the numerator; it is not part of the normal vector’s length. Keeping those roles separate helps prevent sign and denominator errors.

Worked Example: A Point on the Opposite Side

Find the signed distance and ordinary distance from \(P=(1,2,-1)\) to \(3x-4y+12z+5=0\).

The normal is \(\mathbf{n}=\langle3,-4,12\rangle\), so \[ \|\mathbf{n}\|=\sqrt{3^2+(-4)^2+12^2} =\sqrt{9+16+144} =\sqrt{169}=13. \] Evaluate the plane’s left side at \(P\): \[ 3(1)-4(2)+12(-1)+5=3-8-12+5=-12. \] Thus the signed distance is \(-12/13\), while the ordinary distance is \(12/13\).

As a check, the signed-distance formula gives the same numerator and denominator, \(-12\) and \(13\). The negative sign indicates that \(P\) is opposite the direction of the normal from the plane. Reversing the normal would change the signed distance to \(12/13\), but not the ordinary distance.

The Perpendicular Foot

The same projection that gives the distance can locate the nearest point on the plane. If \(\mathbf{p}\) is the position vector of \(P\), the signed distance is \(s=\mathbf{n}\cdot(\mathbf{p}-\mathbf{r}_0)/\|\mathbf{n}\|\). The displacement from \(P\) to its perpendicular foot is \(-s\mathbf{n}/\|\mathbf{n}\|\). Equivalently, use the following formula, which works directly with a non-unit normal.

Formula: The perpendicular foot \(H\) from \(P\) to the plane has position vector \[ \mathbf{h} = \mathbf{p} - \frac{\mathbf{n}\cdot(\mathbf{p}-\mathbf{r}_0)}{\|\mathbf{n}\|^2}\mathbf{n}. \] The denominator is \(\|\mathbf{n}\|^2\), not \(\|\mathbf{n}\|\), because the correction is a vector parallel to \(\mathbf{n}\).

Indeed, the subtracted vector removes the component of \(\mathbf{p}-\mathbf{r}_0\) normal to the plane. Substituting \(\mathbf{h}\) into the plane condition gives \[ \mathbf{n}\cdot(\mathbf{h}-\mathbf{r}_0) = \mathbf{n}\cdot(\mathbf{p}-\mathbf{r}_0) - \frac{\mathbf{n}\cdot(\mathbf{p}-\mathbf{r}_0)}{\|\mathbf{n}\|^2} \mathbf{n}\cdot\mathbf{n} =0, \] so \(H\) lies on the plane.

Worked Example: Finding and Checking the Perpendicular Foot

Find the perpendicular foot from \(P=(2,0,3)\) to the plane \(x+2y+2z-6=0\), then verify its distance from \(P\).

Here \(\mathbf{n}=\langle1,2,2\rangle\). A point on the plane is \(Q=(6,0,0)\), since \(6+0+0-6=0\). Thus \(\mathbf{p}-\mathbf{q}=\langle-4,0,3\rangle\), and \[ \mathbf{n}\cdot(\mathbf{p}-\mathbf{q})=-4+0+6=2, \qquad \|\mathbf{n}\|^2=1+4+4=9. \] The foot is \[ \mathbf{h} =\langle2,0,3\rangle-\frac{2}{9}\langle1,2,2\rangle =\left\langle\frac{16}{9},-\frac{4}{9},\frac{23}{9}\right\rangle. \] Check that it is on the plane: \[ \frac{16}{9}+2\left(-\frac{4}{9}\right) +2\left(\frac{23}{9}\right)-6 =\frac{16-8+46}{9}-\frac{54}{9}=0. \] The displacement from the foot to \(P\) is \[ \mathbf{p}-\mathbf{h} =\left\langle\frac{2}{9},\frac{4}{9},\frac{4}{9}\right\rangle =\frac{2}{9}\mathbf{n}. \] It is parallel to the normal, confirming perpendicularity. Its length is \[ \|\mathbf{p}-\mathbf{h}\| =\sqrt{\frac{4+16+16}{81}} =\sqrt{\frac{36}{81}}=\frac{2}{3}. \] Finally, the plane expression at \(P\) is \(2+0+6-6=2\), so the distance formula gives \(|2|/\sqrt{9}=2/3\), matching the foot calculation.

A Common Error: Forgetting to Normalize

A frequent mistake is to use the plane expression itself as the distance. Its value depends on the scale chosen for the equation, whereas geometric distance does not. For example, multiplying a plane equation by a nonzero constant changes the expression’s value but does not move the plane.

Worked Example: Correcting a Missing Denominator

For \(2x+2y+z-3=0\), find the distance from \(P=(1,1,1)\). A wrong calculation evaluates the left side as \(2+2+1-3=2\) and reports distance \(2\). That is not the distance: the normal has length \[ \sqrt{2^2+2^2+1^2}=\sqrt{9}=3. \] The correct distance is therefore \(|2|/3=2/3\).

Check the scale issue by rewriting the same plane as \(4x+4y+2z-6=0\). The expression at \(P\) is now \(4+4+2-6=4\), but the normal length is \(\sqrt{16+16+4}=6\). The quotient is again \(4/6=2/3\). Both equations describe the same plane and give the same distance only when the normal length is included.

Choosing the Correct Result

Use signed distance when the side of the plane matters, such as when comparing points against a fixed normal orientation. Use absolute distance when only the shortest nonnegative length is requested. If the point is on the plane, the numerator vanishes and both results are zero. A quick check is to confirm that the denominator is the length of the three normal coefficients and that the constant term appears only in the numerator.

Identify the normal.
For \(ax+by+cz+d=0\), use \(\langle a,b,c\rangle\).

Evaluate the plane expression at the point.
For the general form, this is \(ax_0+by_0+cz_0+d\).

Divide by the normal’s length.
Use \(\sqrt{a^2+b^2+c^2}\), then take absolute value if ordinary distance is required.

Check Your Understanding

For each question, show the normal length and distinguish signed distance from ordinary distance where relevant.

Find the distance from \((0,0,0)\) to \(2x-y+2z-6=0\).
For the plane through \((1,0,0)\) with normal \(\langle1,2,2\rangle\), find the signed distance from \((2,2,1)\).
Explain why multiplying every coefficient in a plane equation by the same nonzero constant does not change the distance formula’s result.
What does a negative signed distance indicate, and what happens to its sign if the normal is reversed?
State the formula for the perpendicular foot using a non-unit normal, and explain why its denominator is the squared normal length.