Equation of a plane

A line is one-dimensional; a plane is the next rung up — a flat surface with no thickness, stretching infinitely in every direction it lies in. In ISC Class 12's Three-Dimensional Geometry unit, a plane is pinned down the same way a line is: by combining a point on it with a direction — except now the direction is a normal vector, perpendicular to the whole surface rather than lying along it. From that single idea — "the normal is perpendicular to every vector inside the plane" — fall four equivalent ways to write a plane's equation (one-point, general Cartesian, normal, and intercept form), a determinant test for three points to define a plane, and a formula for how far any point sits from it. This article walks through all four forms, derives the distance formula, and reworks both Pass-1 examples end to end.

The core insight

A plane is the set of points whose position vector, measured from any point already on the plane, is perpendicular to the plane's fixed normal vector.

The core idea

One-point (normal) form. Take a point on the plane with position vector $\vec a$ and a normal vector $\vec n$ (direction ratios $a,b,c$ ). Any other point $\vec r$ lies on the plane exactly when the vector $\vec r-\vec a$ , which lies in the plane, is perpendicular to $\vec n$ :

(\vec r-\vec a)\cdot\vec n = 0.

where: $\vec a$ is a known point on the plane, $\vec r=(x,y,z)$ is a general point on it, and $\vec n=(a,b,c)$ is the normal direction — not a direction lying in the plane.

General Cartesian form. Expanding the dot product above with $\vec a=(x_1,y_1,z_1)$ gives $a(x-x_1)+b(y-y_1)+c(z-z_1)=0$ , which simplifies to

ax+by+cz+d=0,

where $d=-(ax_1+by_1+cz_1)$ . This is the form you will manipulate most often — $a,b,c$ are always the normal's direction ratios, never a direction ratio of a line drawn inside the plane.

Normal form and intercept form. If $\hat n=(l,m,n)$ is the unit normal and $p\geq0$ is the plane's perpendicular distance from the origin, $\vec r\cdot\hat n=p$ , i.e. $lx+my+nz=p$ . If instead the plane cuts the axes at $(a,0,0),(0,b,0),(0,0,c)$ ,

\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1.

Three non-collinear points. For $P_1(x_1,y_1,z_1)$ , $P_2$ , $P_3$ , a point $(x,y,z)$ lies on their plane exactly when the vector from $P_1$ to it is coplanar with $P_1P_2$ and $P_1P_3$ — a condition a $3\times3$ determinant captures directly:

\begin{vmatrix} x-x_1 & y-y_1 & z-z_1\\ x_2-x_1 & y_2-y_1 & z_2-z_1\\ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0.

Equivalently, the cross product $P_1P_2\times P_1P_3$ is a normal to the plane, and expanding that determinant along the first row reproduces exactly the same coefficients — the two routes are the same calculation in different clothing.

Figure

A plane shown edge-on as a tilted line with its normal vector drawn perpendicular to it from a point on the plane, and a second point off the plane connected by a dashed segment marked D for the perpendicular distance

Distance of a point from a plane

The core derivation of this article is the formula for how far a given point sits from a plane whose Cartesian equation you already know, since it is the calculation that ties every other form together and appears in almost every exam problem on planes.

Derivation ADistance of a point from a plane

The goal is to find, purely from the Cartesian equation $ax+by+cz+d=0$ of a plane and the coordinates $(x_1,y_1,z_1)$ of any point in space, a formula for the shortest (perpendicular) distance $D$ between that point and the plane, using only the projection idea already familiar from the dot product — no new machinery beyond vectors already covered in Vector Algebra.

Step 1. Pick any point $Q(x_0,y_0,z_0)$ that actually lies on the plane, so $ax_0+by_0+cz_0+d=0$ , which means $d=-(ax_0+by_0+cz_0)$ . This point exists because the plane is nonempty, and its exact location will drop out of the final answer.

Step 2. Form the vector from $Q$ to the given point $P(x_1,y_1,z_1)$ : $\vec{QP}=(x_1-x_0,\,y_1-y_0,\,z_1-z_0)$ . The shortest distance from $P$ to the plane is the length of the component of $\vec{QP}$ along the unit normal $\hat n=(a,b,c)/\sqrt{a^2+b^2+c^2}$ , since the normal is the only direction perpendicular to the plane, and any shorter path from $P$ to the plane would have to cross it at an angle, making it longer.

Step 3. That component is the dot product $\vec{QP}\cdot\hat n$ , so

D = \left|\vec{QP}\cdot\hat n\right| = \frac{\left|a(x_1-x_0)+b(y_1-y_0)+c(z_1-z_0)\right|}{\sqrt{a^2+b^2+c^2}}.

Expand the numerator: $a x_1+b y_1+c z_1 -(ax_0+by_0+cz_0)$ . But Step 1 showed $ax_0+by_0+cz_0=-d$ , so the numerator becomes $ax_1+by_1+cz_1+d$ — the point $Q$ has vanished entirely, exactly as it should, since the distance cannot depend on which point of the plane you happened to pick.

Step 4. Read off the result:

D = \frac{|ax_1+by_1+cz_1+d|}{\sqrt{a^2+b^2+c^2}}.

This is why the coefficients $a,b,c,d$ of the general Cartesian form are so useful for distance problems: you can substitute the point directly into the plane's own equation, take the absolute value, and divide by the length of the normal — no need to locate a specific point on the plane at all.

Worked example

Find the Cartesian equation of the plane through the point $(2,3,-1)$ with normal direction ratios $(1,-2,3)$ , then find the distance of $(4,-1,2)$ from it, and finally check the same plane also passes through $(-3,7,-4)$ .

Step 1. One-point form: $1\cdot(x-2)-2\cdot(y-3)+3\cdot(z+1)=0$ .

Step 2. Expand: $x-2-2y+6+3z+3=0 \implies x-2y+3z+7=0$ , so $a=1,b=-2,c=3,d=7$ .

Step 3. Distance of $(4,-1,2)$ using the boxed formula from the derivation:

D=\frac{|1(4)-2(-1)+3(2)+7|}{\sqrt{1+4+9}}=\frac{|4+2+6+7|}{\sqrt{14}}=\frac{19}{\sqrt{14}}\text{ units}\approx 5.08\text{ units}.

Step 4. Check the plane passes through $(2,3,-1)$ : $2-6-3+7=0$ ✓. Check the extra point $(-3,7,-4)$ : $-3-14-12+7=-22\neq0$ , so this point does not lie on the plane — a reminder that only substituting into $ax+by+cz+d$ and getting exactly zero confirms membership; a nonzero value of $22$ (divided by $\sqrt{14}$ ) would in fact be that point's own distance from the plane.

Answer: the plane is $x-2y+3z+7=0$ , the distance of $(4,-1,2)$ from it is $\dfrac{19}{\sqrt{14}}\approx5.08$ units, and $(-3,7,-4)$ is not on the plane.

Common traps

Confusing direction ratios of the plane's normal ( $a,b,c$ in $ax+by+cz+d=0$ ) with direction ratios of a line lying inside the plane — only the normal's ratios go into the plane equation and the distance formula.
Forgetting the absolute value in the distance formula, or forgetting to divide by $\sqrt{a^2+b^2+c^2}$ — both mistakes give a wrong magnitude, not just a wrong sign.
Skipping normalisation when converting a general Cartesian equation to normal form $lx+my+nz=p$ : you must divide every term by $\sqrt{a^2+b^2+c^2}$ and choose signs so $p\geq0$ .
Assuming intercept form always exists — it fails whenever the plane passes through the origin or runs parallel to a coordinate axis, since one of $a,b,c$ in $\tfrac{x}{a}+\tfrac{y}{b}+\tfrac{z}{c}=1$ would then require division by zero.

Ask the tutor: stuck converting a plane between one-point, general Cartesian, normal, and intercept form, or want another distance-of-a-point example worked through? Ask the AI tutor to take your specific plane through all the steps.

Why does the point Q used in the distance derivation not appear in the final formula?

Because the perpendicular distance from a fixed external point to a flat plane cannot depend on which point of that plane you happened to use as a reference — the plane is the same object no matter where you stand on it. Algebraically, Step 3 of the derivation showed that $ax_0+by_0+cz_0$ always equals $-d$ for any point $(x_0,y_0,z_0)$ on the plane, so it cancels out and only $d$ itself survives. That cancellation is the whole reason the general Cartesian form $ax+by+cz+d=0$ is so convenient for distance problems.

How is the three-point determinant method related to using direction ratios directly?

They are the same computation viewed two ways. Given three points, the cross product of two vectors lying in the plane, say $P_1P_2\times P_1P_3$ , gives a vector perpendicular to both, which is exactly the plane's normal; you can then plug those direction ratios into the one-point form. Expanding the $3\times3$ determinant along its first row produces those same normal direction ratios as coefficients of $x,y,z$ . Use whichever method you find less error-prone in an exam — both earn full marks.

If a plane's equation comes out as 2x + 3y minus 3z equals 5, how do I convert it to normal form?

First write it as $ax+by+cz+d=0$ , here $2x+3y-3z-5=0$ , so $a=2,b=3,c=-3,d=-5$ . Compute $\sqrt{a^2+b^2+c^2}=\sqrt{4+9+9}=\sqrt{22}$ , then divide every term by that value: $\tfrac{2}{\sqrt{22}}x+\tfrac{3}{\sqrt{22}}y-\tfrac{3}{\sqrt{22}}z=\tfrac{5}{\sqrt{22}}$ . Since the right-hand side is already positive, no sign flip is needed, and this is the normal form $lx+my+nz=p$ with $p=5/\sqrt{22}$ , the plane's perpendicular distance from the origin.

Equation of a plane

The core idea

Distance of a point from a plane

Worked example

Common traps

Before this

Where this leads