Equation of a Plane

A normal vector $\mathbf{n}$ to a plane is any vector that is perpendicular to every vector lying in the plane. If a plane is tilted, its normal vector points 'straight out' of the surface. For example, the xy-plane has normal vector $\mathbf{n} = (0, 0, 1)$ because every vector in that plane has zero z-component, making it perpendicular to the z-axis direction.

Let the plane pass through a known point $A$ with position vector $\mathbf{a}$, and let $\mathbf{r}$ be the position vector of any general point $P$ on the plane. The vector $\overrightarrow{AP} = \mathbf{r} - \mathbf{a}$ lies in the plane, so it must be perpendicular to $\mathbf{n}$. Using the dot product condition for perpendicularity gives the vector form: $\mathbf{n} \cdot (\mathbf{r} - \mathbf{a}) = 0$, which can be rewritten as $\mathbf{n} \cdot \mathbf{r} = \mathbf{n} \cdot \mathbf{a}$. Since $\mathbf{n} \cdot \mathbf{a}$ is a fixed scalar $d$, the vector form is $\mathbf{n} \cdot \mathbf{r} = d$.

Write $\mathbf{n} = (a, b, c)$ and $\mathbf{r} = (x, y, z)$. Computing the dot product $\mathbf{n} \cdot \mathbf{r} = ax + by + cz$ gives the Cartesian (or scalar) form of the plane: $ax + by + cz = d$. The constants $a$, $b$, $c$ are exactly the components of the normal vector, which is why you can read off $\mathbf{n}$ immediately from the Cartesian equation. For instance, the plane $2x - y + 3z = 7$ has normal vector $\mathbf{n} = (2, -1, 3)$.

Sometimes it is useful to use a unit normal vector $\hat{\mathbf{n}} = \frac{\mathbf{n}}{|\mathbf{n}|}$. The equation becomes $\hat{\mathbf{n}} \cdot \mathbf{r} = p$, where $p = \hat{\mathbf{n}} \cdot \mathbf{a}$ is the perpendicular distance from the origin to the plane. This is the normal form, and $p \geq 0$ by convention. For example, for $2x - y + 3z = 7$, the magnitude is $|\mathbf{n}| = \sqrt{4 + 1 + 9} = \sqrt{14}$, so $p = \frac{7}{\sqrt{14}} = \frac{\sqrt{14}}{2}$.

Given three non-collinear points $P_1, P_2, P_3$ on the plane, form two vectors in the plane: $\mathbf{u} = \overrightarrow{P_1 P_2}$ and $\mathbf{v} = \overrightarrow{P_1 P_3}$. The normal vector is their cross product: $\mathbf{n} = \mathbf{u} \times \mathbf{v}$. Then substitute any of the three known points into $\mathbf{n} \cdot \mathbf{r} = d$ to find $d$. This technique connects the equation of a plane directly to cross products.

• a, b, c — Normal vector components (dimensionless): The components of the normal vector n perpendicular to the plane; they determine the plane's orientation in space.
• x, y, z — Coordinates of a general point (length units): The coordinates of any point P lying on the plane; they are the variables in the equation.
• d — Scalar constant (dimensionless): Equals n · a, the dot product of the normal vector with the position vector of any known point on the plane. It fixes the plane's position relative to the origin.

Problem. Find the Cartesian equation of the plane that passes through the points P₁ = (1, 2, 3), P₂ = (2, 0, 1), and P₃ = (0, 1, 4). Then find the perpendicular distance from the origin to this plane.

Answer. Cartesian equation: 4x − y + 3z = 11; perpendicular distance from origin = 11/√26 = 11√26/26 ≈ 2.16 units.

• Vectors in Three Dimensions — The vector form of the plane equation relies entirely on vector notation and the dot product; a solid understanding of 3D vectors is prerequisite.
• The Dot Product — The condition n · (r − a) = 0 that defines the plane is a direct application of the dot product's geometric meaning: two vectors are perpendicular when their dot product is zero.
• The Cross Product — The cross product of two vectors lying in a plane produces the normal vector, which is the key method for finding a plane's equation from three given points.
• Distance from a Point to a Plane — Once the equation of a plane is known, the formula for the perpendicular distance from a point to the plane follows directly and is a major application of the normal form.
• Lines in Three-Dimensional Space — Lines and planes are the fundamental objects of 3D geometry; understanding their equations together is essential for solving intersection and angle problems.