Point-Normal Equation of a Plane¶

Definition¶

The point-normal equation of a plane through $(x_0,y_0,z_0)$ with normal vector $\boldsymbol{n}=(a,b,c)$ is $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ $0$ .

Interpretation¶

The point-normal equation describes a plane by a test rather than a recipe: a point $\boldsymbol{P}=(x,y,z)$ lies on the plane exactly when the displacement $\boldsymbol{P}-\boldsymbol{P}_0$ from the fixed point is orthogonal to the normal vector $\boldsymbol{n}$ .

A shaded plane with a fixed point P0, another point P in it, and a normal vector n perpendicular to the displacement from P0 to P.

In the figure the displacement $\boldsymbol{P}-\boldsymbol{P}_0$ lies in the plane and $\boldsymbol{n}=(a,b,c)$ stands perpendicular to it. Writing this with the inner product and expanding the coordinate sum gives the defining equation: $\boldsymbol{n}\cdot(\boldsymbol{P}-\boldsymbol{P}_0)=a(x-x_0)+b(y-y_0)+c(z-z_0)=0$ $0$ . For a point off the plane, the displacement has a component along $\boldsymbol{n}$ and the left-hand side is nonzero.

Expanding further gives $ax+by+cz=d$ with $d=\boldsymbol{n}\cdot\boldsymbol{P}_0$ $P_{0}$ . The coefficients of $x,y,z$ form the normal vector, and the constant fixes the position: changing $d$ while keeping $(a,b,c)$ shifts the plane parallel to itself. One equation in three unknowns leaves two free variables, matching the two parameters of the parametric description. For the plane $3x-7y+z=0$ of the Examples, $z=-3x+7y$ $7 y$ ; choosing $x=s$ and $y=t$ gives $s(1,0,-3)+t(0,1,7)$ $t (0, 1, 7)$ , which recovers $\boldsymbol{B}$ and $\boldsymbol{C}$ as direction vectors.

The same pattern in $\mathbb{R}^2$ , $a(x-x_0)+b(y-y_0)=0$ $0$ , describes a line with normal $(a,b)$ , and in $\mathbb{R}^n$ a hyperplane. In linear programming, the boundary of a constraint $a_1x_1+a_2x_2+a_3x_3\le d$ is such a plane, with the coefficient vector as its normal.

A common slip is to treat the equation as unique. Multiplying by a nonzero number, as in $6x-14y+2z=0$ $0$ , describes the same plane, because the normal vector is only fixed up to a nonzero multiple.

Key Results¶

It can be rewritten as $ax+by+cz=d$ $d$ , where $d=ax_0+by_0+cz_0$ $c z_{0}$ .
The normal vector is read from the coefficients of $x,y,z$ .

Examples¶

A plane through $(0,0,0)$ with normal $(0,0,1)$ has equation $z=0$ $0$ .
For $\boldsymbol{A}=(0,0,0)$ $(0, 0, 0)$ , $\boldsymbol{B}=(1,0,-3)$ and $\boldsymbol{C}=(0,1,7)$ $(0, 1, 7)$ , a normal $\boldsymbol{n}=(a,b,c)$ satisfies $\boldsymbol{n}\cdot(\boldsymbol{B}-\boldsymbol{A})=a-3c=0$ and $\boldsymbol{n}\cdot(\boldsymbol{C}-\boldsymbol{A})=b+7c=0$ $0$ . Taking $c=1$ gives $\boldsymbol{n}=(3,-7,1)$ and the plane $3x-7y+z=0$ $0$ .

Prerequisites¶

Sources¶

Math II: Linear Algebra · Week 5

Aliases: point-normal form, normal form of a plane, point-normal equation Symbols: real numbers

Last updated: 20 October 2026 03:02