Plane Equations in R3

From a Linear Equation to a Plane

A point in three-dimensional space is recorded by three coordinates, \((x,y,z)\). A linear equation in these coordinates imposes one condition on which points are allowed. For example, \(2x-y+2z=5\) does not usually specify a single point: many different triples satisfy it. Geometrically, those triples make up a plane in \(\mathbb{R}^3\).

Throughout, the scalars are real. A triple \((x,y,z)\) is a solution triple of an equation when substituting its coordinates makes the equation true. The set of all its solution triples is its solution set. This is the same condition as saying that the corresponding point lies on the geometric object described by the equation.

Here, \(X=(x,y,z)\) and \(P=(x_0,y_0,z_0)\) are written as coordinate triples, and the dot product is \((a,b,c)\cdot(u,v,w)=au+bv+cw\). Thus, the normal vector gives a direction perpendicular to the plane. The equation says that the displacement from a point on the plane has zero dot product with this normal.

Why Every Nontrivial Equation Gives a Plane

Proof. Assume \(a,b,c\) are not all zero. First, the equation has a solution. If \(a\ne0\), take \(P=(d/a,0,0)\). If \(a=0\) and \(b\ne0\), take \(P=(0,d/b,0)\). If \(a=b=0\), then \(c\ne0\), and we take \(P=(0,0,d/c)\). In each case, substitution gives \(ax_0+by_0+cz_0=d\). Now write \(\mathbf{n}=(a,b,c)\) and \(X=(x,y,z)\). Since \(\mathbf{n}\cdot P=d\), the original equation is equivalent to \[ \mathbf{n}\cdot X=d \quad\Longleftrightarrow\quad \mathbf{n}\cdot(X-P)=0. \] The vector \(\mathbf{n}\) is nonzero by assumption, so this is exactly the definition of the plane through \(P\) with normal \(\mathbf{n}\). \(\square\)

The word nontrivial matters. If \(a=b=c=0\), the equation becomes \(0=d\). When \(d\ne0\), no point satisfies it; when \(d=0\), every point in \(\mathbb{R}^3\) satisfies it. Neither case describes a single plane.

The equation’s coefficients also help describe the plane’s position. If \(d=0\), the origin \((0,0,0)\) is a solution, so the plane passes through the origin. If \(d\ne0\), the origin is not a solution. Multiplying all four coefficients \(a,b,c,d\) by the same nonzero real number does not change the plane: both sides of the equation are scaled equally, so the equation remains true for exactly the same triples.

Reading a Plane and Checking Its Points

Displacements Within the Plane

Proof. Write \(P=(x_0,y_0,z_0)\) and \(X=(x,y,z)\). Since \(P\) lies on the plane, \(ax_0+by_0+cz_0=d\). The displacement \(X-P=(x-x_0,y-y_0,z-z_0)\) has dot product \[ \mathbf{n}\cdot(X-P) =a(x-x_0)+b(y-y_0)+c(z-z_0) =(ax+by+cz)-(ax_0+by_0+cz_0). \] The second parenthesis equals \(d\). Therefore this dot product is zero exactly when \(ax+by+cz=d\). By the definition of perpendicular vectors, dot product zero means \(X-P\) is perpendicular to \(\mathbf{n}\). This proves both directions. \(\square\)

This result describes how to move from one point on the plane to another: the displacement must be perpendicular to the normal. It also explains why testing a point by substitution is decisive. The point belongs to the plane precisely when its equation holds; being near a point that lies on the plane is not enough.

Common Mistakes

Rejecting a point without substituting all three coordinates. For \(2x-y+2z=5\), a mistaken check might say that \((1,1,1)\) lies on the plane because its coordinates are all \(1\). Substitution gives \(2(1)-1+2(1)=3\), not \(5\), so the point is not on the plane. By contrast, substituting \((2,1,1)\) gives \(4-1+2=5\), so that point does lie on it.

Treating a missing variable as a missing coordinate. In \(2x+3y=6\), the coefficient of \(z\) is zero; \(z\) is still a coordinate of points in \(\mathbb{R}^3\). For instance, \((3,0,0)\) and \((3,0,4)\) both satisfy the equation. The equation describes a plane extending in the \(z\)-direction, not just a line drawn in the \(xy\)-plane.

Confusing the plane with its normal. For \(2x-y+2z=5\), the vector \((2,-1,2)\) is not a point on the plane. It is a direction perpendicular to the plane. To check a point, substitute its three coordinates into the equation; to identify the normal, read off the three variable coefficients.

When interpreting a single linear equation in three variables, keep the roles distinct: the solutions are points on the plane, the coefficient vector is its normal, and the right-hand side determines whether the origin lies on it. A zero coefficient indicates that the corresponding coordinate does not affect the equation, not that points lack that coordinate.