Planes Containing a Line or Two Directions

Introduction

A plane is determined by a point and a normal vector: the normal is perpendicular to every direction that lies in the plane. This makes it possible to construct a plane from geometric information that does not directly give a normal. For example, a line supplies one direction, and a second point off the line supplies another direction from the line to that point. A cross product of those two directions gives a normal.

The essential task is not just to calculate a cross product. It is to check that the resulting plane satisfies all the conditions. In particular, a plane containing a line must contain both a point on the line and its direction.

The Conditions for a Line to Lie in a Plane

Let a line be given in vector form by \(\mathbf{r}=\mathbf{a}+t\mathbf{d}\), where \(\mathbf{a}\) is the position vector of a point \(A\) on the line, \(\mathbf{d}\ne\mathbf{0}\) is its direction vector, and \(t\in\mathbb{R}\). A plane with normal vector \(\mathbf{n}\ne\mathbf{0}\) passing through \(A\) has equation \(\mathbf{n}\cdot(\mathbf{x}-\mathbf{a})=0\). For the entire line to lie in this plane, the line’s direction must also be perpendicular to the normal.

If coordinates are used, a plane through \(A=(x_0,y_0,z_0)\) with normal \(\mathbf{n}=(a,b,c)\) has equation \(a(x-x_0)+b(y-y_0)+c(z-z_0)=0\). The coefficients \(a,b,c\) must not all be zero.

Constructing a Plane from a Line and an Off-Line Point

Suppose the line passes through \(A\) with direction \(\mathbf{d}\), and the specified point is \(Q\), not on the line. The displacement vector \(\mathbf{w}=\overrightarrow{AQ}\) gives a second direction in the desired plane. Since \(Q\) is not on the line, \(\mathbf{w}\) is not a scalar multiple of \(\mathbf{d}\). Thus, \(\mathbf{d}\) and \(\mathbf{w}\) are independent, and their cross product is nonzero and perpendicular to both.

Taking \(\mathbf{n}=\mathbf{d}\times\mathbf{w}\), the plane through \(A\) with normal \(\mathbf{n}\) contains the line direction and the displacement to \(Q\). This constructs the plane; the conditions can also be verified directly by substitution and dot products.

Two Independent Directions at a Common Point

Now suppose two direction vectors \(\mathbf{u}\) and \(\mathbf{v}\) are based at the same point \(P\). If they are not parallel, they span the directions within a plane through \(P\). A normal to that plane is \(\mathbf{u}\times\mathbf{v}\). The point-normal equation then gives the plane. Reversing the order of the cross product reverses the normal but does not change the plane, because a normal and any nonzero scalar multiple of it describe the same plane.

Why the Cross Product Gives the Required Plane

The construction follows from the defining properties of the cross product. For vectors \(\mathbf{u}\) and \(\mathbf{v}\), \(\mathbf{u}\times\mathbf{v}\) is perpendicular to each. If the vectors are independent, their cross product is nonzero. These facts give both the normal condition and the necessary nondegeneracy for a unique plane.

For uniqueness, any plane satisfying the conditions has a normal perpendicular to both independent vectors \(\mathbf{d}\) and \(\mathbf{w}\). In three-dimensional space, the vectors perpendicular to two independent directions form a one-dimensional set: all are scalar multiples of \(\mathbf{d}\times\mathbf{w}\). Thus any such plane has a normal parallel to the constructed normal. Since it must also contain \(A\), its equation is the same plane.

The same reasoning applies when the two given directions are \(\mathbf{u}\) and \(\mathbf{v}\) at a common point \(P\). If they are independent, \(\mathbf{u}\times\mathbf{v}\ne\mathbf{0}\) gives a normal, and the plane through \(P\) is unique. Independence is essential to this uniqueness claim.

When the Conditions Do Not Determine a Unique Plane

If the specified point \(Q\) lies on the given line, then \(Q-A\) is a scalar multiple of the line direction. The cross product is zero, so it cannot be used as a normal. This is not a failed calculation: the data do not determine a unique plane. Every plane containing the line also contains \(Q\), so there are infinitely many possibilities.

Likewise, two parallel or zero direction vectors do not determine a unique plane through a common point. If the vectors are parallel and nonzero, they give only one direction; many planes through that point contain it. If one vector is zero, it contributes no direction at all. A unique plane from two directions requires two independent vectors.

A Common Mistake: Checking Only One Point

A frequent error is to check that a plane contains a line’s base point and then assume it contains the whole line. A line has infinitely many points; the direction condition must be checked as well. For the line \((x,y,z)=(1+2t,-2+t,-t)\), consider the proposed plane \(2x+y-z=0\). Substitution yields

This expression is not zero for every \(t\). At \(t=0\), it is \(0\), so the base point \((1,-2,0)\) is indeed in the proposed plane. But at \(t=1\), it is \(6\), so the point \((3,-1,-1)\) on the line is not in the plane. The same failure appears from the direction test: the proposed normal \((2,1,-1)\) has dot product \(2\cdot2+1\cdot1+(-1)(-1)=6\) with the line direction \((2,1,-1)\), not zero. A valid check must cover the base point and the direction.

Summary and Practical Checks

When constructing a plane, first identify the points and directions that must belong to it. For a line and a point off it, use the line direction and displacement from a point on the line to the specified point. For two independent directions at a common point, use those directions directly. In both cases, the cross product supplies a normal, and the point-normal equation supplies the plane.

• Confirm that the cross product is nonzero. A zero result signals dependent directions and no unique plane from those data.
• Substitute a point that must lie in the plane.
• For a line, check that its direction has zero dot product with the normal.
• For two directions, check that both have zero dot product with the normal.
• If the supplied point lies on the line, recognize that infinitely many planes contain the given data.