Intersections of a Line and a Plane

How a Line Meets a Plane

A line in three-dimensional space can meet a plane at one point, lie entirely within it, or be parallel to it without meeting it. The key is to substitute the line’s parametric coordinates into the plane equation. This turns a geometric question into a single equation in one parameter.

The result is especially useful because the cases are determined by two calculations: the dot product of the plane’s normal vector and the line’s direction vector, and whether a point on the line satisfies the plane equation.

Definition: Write a line as \(\mathbf{x}=\mathbf{p}+t\mathbf{v}\), where \(\mathbf{p}\) is a point on the line, \(\mathbf{v}\ne\mathbf{0}\) is its direction vector, and \(t\in\mathbb{R}\). Write a plane as \(\mathbf{n}\cdot\mathbf{x}=d\), where \(\mathbf{n}\ne\mathbf{0}\) is a normal vector and \(d\) is a constant.

In coordinates, if \(\mathbf{p}=(x_0,y_0,z_0)\) and \(\mathbf{v}=(a,b,c)\), the line has parametric equations \(x=x_0+at\), \(y=y_0+bt\), and \(z=z_0+ct\). If \(\mathbf{n}=(A,B,C)\), the plane equation is \(Ax+By+Cz=d\). A point belongs to the line–plane intersection precisely when its coordinates satisfy both descriptions.

The Substitution Test

For the line \(\mathbf{x}=\mathbf{p}+t\mathbf{v}\) to lie in the plane \(\mathbf{n}\cdot\mathbf{x}=d\), its parameter must satisfy

\mathbf{n}\cdot(\mathbf{p}+t\mathbf{v})=d.

Using the distributive property of the dot product gives a linear equation in \(t\):

\mathbf{n}\cdot\mathbf{p}+t(\mathbf{n}\cdot\mathbf{v})=d, \qquad t(\mathbf{n}\cdot\mathbf{v})=d-\mathbf{n}\cdot\mathbf{p}.

The coefficient \(\mathbf{n}\cdot\mathbf{v}\) tells us whether the line’s direction has a component along the plane’s normal. The right-hand side tells us whether the starting point \(\mathbf{p}\) is already on the plane.

Theorem: Let \(\mathbf{x}=\mathbf{p}+t\mathbf{v}\), with \(\mathbf{v}\ne\mathbf{0}\), be a line, and let \(\mathbf{n}\cdot\mathbf{x}=d\), with \(\mathbf{n}\ne\mathbf{0}\), be a plane.

If \(\mathbf{n}\cdot\mathbf{v}\ne 0\), the line intersects the plane at exactly one point.
If \(\mathbf{n}\cdot\mathbf{v}=0\) and \(\mathbf{n}\cdot\mathbf{p}=d\), the line lies entirely in the plane.
If \(\mathbf{n}\cdot\mathbf{v}=0\) and \(\mathbf{n}\cdot\mathbf{p}\ne d\), the line is parallel to and disjoint from the plane.

When the intersection is unique, the parameter and intersection point are

t=\frac{d-\mathbf{n}\cdot\mathbf{p}}{\mathbf{n}\cdot\mathbf{v}}, \qquad \mathbf{x}=\mathbf{p}+t\mathbf{v}.

This formula requires \(\mathbf{n}\cdot\mathbf{v}\ne 0\). If that denominator is zero, do not divide by it; check separately whether \(\mathbf{p}\) satisfies the plane equation.

Why the Three Cases Are Exhaustive

Substitution reduces the problem to one scalar equation, so its possible outcomes explain the entire classification. The following argument also justifies why a nonzero dot product gives exactly one intersection, rather than merely showing how to calculate a candidate.

Substitute the line into the plane.
Every point on the line has the form \(\mathbf{p}+t\mathbf{v}\). It is in the plane exactly when \(\mathbf{n}\cdot\mathbf{p}+t(\mathbf{n}\cdot\mathbf{v})=d\).

Consider a nonzero coefficient.
If \(\mathbf{n}\cdot\mathbf{v}\ne0\), the equation has the single solution \(t=(d-\mathbf{n}\cdot\mathbf{p})/(\mathbf{n}\cdot\mathbf{v})\). A line has exactly one point for each parameter value, so this produces exactly one intersection point.

Consider a zero coefficient.
If \(\mathbf{n}\cdot\mathbf{v}=0\), the equation reduces to \(\mathbf{n}\cdot\mathbf{p}=d\). If this equality holds, every \(t\) satisfies the plane equation, so every point on the line lies in the plane. If it fails, no \(t\) satisfies it, so the line and plane are disjoint. In this case the line’s direction is parallel to the plane because it is perpendicular to the plane’s normal.

These cases cover all possibilities: the coefficient is either zero or nonzero, and when it is zero the remaining equality either holds or fails.

Worked Examples

Worked Example: A Unique Intersection

Find where the line \(x=1+2t,\ y=-1+t,\ z=3-t\) intersects the plane \(2x-y+z=7\), if it does.

Substitute each coordinate expression into the plane equation. Keep the negative sign in front of \(y\):

$$ 2(1+2t)-(-1+t)+(3-t)=7. $$

Expand and combine terms:

2+4t+1-t+3-t=7, $$ $$ 6+2t=7, \qquad 2t=1, \qquad t=\frac{1}{2}.

Substitute this parameter value into the line:

x=1+2\left(\frac{1}{2}\right)=2, \qquad y=-1+\frac{1}{2}=-\frac{1}{2}, \qquad z=3-\frac{1}{2}=\frac{5}{2}.

The candidate intersection is \(\left(2,-\frac{1}{2},\frac{5}{2}\right)\). Check it in the plane:

2(2)-\left(-\frac{1}{2}\right)+\frac{5}{2} =4+\frac{1}{2}+\frac{5}{2} =7.

It satisfies the plane equation, so the line intersects the plane at \(\left(2,-\frac{1}{2},\frac{5}{2}\right)\).

Worked Example: A Parallel Line That Does Not Meet the Plane

Determine the relationship between the line \(\mathbf{x}=(0,0,0)+t(1,1,3)\) and the plane \(x+2y-z=4\).

The plane has normal vector \(\mathbf{n}=(1,2,-1)\), and the line has direction vector \(\mathbf{v}=(1,1,3)\). Their dot product is

\mathbf{n}\cdot\mathbf{v} =(1)(1)+(2)(1)+(-1)(3) =1+2-3 =0.

Therefore the line’s direction is parallel to the plane. Now check the point \(\mathbf{p}=(0,0,0)\):

\mathbf{n}\cdot\mathbf{p} =(1)(0)+(2)(0)+(-1)(0) =0\ne4.

The starting point is not on the plane. More directly, points on the line have coordinates \((t,t,3t)\), and substituting them gives \(t+2t-3t=0\), which never equals \(4\). The line is parallel to and disjoint from the plane.

Worked Example: A Line Contained in the Plane

Determine whether the line \(\mathbf{x}=(3,0,0)+t(2,-1,0)\) intersects the plane \(x+2y-z=3\) at one point, or lies in it.

The plane normal is \(\mathbf{n}=(1,2,-1)\), and the direction vector is \(\mathbf{v}=(2,-1,0)\). First calculate

\mathbf{n}\cdot\mathbf{v} =(1)(2)+(2)(-1)+(-1)(0) =2-2+0 =0.

The line’s direction is parallel to the plane. Its starting point satisfies the plane equation because

\mathbf{n}\cdot(3,0,0) =(1)(3)+(2)(0)+(-1)(0) =3.

This equals the plane’s constant, so the line lies in the plane. Check directly for every parameter value: the line has coordinates \((3+2t,-t,0)\), and

$$ (3+2t)+2(-t)-0 =3+2t-2t =3. $$

The plane equation holds for every \(t\), confirming that the entire line is contained in the plane.

A Common Sign Error

When substituting coordinates, a minus sign before a variable applies to the entire expression for that variable. In the first worked example, \(y=-1+t\), so the term \(-y\) becomes \(-(-1+t)=1-t\).

A mistaken substitution would be \(-y=-1+t\), leading to the incorrect equation

2+4t-1+t+3-t=7, \qquad 4+4t=7, \qquad t=\frac{3}{4}.

This parameter gives the point \(\left(\frac{5}{2},-\frac{1}{4},\frac{9}{4}\right)\), but substituting it into the plane shows the error:

2\left(\frac{5}{2}\right) -\left(-\frac{1}{4}\right) +\frac{9}{4} =5+\frac{1}{4}+\frac{9}{4} =\frac{15}{2}\ne7.

The correction is to distribute the negative sign correctly: \(-(-1+t)=1-t\). Then the equation is \(6+2t=7\), giving \(t=\frac{1}{2}\), and the resulting point does satisfy the plane equation.

Why This Test Is Useful

A parametric line describes all of its points using one parameter, while a plane equation tests whether any given point lies in the plane. Substitution connects those descriptions directly. It avoids guessing a point of intersection and makes clear when an intersection cannot be unique.

The dot product test also helps organize the calculation. If \(\mathbf{n}\cdot\mathbf{v}\ne0\), solve for the parameter and check the resulting point. If \(\mathbf{n}\cdot\mathbf{v}=0\), do not try to solve by division: test the line’s point in the plane instead. Equality means the entire line is in the plane; inequality means the line and plane do not meet.

Takeaway: Substitute the line into the plane. A nonzero coefficient of the parameter gives one intersection; a zero coefficient requires checking whether the line’s point is on the plane.

Check Your Understanding

For each question, identify the calculation that determines the line–plane relationship.

For the line \(\mathbf{x}=(1,0,0)+t(1,0,0)\) and plane \(x+y+z=3\), find the intersection point.
For the line \(\mathbf{x}=(0,1,0)+t(1,-1,0)\) and plane \(x+y+z=1\), determine whether the line is contained in the plane.
What does \(\mathbf{n}\cdot\mathbf{v}=0\) tell you, and what additional check distinguishes containment from a disjoint parallel line?
Why is division by \(\mathbf{n}\cdot\mathbf{v}\) not allowed when that dot product is zero?
For a proposed intersection point, what substitution verifies that it lies in the plane?