Constructing Lines from Geometric Conditions

From Geometric Conditions to a Line

A line in three-dimensional space can be described by a point on it and a direction in which it runs. Many construction problems give those pieces directly; others describe them indirectly, for example by giving two points on the line or requiring its direction to be perpendicular to specified vectors.

The central task is to translate the conditions into a point and a direction vector, then write the line equation. This tutorial focuses on that construction. It uses the parametric form of a line and the familiar dot-product and cross-product tests for perpendicularity and parallelism.

If \(\mathbf{p}=\langle x_0,y_0,z_0\rangle\) and \(\mathbf{v}=\langle a,b,c\rangle\), the parametric equations are \(x=x_0+at\), \(y=y_0+bt\), and \(z=z_0+ct\). Every value of \(t\) gives a point on the line, and varying \(t\) traces the entire line.

Finding a Direction from Two Points

If a line passes through two distinct points \(P\) and \(Q\), the displacement from \(P\) to \(Q\) is a direction vector. Subtract the coordinates of \(P\) from the corresponding coordinates of \(Q\). The order matters for the sign of the vector, but either subtraction order gives a direction for the same line.

Because the points are distinct, \(\mathbf{q}-\mathbf{p}\ne\mathbf{0}\), so this is a valid line direction. The line through them is \(\mathbf{r}(t)=\mathbf{p}+t(\mathbf{q}-\mathbf{p})\). At \(t=0\) it gives \(P\), and at \(t=1\) it gives \(Q\).

Why This Construction Gives the Required Line

The two-point construction works because the direction vector records exactly how to move from one specified point to the other. It also gives the only line through two distinct points. The reasoning is short but important.

This proves both existence and uniqueness: two distinct points determine exactly one line. If the two given points coincide, their displacement is zero, and the information does not determine a unique line; an additional direction condition would be needed.

Constructing a Line Parallel to a Given Direction

When a problem says that a line is parallel to a vector, that vector can be used as the line's direction, provided it is nonzero. If the given object is itself a line in parametric form, its direction vector is the coefficient of its parameter. A parallel line through a new point keeps that direction and replaces the point vector.

Directions from Perpendicularity Conditions

Some problems do not state a direction vector directly. Instead, they require the line to have a direction perpendicular to one or more given vectors. The dot product supplies the test: vectors \(\mathbf{d}\) and \(\mathbf{u}\) are perpendicular exactly when \(\mathbf{d}\cdot\mathbf{u}=0\). This condition can restrict possible directions, but a single perpendicularity condition usually leaves more than one possible direction in three-dimensional space.

If a direction must be perpendicular to two nonparallel vectors \(\mathbf{u}\) and \(\mathbf{v}\), their cross product provides such a direction. The cross product \(\mathbf{u}\times\mathbf{v}\) is perpendicular to each of its factors. It is nonzero when the two vectors are nonparallel, so it can then serve as a line direction. If the vectors are parallel, their cross product is zero and does not provide a direction.

Choosing a Convenient Direction Vector

A line does not have a unique direction vector. If \(\mathbf{d}\ne\mathbf{0}\), then \(k\mathbf{d}\) is a direction for the same line whenever \(k\ne0\). Indeed, as \(t\) ranges over all real numbers, \(tk\) also ranges over all real numbers. Thus \(\mathbf{p}+t(k\mathbf{d})\) traces the same points as \(\mathbf{p}+t\mathbf{d}\).

This lets you simplify a direction vector when possible. For example, \(\langle 2,-2,-2\rangle\) can be replaced by \(\langle 1,-1,-1\rangle\). A negative multiple is also valid: reversing the direction of travel changes the parameterization but not the geometric line.

When there is more than one condition on a direction, check all of them rather than relying only on a remembered formula. Dot products directly check perpendicularity. For parallelism, the direction vectors must be scalar multiples of one another; equivalently, their cross product is zero. The zero vector itself is not a valid direction for a line.

A Common Mistake: Using the Wrong Difference

Subtracting coordinates in the wrong order is not necessarily an error, but failing to keep the resulting signs consistent is. For example, for the points \(P=(1,-2,3)\) and \(Q=(4,0,-1)\), the vector from \(P\) to \(Q\) is \(\langle 3,2,-4\rangle\). A common incorrect calculation is to use \(\langle 3,2,4\rangle\), changing only the last sign. That vector is not the displacement between the points.

Using the incorrect vector from \(P\) would give \(\mathbf{r}(t)=\langle 1,-2,3\rangle+t\langle 3,2,4\rangle\). At \(t=1\), this produces \((4,0,7)\), not \(Q=(4,0,-1)\). The corrected subtraction gives \(-1-3=-4\), so the direction is \(\langle 3,2,-4\rangle\); substituting \(t=1\) then gives \((4,0,-1)\). Always check the constructed line against the conditions, especially by substituting the parameter value for each specified point.

A Reliable Construction Checklist

For a line described by geometric conditions, separate the information into the point and direction before writing the equation.

• Identify a point the line must pass through.
• Determine a nonzero direction vector: subtract endpoint coordinates, reuse a stated parallel direction, or calculate a vector satisfying perpendicularity conditions.
• Check that the direction meets every stated condition.
• Write \(\mathbf{r}(t)=\mathbf{p}+t\mathbf{d}\), then check that the specified point is obtained for an appropriate parameter value.

The same line may have many valid parametric equations because the chosen point can be any point on the line and the direction can be any nonzero scalar multiple of a direction vector. What matters is that the equation describes the required set of points and satisfies the geometric conditions.