Parametrizing a Line Through Two Points

From Two Points to a Parametrized Line

A line in space can be specified by a point on it and a nonzero direction vector. When two distinct points are given, they supply both pieces of information: either point can serve as the starting point, and the displacement from one to the other gives a direction along the line.

This construction is useful because it turns geometric data—two locations—directly into a vector-valued function. The parameter then records how far, and in which direction, the point moves along the line.

The Displacement Vector

Let \(A\) and \(B\) be points in three-dimensional space, with position vectors \(\mathbf{a}\) and \(\mathbf{b}\). The displacement from \(A\) to \(B\) is the vector obtained by subtracting the starting position vector from the ending position vector:

\mathbf{d}=\mathbf{b}-\mathbf{a}.

If \(A\ne B\), then \(\mathbf{d}\ne\mathbf{0}\), so it can serve as a direction vector for the line. Starting at \(A\), moving \(t\) copies of this displacement gives the position vector \(\mathbf{a}+t\mathbf{d}\).

Definition: If \(A\) and \(B\) are distinct points with position vectors \(\mathbf{a}\) and \(\mathbf{b}\), a parametrization of the full line through them is

\mathbf{r}(t)=\mathbf{a}+t(\mathbf{b}-\mathbf{a}),\qquad t\in\mathbb{R}.

Here \(t\) is a real parameter, and \(\mathbf{b}-\mathbf{a}\) is the displacement from \(A\) to \(B\).

The parameter values \(t=0\) and \(t=1\) have special meanings. At \(t=0\), the displacement term contributes nothing, so the position is \(A\). At \(t=1\), the full displacement from \(A\) to \(B\) has been added, so the position is \(B\). Other real values continue along the same line, including values less than \(0\) and greater than \(1\).

Why the Formula Gives the Line

The formula starts at \(A\) and changes position only by scalar multiples of the fixed vector \(\mathbf{b}-\mathbf{a}\). Thus every point it produces lies in the direction of the line through \(A\) and \(B\). Conversely, any point on that line is reached by moving some real multiple of this direction vector from \(A\). This establishes that the formula traces the whole line, not just a portion of it.

Derivation: A point on the line through \(A\) in the direction \(\mathbf{d}\) has position vector \(\mathbf{a}+t\mathbf{d}\) for some \(t\in\mathbb{R}\). Since \(B\) lies on the line and its displacement from \(A\) is \(\mathbf{b}-\mathbf{a}\), choose \(\mathbf{d}=\mathbf{b}-\mathbf{a}\). Then the line has the stated parametrization. Because \(A\ne B\), this direction vector is nonzero, and its real scalar multiples extend in both directions along the line.

In component form, if \(A=(a_1,a_2,a_3)\) and \(B=(b_1,b_2,b_3)\), then subtracting coordinates gives

\mathbf{b}-\mathbf{a} = \langle b_1-a_1,\ b_2-a_2,\ b_3-a_3\rangle.

Therefore the coordinate functions of the line are \(x=a_1+t(b_1-a_1)\), \(y=a_2+t(b_2-a_2)\), and \(z=a_3+t(b_3-a_3)\). A zero component in the displacement means that coordinate remains constant as the parameter changes.

Worked Examples

Worked Example: A Line Through Two Points in Space

Find a parametrization of the line through \(A=(1,-2,3)\) and \(B=(4,0,-1)\), then verify that it passes through both points.

First find the displacement from \(A\) to \(B\) by subtracting coordinate by coordinate:

\mathbf{b}-\mathbf{a} = \langle 4-1,\ 0-(-2),\ -1-3\rangle = \langle 3,\ 2,\ -4\rangle.

Use the position vector of \(A\) as the starting point and this displacement as the direction:

\mathbf{r}(t) = \langle 1,-2,3\rangle+t\langle 3,2,-4\rangle = \langle 1+3t,\ -2+2t,\ 3-4t\rangle,\qquad t\in\mathbb{R}.

Check \(t=0\):

\mathbf{r}(0) = \langle 1+3(0),\ -2+2(0),\ 3-4(0)\rangle = \langle 1,-2,3\rangle = \mathbf{a}.

Check \(t=1\):

\mathbf{r}(1) = \langle 1+3(1),\ -2+2(1),\ 3-4(1)\rangle = \langle 4,0,-1\rangle = \mathbf{b}.

Both endpoint checks succeed, so the parametrization passes through the two required points. Since \(t\) ranges over all real numbers, it traces the full line.

Worked Example: A Constant Coordinate

Find and check a parametrization of the line through \(C=(-2,5,1)\) and \(D=(3,5,-2)\).

The displacement is

\mathbf{d} = \langle 3-(-2),\ 5-5,\ -2-1\rangle = \langle 5,\ 0,\ -3\rangle.

Thus a parametrization is

\mathbf{r}(t) = \langle -2,5,1\rangle+t\langle 5,0,-3\rangle = \langle -2+5t,\ 5,\ 1-3t\rangle,\qquad t\in\mathbb{R}.

The second displacement component is zero, so the second coordinate remains \(5\). Check the first point at \(t=0\):

\mathbf{r}(0) = \langle -2+5(0),\ 5,\ 1-3(0)\rangle = \langle -2,5,1\rangle = \mathbf{c}.

Check the second point at \(t=1\):

\mathbf{r}(1) = \langle -2+5(1),\ 5,\ 1-3(1)\rangle = \langle 3,5,-2\rangle = \mathbf{d}.

The unchanged coordinate is not an error: both specified points have second coordinate \(5\), and the line’s direction has no second-coordinate component.

Worked Example: Reversing the Order of the Points

Let \(E=(2,-1,4)\) and \(F=(-1,3,0)\). Construct a parametrization starting at \(F\), and check both specified points.

Because the starting point is \(F\), calculate the displacement from \(F\) to \(E\):

\mathbf{e}-\mathbf{f} = \langle 2-(-1),\ -1-3,\ 4-0\rangle = \langle 3,-4,4\rangle.

The parametrization is therefore

\mathbf{q}(s) = \langle -1,3,0\rangle+s\langle 3,-4,4\rangle = \langle -1+3s,\ 3-4s,\ 4s\rangle,\qquad s\in\mathbb{R}.

At \(s=0\),

\mathbf{q}(0) = \langle -1+3(0),\ 3-4(0),\ 4(0)\rangle = \langle -1,3,0\rangle = \mathbf{f}.

At \(s=1\),

\mathbf{q}(1) = \langle -1+3(1),\ 3-4(1),\ 4(1)\rangle = \langle 2,-1,4\rangle = \mathbf{e}.

Changing the starting point reverses the direction of travel as the parameter increases, but it does not change the geometric line. Both parametrizations cover the full line because their parameters range over all real numbers.

Choosing the Order and Checking the Result

Either specified point can be chosen as the starting point. Starting at \(A\) uses the displacement \(\mathbf{b}-\mathbf{a}\); starting at \(B\) uses \(\mathbf{a}-\mathbf{b}\). These direction vectors are negatives of one another, so the parameter runs along the same line in opposite directions. The endpoint checks should match the chosen order: starting at \(A\) gives \(t=0\) at \(A\) and \(t=1\) at \(B\).

A useful routine is to write the starting position vector first, subtract in the same order to form the displacement, and only then substitute parameter values to check. Keeping the subtraction order consistent prevents a sign error from being hidden in the final component formula.

Common Mistake: Subtracting in the Wrong Order

Suppose the points are \(A=(1,2,0)\) and \(B=(4,-1,2)\). The displacement from \(A\) to \(B\) is \(\mathbf{b}-\mathbf{a}\), not \(\mathbf{a}-\mathbf{b}\). A mistaken subtraction gives

\mathbf{a}-\mathbf{b} = \langle 1-4,\ 2-(-1),\ 0-2\rangle = \langle -3,3,-2\rangle.

If this incorrect vector is added to the starting point \(A\), the claimed parametrization gives, at \(t=1\),

\langle 1,2,0\rangle+\langle -3,3,-2\rangle = \langle -2,5,-2\rangle \ne \langle 4,-1,2\rangle.

The endpoint check exposes the error immediately. Correcting the subtraction gives

\mathbf{b}-\mathbf{a} = \langle 4-1,\ -1-2,\ 2-0\rangle = \langle 3,-3,2\rangle,

and now the check at \(t=1\) succeeds:

\langle 1,2,0\rangle+\langle 3,-3,2\rangle = \langle 4,-1,2\rangle = \mathbf{b}.

The reverse displacement is not inherently wrong. It is wrong only when paired with the other point as the starting position without adjusting the parametrization. Always check the point produced at \(t=0\) and at \(t=1\).

When the Construction Applies

The two points must be distinct to determine a unique line. If \(A=B\), then \(\mathbf{b}-\mathbf{a}=\mathbf{0}\), and the formula \(\mathbf{a}+t(\mathbf{b}-\mathbf{a})\) gives the same point for every \(t\). It does not trace a line, because no direction has been specified. In that case, the given data identify only one point, and additional information is needed to choose a line through it.

For distinct points, allowing every real value of the parameter is essential to the full-line construction. The endpoint checks verify that the specified points lie on the parametrized line; the nonzero displacement and unrestricted real parameter ensure that the formula extends beyond those endpoints in both directions. Restricting the parameter changes which part of the line is traced, so it should not be done when the requested object is the full line.

Check Your Understanding

Use the displacement-vector construction and verify your results by substituting the relevant parameter values.

Find a parametrization of the line through \(P=(0,2,-1)\) and \(Q=(3,-2,5)\). What are the position vectors at \(t=0\) and \(t=1\)?
For points \(R=(4,1,2)\) and \(S=(4,-3,0)\), identify the displacement vector and the coordinate that remains constant along the line.
Explain why a parametrization of the full line must allow parameter values below \(0\) and above \(1\).
What does the formula based on two points produce if the points coincide? Why does it fail to specify a line?
When the starting point is \(B\) instead of \(A\), what displacement vector should be used, and how does the direction of travel change?