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How do you write the equation of a line in three dimensions, and how do you find where two lines meet?

The vector and Cartesian equations of a straight line in three dimensions, the direction vector, and finding the intersection of two lines or showing that they are parallel or skew.

A focused answer to the OCR A-Level Further Mathematics A content on the equations of lines in three dimensions, covering the vector equation with a point and a direction vector, the Cartesian (symmetric) form, and finding the intersection of two lines or determining that they are parallel, intersecting or skew.

Generated by Claude Opus 4.811 min answer

Reviewed by: AI editorial process; not yet individually human-reviewed

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  1. What this dot point is asking
  2. The vector equation of a line
  3. The Cartesian (symmetric) form
  4. Intersecting, parallel and skew lines
  5. Try this

What this dot point is asking

OCR wants you to write the equation of a straight line in three dimensions in vector form (a point plus a parameter times a direction vector) and in Cartesian (symmetric) form, to identify the direction vector, and to determine the relationship between two lines: whether they are parallel, intersect at a point, or are skew (neither parallel nor intersecting).

The vector equation of a line

A line is fixed by one point on it and a direction. Starting at the point and moving any multiple of the direction vector traces out the whole line.

The Cartesian (symmetric) form

Writing each coordinate in terms of λ\lambda and then equating the expressions for λ\lambda removes the parameter and gives the symmetric Cartesian form. This is useful for reading off a point and direction, and for comparison with a plane equation.

Intersecting, parallel and skew lines

In three dimensions two lines need not meet even when not parallel; such lines are skew. The test has two stages: first compare directions, then, if not parallel, look for a common point.

  • Parallel. The direction vectors are scalar multiples.
  • Intersecting. Not parallel, and the position vectors can be made equal: solving two components for λ,μ\lambda, \mu gives values that also satisfy the third.
  • Skew. Not parallel, and the third component is inconsistent, so no common point exists.

Lines in three dimensions lead directly into planes, where a line can lie in, be parallel to, or pierce a plane.

Try this

Q1. Write the direction vector of the line x−12=y+3−1=z4\dfrac{x - 1}{2} = \dfrac{y + 3}{-1} = \dfrac{z}{4}. [1 mark]

  • Cue. Read the denominators: direction 2i−j+4k2\mathbf{i} - \mathbf{j} + 4\mathbf{k}.

Q2. Are the lines with directions (246)\begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix} and (123)\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} parallel? [1 mark]

  • Cue. Yes; the first is twice the second, so the directions are scalar multiples.

Exam-style practice questions

Practice questions written in the style of OCR exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

OCR 20185 marksA line passes through A(1,2,−1)A(1, 2, -1) and B(3,−1,2)B(3, -1, 2). Find a vector equation of the line and its Cartesian equation.
Show worked answer →

Direction vector (M1): AB→=b−a=(3−1)i+(−1−2)j+(2−(−1))k=2i−3j+3k\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (3 - 1)\mathbf{i} + (-1 - 2)\mathbf{j} + (2 - (-1))\mathbf{k} = 2\mathbf{i} - 3\mathbf{j} + 3\mathbf{k} (A1).

Vector equation through AA (M1): r=(12−1)+λ(2−33)\mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ -1 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -3 \\ 3 \end{pmatrix} (A1).

Cartesian form by equating λ\lambda (A1): x−12=y−2−3=z+13\dfrac{x - 1}{2} = \dfrac{y - 2}{-3} = \dfrac{z + 1}{3}.

Markers reward the direction vector, the vector equation through a point, and the symmetric Cartesian form.

OCR 20226 marksThe lines r1=(102)+λ(111)\mathbf{r}_1 = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} and r2=(231)+μ(21−1)\mathbf{r}_2 = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix}. Determine whether they intersect, are parallel or are skew.
Show worked answer →

The direction vectors (111)\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} and (21−1)\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} are not multiples of each other, so the lines are not parallel (M1).

Set the position vectors equal and solve two of the three component equations (M1): from xx: 1+λ=2+2μ1 + \lambda = 2 + 2\mu; from yy: λ=3+μ\lambda = 3 + \mu. Substituting λ=3+μ\lambda = 3 + \mu into the first gives 1+3+μ=2+2μ1 + 3 + \mu = 2 + 2\mu, so μ=2\mu = 2 and λ=5\lambda = 5 (A1).

Check the third (z) component (M1): line 1 gives 2+λ=72 + \lambda = 7; line 2 gives 1−μ=−11 - \mu = -1. These disagree (7≠−17 \ne -1), so the equations are inconsistent (A1).

The lines are not parallel and do not meet, so they are skew (A1).

Markers reward checking parallelism, solving two components, testing the third for consistency, and the skew conclusion.

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