Paper 3 only. In three dimensions a line has no gradient and no \(y = mx + c\). Instead you describe it as a journey: start at a known point, then travel any multiple of a fixed direction.
The line through the point with position vector \(\mathbf{a}\), in the direction of \(\mathbf{b}\), is \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\). Each value of the parameter \(t\) gives one point on the line, and every point on the line comes from exactly one \(t\).
Fig. 1: start at A (position vector a) and step along the direction b; the parameter t counts the steps, so t = 2 reaches the point R with r = a + 2b, and negative t goes backwards past A
Section 01
What every symbol means.
Read \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\) as a set of directions: from the origin go to \(A\), then walk \(t\) lots of \(\mathbf{b}\). As \(t\) runs through all real numbers, the end point \(R\) traces out the whole line (Fig. 1).
\(\mathbf{r}\) is the position vector of a general point \(R\) on the line: \(\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}\).
\(\mathbf{a}\) is the position vector of one known point on the line. Any point on the line will do.
\(\mathbf{b}\) is a direction vector. Any non-zero multiple of it describes the same direction.
\(t\) (or \(\lambda\), \(s\), \(\mu\)) is a scalar parameter. Use different letters for different lines.
Worked example Points from parameters
The line \(l\) passes through \(A(1, -2, 3)\) and is parallel to \(2\mathbf{i} + \mathbf{j} - \mathbf{k}\). Write down a vector equation of \(l\), and find the point on \(l\) where \(z = 0\).
Point plus multiple of direction: \(\mathbf{r} = \begin{pmatrix}1\\-2\\3\end{pmatrix} + t\begin{pmatrix}2\\1\\-1\end{pmatrix}\).
A general point is \((1 + 2t,\ -2 + t,\ 3 - t)\).
\(z = 0\): \(3 - t = 0\), so \(t = 3\).
Substitute \(t = 3\): \((1 + 6,\ -2 + 3,\ 0)\).
Answer: \(\mathbf{r} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k} + t(2\mathbf{i} + \mathbf{j} - \mathbf{k})\); the point is \((7, 1, 0)\).
Worked example Is the point on the line?
Determine whether the points \(P(5, 0, 1)\) and \(Q(-3, -4, 6)\) lie on the line \(l\) above.
For \(P\): the \(x\)-coordinate gives \(1 + 2t = 5\), so \(t = 2\).
Check the other two with \(t = 2\): \(y = -2 + 2 = 0\) ✓ and \(z = 3 - 2 = 1\) ✓. One \(t\) fits all three, so \(P\) is on \(l\).
Answer: \(P\) lies on \(l\) (with \(t = 2\)); \(Q\) does not.
Examiner trap
Finding \(t\) from ONE component proves nothing. A point lies on the line only if the SAME \(t\) works in all three components; always check the remaining two.
Section 02
A line through two points.
Two points fix a line. Use either point as \(\mathbf{a}\) and the displacement between them as \(\mathbf{b}\). Different choices give equations that look different but describe the same line.
Worked example Two points
Find a vector equation of the line through \(A(2, 0, -1)\) and \(B(5, 4, 1)\).
The direction is \(\mathbf{b} - \mathbf{a}\), never the position vector of the second point. Writing \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\) with \(\mathbf{b} = \overrightarrow{OB}\) gives a line through \(A\) parallel to \(OB\), which usually misses \(B\) completely.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
Quick check+10 XP
In \(\mathbf{r} = (3, -1, 2) + t(1, 4, -2)\), which is a direction vector of the line?
Quick check+10 XP
Find the point with parameter \(t = -2\) on \(\mathbf{r} = (3, -1, 2) + t(1, 4, -2)\).
Quick check+10 XP
Which is a vector equation of the line through \(A(1, 0, 4)\) and \(B(3, 2, 1)\)?
Quick check+10 XP
Which point lies on \(\mathbf{r} = (2, 1, 0) + t(1, -1, 3)\)?
Section 03
Distances along a line.
Moving from \(t = 0\) to \(t = k\) covers a distance of \(|k|\,|\mathbf{b}|\). This turns distance questions into equations in \(t\).
Worked example Points a given distance away
For the line \(\mathbf{r} = (1, -2, 3) + t(2, 1, -1)\), find the two points on the line at a distance \(2\sqrt{6}\) from \(A(1, -2, 3)\).
\(|\mathbf{b}| = \sqrt{4 + 1 + 1} = \sqrt{6}\), so a parameter change of \(t\) moves \(|t|\sqrt{6}\).
A laser at \(S(0, 0, 10)\) shines through a small window at \(W(4, 2, 8)\) and hits the ground, the plane \(z = 0\), at \(G\). Find \(G\) and the length of the beam \(SG\), giving your answer to 3 significant figures.
The parameter is not a distance unless \(\mathbf{b}\) is a unit vector. With \(\mathbf{b} = (4, 2, -2)\), \(t = 5\) means \(5\sqrt{24}\) units, not 5 units.
Stage 2 · Exam
Exam-style questions
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The line \(l\) passes through \(A(2, -1, 5)\) and \(B(4, 3, -1)\). Find the point on \(l\) with \(x\)-coordinate 8.
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The point \((a, 7, -1)\) lies on the line \(\mathbf{r} = (1, 1, 3) + t(2, 3, -2)\). Find \(a\).
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\(A\) is \((1, 2, 3)\) and \(B\) is \((3, 6, 1)\). Which of these is NOT an equation of the line \(AB\)?
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The line \(m\) passes through \((4, -1, 2)\) and is parallel to the line through \(P(1, 3, -2)\) and \(Q(-1, 4, 2)\). Find the point on \(m\) where \(y = 2\).
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Lines in space, sorted.
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