AS & A Level · Practice questions · The vector equation of a line

Start here, travel this way.

Six original Cambridge-style questions on the vector equation of a line: forming it, using the parameter, testing points and working with distances along the line.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

The method

A point, a direction, a parameter.

Form: \(\mathbf{r} = \mathbf{a} + t\mathbf{b}\), with \(\mathbf{a}\) a point on the line and \(\mathbf{b}\) a direction.
Two points: \(\mathbf{b} = \overrightarrow{AB} = \mathbf{b} - \mathbf{a}\); either point can be \(\mathbf{a}\).
Point on line? Find \(t\) from one component, then check it in the other two.
Same line? Parallel directions plus one shared point.
Distance: a parameter change of \(t\) moves \(|t|\,|\mathbf{b}|\).

Recall

[2 marks]

The line \(l\) passes through \(A(3, 0, -2)\) and is parallel to \(\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}\). Write down a vector equation for \(l\) and find the point on \(l\) with parameter \(-3\).

\(\mathbf{r} = 3\mathbf{i} - 2\mathbf{k} + t(\mathbf{i} - 2\mathbf{j} + 4\mathbf{k})\). ✓
\(t = -3\): \((3 - 3,\ 0 + 6,\ -2 - 12) = (0, 6, -14)\). ✓

Calculation

[4 marks]

Find a vector equation of the line through \(A(1, 4, -2)\) and \(B(3, 1, 2)\). Determine whether each of the points \(C(7, -5, 10)\) and \(D(-1, 7, -5)\) lies on the line.

\(\overrightarrow{AB} = (2, -3, 4)\), so \(\mathbf{r} = (1, 4, -2) + t(2, -3, 4)\). ✓
\(C\): \(1 + 2t = 7\) gives \(t = 3\); \(y = 4 - 9 = -5\) ✓, \(z = -2 + 12 = 10\) ✓. \(C\) lies on the line. ✓
\(D\): \(1 + 2t = -1\) gives \(t = -1\); \(y = 4 + 3 = 7\) ✓, but \(z = -2 - 4 = -6 \ne -5\). \(D\) does not lie on the line. ✓

Proof

[3 marks]

Show that \(\mathbf{r} = (2, -1, 3) + s(1, 2, -1)\) and \(\mathbf{r} = (5, 5, 0) + t(-2, -4, 2)\) are equations of the same line.

\((-2, -4, 2) = -2(1, 2, -1)\), so the lines are parallel. ✓
\((5, 5, 0)\) on the first line: \(2 + s = 5\) gives \(s = 3\); \(y = -1 + 6 = 5\) ✓, \(z = 3 - 3 = 0\) ✓. ✓
Parallel lines with a common point coincide, so they are the same line. ✓

Calculation

[3 marks]

The line \(l\) has equation \(\mathbf{r} = (1, 2, -1) + t(1, -2, 2)\). The point \((p, q, 7)\) lies on \(l\). Find \(p\) and \(q\).

\(z\): \(-1 + 2t = 7\), so \(t = 4\). ✓
\(p = 1 + 4 = 5\) and \(q = 2 - 8 = -6\). ✓

Analysis

[4 marks]

The line \(l\) has equation \(\mathbf{r} = (3, 1, -2) + t(2, -2, 1)\) and \(A\) is the point \((3, 1, -2)\). Find the coordinates of the two points on \(l\) whose distance from \(A\) is 9.

\(|(2, -2, 1)| = \sqrt{4 + 4 + 1} = 3\), so the distance from \(A\) is \(3|t|\). ✓
\(3|t| = 9\) gives \(t = 3\) or \(t = -3\). ✓
\(t = 3\): \((9, -5, 1)\). \(t = -3\): \((-3, 7, -5)\). ✓

Analysis

[5 marks]

Relative to an origin \(O\) on horizontal ground, with \(z\) measured vertically upwards in metres, a straight cable runs from the top of a mast at \(T(2, -1, 12)\) through a ring at \(R(4, 0, 9)\) and is anchored to the ground at \(G\).

(a) Find a vector equation for the line of the cable. (b) Find the coordinates of \(G\). (c) Find the length of cable from \(T\) to \(G\), correct to 3 significant figures.

(a) \(\overrightarrow{TR} = (2, 1, -3)\), so \(\mathbf{r} = (2, -1, 12) + t(2, 1, -3)\). ✓
(b) Ground: \(12 - 3t = 0\), so \(t = 4\) and \(G = (2 + 8,\ -1 + 4,\ 0) = (10, 3, 0)\). ✓
(c) \(TG = 4 \times |(2, 1, -3)| = 4\sqrt{14} = 15.0\) m (3 s.f.). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.