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P3 7 · P3

Vectors

Pure Mathematics 3 · Paper 3 only

This topic is on Paper 3 only. If you are sitting AS Level with Paper 2, you do not need it.

🎯What you need to be able to do

  • Use component and \( \mathbf{i}, \mathbf{j}, \mathbf{k} \) notation, and convert between them.
  • Find the magnitude of a vector and the unit vector in its direction.
  • Find the displacement vector between two points, and the distance between them.
  • Calculate the scalar product and use it to find the angle between two vectors.
  • Test for perpendicular and parallel vectors.
  • Write the vector equation of a line, and find a point on it for a given parameter.
  • Determine whether two lines intersect, are parallel, or are skew — and find the point of intersection when it exists.
  • Find the angle between two lines.

📚The mathematics

Notation and magnitude

A vector has both magnitude and direction. Two equivalent notations:

\[ \mathbf{v} = \begin{pmatrix} 3 \\ -1 \\ 2 \end{pmatrix} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k} \]

The magnitude extends Pythagoras into three dimensions:

\[ |\mathbf{v}| = \sqrt{v_1^{2} + v_2^{2} + v_3^{2}} \]

and the unit vector in the direction of \( \mathbf{v} \) is \( \dfrac{\mathbf{v}}{|\mathbf{v}|} \) — divide each component by the magnitude. A unit vector always has magnitude 1, which is a free check on your arithmetic.

The displacement vector from \(A\) to \(B\) is

\[ \overrightarrow{AB} = \mathbf{b} - \mathbf{a} \]

— destination minus start. Getting this backwards reverses the direction of everything that follows. The distance \(AB\) is then \( \left|\overrightarrow{AB}\right| \).

The scalar product

\[ \mathbf{a}\cdot\mathbf{b} = a_1b_1 + a_2b_2 + a_3b_3 = |\mathbf{a}||\mathbf{b}|\cos\theta \]

Two forms, one number. Multiply corresponding components and add — the result is a scalar, not a vector, which is where the name comes from. Equating the two forms gives the angle:

\[ \cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|} \]

Two consequences are used constantly:

  • Perpendicular \( \iff \mathbf{a}\cdot\mathbf{b} = 0 \), since \( \cos 90^{\circ} = 0 \). This is the standard test, and it is usually the fastest route into a question asking you to prove a right angle.
  • Parallel \( \iff \) one vector is a scalar multiple of the other. Note this is not a scalar-product test — check whether the components are in a constant ratio.

The sign of the scalar product tells you the angle before you compute it: positive means acute, zero means right, negative means obtuse. Use that as a check on the final answer.

The vector equation of a line

\[ \mathbf{r} = \mathbf{a} + \lambda\mathbf{d} \]

A point you know, plus a parameter times a direction you travel in. Given two points \(A\) and \(B\) on the line, take \( \mathbf{a} \) as either position vector and \( \mathbf{d} = \overrightarrow{AB} \).

The same line has infinitely many valid equations: any point on it will do for \( \mathbf{a} \), and any non-zero multiple of \( \mathbf{d} \) will do for the direction. So an answer that differs from the mark scheme is not automatically wrong — check whether it describes the same line.

The angle between two lines is the angle between their direction vectors, found with the scalar product. The lines need not intersect for this to make sense. If the scalar product comes out negative you will get an obtuse angle; questions usually want the acute one, so take \( 180^{\circ} \) minus it, and say that you have.

Do two lines meet?

This is the classic Paper 3 question, and it has a fixed procedure.

  1. Check for parallel first. If the direction vectors are scalar multiples of each other, the lines are parallel — and then either identical or never meeting. No further work needed.
  2. If not parallel, set the two position expressions equal component by component. Use different parameters for the two lines, say \( \lambda \) and \( \mu \) — using the same letter for both is the single most common error and it forces a false conclusion.
  3. Solve two of the three equations simultaneously for \( \lambda \) and \( \mu \).
  4. Substitute into the third equation and check. If it is satisfied, the lines intersect; substitute back to find the point. If it is not, the lines are skew.
Step 4 is the question. Skew lines — non-parallel and non-intersecting — have no two-dimensional analogue, so intuition says any two non-parallel lines must cross. In three dimensions they usually do not. Solving two equations always gives you values for \( \lambda \) and \( \mu \); only the third equation reveals whether those values describe the same point. Omitting that check means you cannot answer the question at all, however good the algebra.

✏️Worked example

Lines \( l_1 \) and \( l_2 \) have equations \( \mathbf{r} = \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \) and \( \mathbf{r} = \begin{pmatrix} 4 \\ 3 \\ -2 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \). (a) Show that the lines intersect, and find the point of intersection. (b) Find the acute angle between the lines.

(a) The direction vectors are not scalar multiples of each other, so the lines are not parallel. Setting the position vectors equal component by component:

\( 1 + 2\lambda = 4 + \mu \)
\( \lambda = 3 - \mu \)
\( 2 - \lambda = -2 + 2\mu \)

From the second equation, \( \lambda = 3 - \mu \). Substituting into the first: \( 1 + 2(3 - \mu) = 4 + \mu \), so \( 7 - 2\mu = 4 + \mu \), giving \( 3\mu = 3 \) and \( \mu = 1 \). Then \( \lambda = 2 \).

Now check the third equation, which has not been used: LHS \( = 2 - 2 = 0 \), RHS \( = -2 + 2(1) = 0 \). They agree, so the lines do intersect.

Substituting \( \lambda = 2 \) into \( l_1 \) gives the point of intersection:

\[ \begin{pmatrix} 1 \\ 0 \\ 2 \end{pmatrix} + 2\begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \\ 0 \end{pmatrix} \]

(b) The angle between the lines is the angle between their directions. With \( \mathbf{d}_1 = \begin{pmatrix} 2 \\ 1 \\ -1 \end{pmatrix} \) and \( \mathbf{d}_2 = \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} \):

\[ \mathbf{d}_1\cdot\mathbf{d}_2 = (2)(1) + (1)(-1) + (-1)(2) = 2 - 1 - 2 = -1 \]

Both magnitudes are \( \sqrt{4+1+1} = \sqrt{6} \), so

\[ \cos\theta = \frac{-1}{\sqrt6 \times \sqrt6} = -\frac{1}{6} \;\Longrightarrow\; \theta = 99.6^{\circ} \]

That is obtuse, so the acute angle between the lines is \( 180^{\circ} - 99.6^{\circ} = 80.4^{\circ} \).

Check it. Substitute \( \mu = 1 \) into \( l_2 \): \( \begin{pmatrix} 4 \\ 3 \\ -2 \end{pmatrix} + \begin{pmatrix} 1 \\ -1 \\ 2 \end{pmatrix} = \begin{pmatrix} 5 \\ 2 \\ 0 \end{pmatrix} \) — the same point reached from \( l_1 \). ✓ Always confirm the intersection from both lines; it costs one line of working and catches any slip in solving for the parameters. And in (b), the negative scalar product predicted an obtuse angle before any inverse cosine was taken, so the subtraction to reach the acute angle was genuinely needed.
Two parameters, and the third equation. Using \( \lambda \) for both lines forces a false conclusion, because the two lines reach the meeting point at different parameter values — here \( \lambda = 2 \) but \( \mu = 1 \). And the third component equation is what decides the question: had it not balanced, the lines would have been skew despite the first two equations solving perfectly well. Never stop at step 3.

📝Practise

Work through these, then reveal the answer. Each question targets a different objective from the list above.

1. Find the unit vector in the direction of \( \mathbf{v} = 4\mathbf{i} - 2\mathbf{j} + 4\mathbf{k} \).
\( |\mathbf{v}| = \sqrt{16 + 4 + 16} = \sqrt{36} = 6 \). The unit vector is \( \tfrac{1}{6}\left(4\mathbf{i} - 2\mathbf{j} + 4\mathbf{k}\right) = \tfrac{2}{3}\mathbf{i} - \tfrac{1}{3}\mathbf{j} + \tfrac{2}{3}\mathbf{k} \). Check: \( \sqrt{\tfrac49 + \tfrac19 + \tfrac49} = \sqrt{1} = 1 \) ✓.
2. Points \( A(2, -1, 3) \) and \( B(5, 1, -1) \). Find \( \overrightarrow{AB} \) and the distance \(AB\).
\( \overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix} 5-2 \\ 1-(-1) \\ -1-3 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \\ -4 \end{pmatrix} \). Distance \( = \sqrt{9 + 4 + 16} = \sqrt{29} = 5.39 \).
3. Find the value of \(k\) for which \( \begin{pmatrix} 3 \\ k \\ -2 \end{pmatrix} \) and \( \begin{pmatrix} 1 \\ 4 \\ 5 \end{pmatrix} \) are perpendicular.
Perpendicular means the scalar product is zero: \( 3(1) + k(4) + (-2)(5) = 0 \), so \( 3 + 4k - 10 = 0 \), giving \( 4k = 7 \) and \( k = 1.75 \).
4. Find the angle between \( \mathbf{a} = 2\mathbf{i} + \mathbf{j} + 2\mathbf{k} \) and \( \mathbf{b} = 3\mathbf{i} - 4\mathbf{k} \).
\( \mathbf{a}\cdot\mathbf{b} = (2)(3) + (1)(0) + (2)(-4) = 6 + 0 - 8 = -2 \). Magnitudes: \( |\mathbf{a}| = \sqrt{4+1+4} = 3 \), \( |\mathbf{b}| = \sqrt{9+0+16} = 5 \). So \( \cos\theta = \dfrac{-2}{15} = -0.1333 \), giving \( \theta = 97.7^{\circ} \). The negative scalar product correctly predicted an obtuse angle.
5. Show that the lines \( \mathbf{r} = \begin{pmatrix} 1 \\ 2 \\ 0 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \) and \( \mathbf{r} = \begin{pmatrix} 3 \\ 0 \\ 1 \end{pmatrix} + \mu\begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} \) are parallel but not identical.
The direction vectors satisfy \( \begin{pmatrix} 2 \\ 2 \\ 2 \end{pmatrix} = 2\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \), so the lines are parallel. To show they are not the same line, test whether \( (3,0,1) \) lies on the first: it would need \( 1 + \lambda = 3 \) so \( \lambda = 2 \), but then the second component is \( 2 + 2 = 4 \ne 0 \). So the point is not on the first line and the lines are distinct parallel lines.
6. Lines \( \mathbf{r} = \begin{pmatrix} 0 \\ 1 \\ 2 \end{pmatrix} + \lambda\begin{pmatrix} 1 \\ 2 \\ 1 \end{pmatrix} \) and \( \mathbf{r} = \begin{pmatrix} 2 \\ 3 \\ 1 \end{pmatrix} + \mu\begin{pmatrix} -1 \\ 1 \\ 3 \end{pmatrix} \). Determine whether they intersect.
Not parallel (the directions are not multiples). Equating components: \( \lambda = 2 - \mu \); \( 1 + 2\lambda = 3 + \mu \); \( 2 + \lambda = 1 + 3\mu \).
From the first, \( \lambda = 2 - \mu \). Substituting into the second: \( 1 + 4 - 2\mu = 3 + \mu \), so \( 2 = 3\mu \) and \( \mu = \tfrac23 \), giving \( \lambda = \tfrac43 \).
Check the third: LHS \( = 2 + \tfrac43 = \tfrac{10}{3} \); RHS \( = 1 + 2 = 3 \). Not equal, so the lines do not intersect — they are skew. The third-equation check is the entire answer here.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • 3Blue1Brown — Essence of Linear Algebra, especially the episode on the dot product
  • GeoGebra 3D — plot two lines in space and rotate the view until you can see they miss each other
  • Khan Academy — vectors and the dot product