Paper 3 only. The scalar product turns two vectors into a single number that knows the angle between them. Zero means perpendicular, and that one fact unlocks angles in solids, perpendicular lines and the shortest distance from a point to a line.
\(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}|\,|\mathbf{b}|\cos\theta = a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}\). So \(\cos\theta = \dfrac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}|\,|\mathbf{b}|}\), and non-zero vectors are perpendicular exactly when \(\mathbf{a} \cdot \mathbf{b} = 0\).
Fig. 1: cuboid OABCDEFG with OA = 4, OC = 3 and OD = 2; the angle ODB between the edge DO and the diagonal DB is 68.2°
Section 01
Two formulas, one number.
The scalar (dot) product has two forms. The component form is how you calculate it; the \(\cos\theta\) form is what it means. Putting them equal gives the angle between two vectors.
Components: \(\begin{pmatrix}a_{1}\\a_{2}\\a_{3}\end{pmatrix} \cdot \begin{pmatrix}b_{1}\\b_{2}\\b_{3}\end{pmatrix} = a_{1}b_{1} + a_{2}b_{2} + a_{3}b_{3}\), a number, not a vector.
Geometry: \(\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}|\,|\mathbf{b}|\cos\theta\), where \(\theta\) is the angle between the vectors when they point AWAY from a common point.
Sign: positive means \(\theta\) is acute, negative means obtuse, zero means \(\theta = 90^\circ\).
\(\theta = \cos^{-1}\left(-\tfrac{2}{3}\right) = 131.8^\circ\). Obtuse, as the negative product predicted.
Answer: \(\theta = 131.8^\circ\) (1 d.p.)
Worked example Perpendicular vectors
Find \(p\) given that \(p\mathbf{i} + 3\mathbf{j} - 2\mathbf{k}\) and \(2\mathbf{i} + p\mathbf{j} + 5\mathbf{k}\) are perpendicular.
Perpendicular, so the scalar product is zero: \(2p + 3p - 10 = 0\).
\(5p = 10\).
Answer: \(p = 2\)
Your turn
Show that \(3\mathbf{i} - \mathbf{j} + 2\mathbf{k}\) and \(\mathbf{i} + 5\mathbf{j} + \mathbf{k}\) are perpendicular.
\(3(1) + (-1)(5) + 2(1) = 3 - 5 + 2 = 0\), and neither vector is zero, so they are perpendicular.
Examiner trap
The angle in the formula is between vectors pointing away from the same point. For the angle \(ABC\) in a triangle use \(\overrightarrow{BA}\) and \(\overrightarrow{BC}\); using \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) gives \(180^\circ\) minus the angle you want.
Section 02
Angles between lines.
The angle between two lines is the angle between their direction vectors, and by convention it is the ACUTE angle. If the cosine comes out negative, the acute angle is \(180^\circ\) minus what your calculator gives (equivalently, use the modulus of the scalar product).
Worked example Acute angle between lines
Find the acute angle between the lines \(\mathbf{r} = (1, 2, 3) + s(1, -1, 2)\) and \(\mathbf{r} = (5, 1, 4) + t(2, 1, -3)\).
Use only the directions: \((1, -1, 2) \cdot (2, 1, -3) = 2 - 1 - 6 = -5\).
Magnitudes \(\sqrt{6}\) and \(\sqrt{14}\), so \(\cos\theta = \dfrac{-5}{\sqrt{6}\sqrt{14}} = \dfrac{-5}{\sqrt{84}}\), giving \(\theta = 123.1^\circ\).
The lines make angles of \(123.1^\circ\) and \(180^\circ - 123.1^\circ\) with each other; the acute one is wanted.
Answer: \(56.9^\circ\) (1 d.p.)
Worked example Angle in a cuboid
The cuboid \(OABCDEFG\) in Fig. 1 has \(\overrightarrow{OA} = 4\mathbf{i}\), \(\overrightarrow{OC} = 3\mathbf{j}\), \(\overrightarrow{OD} = 2\mathbf{k}\), with \(B\) the vertex opposite \(O\) on the base. Find angle \(ODB\).
\(\cos\theta = \dfrac{4}{2\sqrt{29}} = \dfrac{2}{\sqrt{29}}\), so \(\theta = 68.2^\circ\).
Check: triangle \(ODB\) has a right angle at \(O\) with \(OB = 5\), \(OD = 2\), and \(\tan\theta = \tfrac{5}{2}\) gives \(68.2^\circ\) ✓.
Answer: Angle \(ODB = 68.2^\circ\)
Examiner trap
Use the DIRECTION vectors of the lines, not the position vectors of the points on them. The position vectors depend on where the origin is; the angle between the lines does not.
Stage 1 · Learn
Check your understanding
Four quick checks. Each correct answer earns XP and records this skill on your profile.
The shortest route from a point \(P\) to a line meets the line at right angles, at the foot of the perpendicular \(N\). Write \(N\) as the general point of the line, then demand that \(\overrightarrow{PN}\) is perpendicular to the direction.
Worked example Foot of the perpendicular and distance
Find the foot of the perpendicular from \(P(1, -2, 4)\) to the line \(\mathbf{r} = (1, 2, 3) + s(1, -1, 2)\), and hence the perpendicular distance from \(P\) to the line.
General point: \(N = (1 + s,\ 2 - s,\ 3 + 2s)\).
\(\overrightarrow{PN} = N - P = (s,\ 4 - s,\ -1 + 2s)\).
Perpendicular to the direction: \((s, 4 - s, -1 + 2s) \cdot (1, -1, 2) = s - 4 + s - 2 + 4s = 6s - 6 = 0\), so \(s = 1\).
Dot \(\overrightarrow{PN}\) with the DIRECTION of the line, not with the position vector of \(N\) or of \(P\). And remember to finish: the question usually wants \(N\) or \(PN\), not just the parameter.
Stage 2 · Exam
Exam-style questions
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Find the angle between the vectors \(3\mathbf{i} + 4\mathbf{k}\) and \(\mathbf{i} + 2\mathbf{j} - 2\mathbf{k}\).
Exam style+20 XP
Find the acute angle between the lines \(\mathbf{r} = \mathbf{i} + s(2\mathbf{i} - \mathbf{j} + 2\mathbf{k})\) and \(\mathbf{r} = 3\mathbf{j} + t(\mathbf{i} + 4\mathbf{j} - \mathbf{k})\).
Exam style+20 XP
Find the foot of the perpendicular from \(P(5, 0, 0)\) to the line \(\mathbf{r} = (0, 1, 2) + t(1, 1, -1)\).
Exam style+20 XP
The cuboid \(OABCDEFG\) has \(\overrightarrow{OA} = 4\mathbf{i}\), \(\overrightarrow{OC} = 2\mathbf{j}\), \(\overrightarrow{OD} = 4\mathbf{k}\), with \(E\) above \(A\), \(F\) above \(B\) and \(G\) above \(C\). Find the angle between \(\overrightarrow{OF}\) and \(\overrightarrow{CE}\).
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