A-Level Mathematics – Chapter 8: Vectors
Vectors provide a rigorous framework for describing quantities that have both magnitude and direction — displacements, velocities, forces, and more. In Pure Mathematics, vectors are used to describe lines and to prove geometric results elegantly. This chapter introduces vectors from first principles, extends them to three dimensions, and develops the essential techniques of the dot product and equations of lines.
Specification Note
Vectors in 2D and 3D, the scalar product, and equations of lines are required for both AS and full A-Level on most specifications. The distance-from-point-to-line technique is primarily developed at A2 depth but may appear in AS questions indirectly.
8.1 Vectors in 2D
Notation and Representation
A vector is a quantity with both magnitude and direction. In two dimensions, vectors can be written in several equivalent ways:
- Column vector: $\mathbf{a} = \begin{pmatrix} 3 \\ -2 \end{pmatrix}$
- Component form: $\mathbf{a} = 3\mathbf{i} - 2\mathbf{j}$, where $\mathbf{i} = \begin{pmatrix}1\\0\end{pmatrix}$ and $\mathbf{j} = \begin{pmatrix}0\\1\end{pmatrix}$
- Bold notation: $\mathbf{a}$ (in print) or $\underline{a}$ (in handwriting)
Key Definitions
Magnitude of $\mathbf{a} = \begin{pmatrix}a_1\\a_2\end{pmatrix}$: $|\mathbf{a}| = \sqrt{a_1^2 + a_2^2}$.
Unit vector in the direction of $\mathbf{a}$: $\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}$, so $|\hat{\mathbf{a}}| = 1$.
Position vector of point $P$ with coordinates $(x, y)$: $\overrightarrow{OP} = \begin{pmatrix}x\\y\end{pmatrix}$.
Displacement vector from $A$ to $B$: $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$, where $\mathbf{a}$, $\mathbf{b}$ are the position vectors of $A$ and $B$ respectively.
Vector Addition and Scalar Multiplication
Vectors add component-wise: $\begin{pmatrix}a_1\\a_2\end{pmatrix} + \begin{pmatrix}b_1\\b_2\end{pmatrix} = \begin{pmatrix}a_1+b_1\\a_2+b_2\end{pmatrix}$.
Scalar multiplication scales the magnitude: $k\begin{pmatrix}a_1\\a_2\end{pmatrix} = \begin{pmatrix}ka_1\\ka_2\end{pmatrix}$. A negative $k$ reverses the direction.
Parallel vectors: $\mathbf{a} \parallel \mathbf{b}$ if and only if $\mathbf{a} = k\mathbf{b}$ for some scalar $k$.
Figure 8.1 — Vector addition in 2D. Vectors $\mathbf{a} = \begin{pmatrix}3\\1\end{pmatrix}$ (blue) and $\mathbf{b} = \begin{pmatrix}1\\2\end{pmatrix}$ (green) combine to give $\mathbf{a}+\mathbf{b} = \begin{pmatrix}4\\3\end{pmatrix}$ (red) via the triangle law. The parallelogram construction is shown with dashed lines.
Example 8.1.1 — Magnitude and Unit Vector
Given $\mathbf{v} = 3\mathbf{i} + 4\mathbf{j}$, find $|\mathbf{v}|$ and the unit vector $\hat{\mathbf{v}}$.
$|\mathbf{v}| = \sqrt{3^2 + 4^2} = \sqrt{25} = 5$.
$\hat{\mathbf{v}} = \dfrac{1}{5}\begin{pmatrix}3\\4\end{pmatrix} = \begin{pmatrix}0.6\\0.8\end{pmatrix}$. Verify: $|\hat{\mathbf{v}}| = \sqrt{0.36 + 0.64} = 1$. ✓
Example 8.1.2 — Displacement Vector and Midpoint
Points $A$ and $B$ have position vectors $\mathbf{a} = \begin{pmatrix}2\\-1\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}6\\3\end{pmatrix}$.
(a) Find $\overrightarrow{AB}$. (b) Find the position vector of the midpoint $M$ of $AB$.
(a) $\overrightarrow{AB} = \mathbf{b} - \mathbf{a} = \begin{pmatrix}6-2\\3-(-1)\end{pmatrix} = \begin{pmatrix}4\\4\end{pmatrix}$.
(b) $\overrightarrow{OM} = \dfrac{\mathbf{a}+\mathbf{b}}{2} = \dfrac{1}{2}\begin{pmatrix}8\\2\end{pmatrix} = \begin{pmatrix}4\\1\end{pmatrix}$.
Example 8.1.3 — Parallel Vectors
Show that $\mathbf{p} = \begin{pmatrix}6\\-4\end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix}-9\\6\end{pmatrix}$ are parallel.
If $\mathbf{p} = k\mathbf{q}$, then $6 = -9k$ gives $k = -\frac{2}{3}$, and $-4 = 6k$ gives $k = -\frac{2}{3}$. Both components give the same $k$, so $\mathbf{p} \parallel \mathbf{q}$.
Example 8.1.4 — Vector Proof of a Geometric Result
$ABCD$ is a parallelogram. $M$ is the midpoint of $AC$. Show that $M$ is also the midpoint of $BD$.
Let position vectors of $A$, $B$, $C$, $D$ be $\mathbf{a}$, $\mathbf{b}$, $\mathbf{c}$, $\mathbf{d}$. Parallelogram condition: $\overrightarrow{AB} = \overrightarrow{DC}$, so $\mathbf{b}-\mathbf{a} = \mathbf{c}-\mathbf{d}$, giving $\mathbf{b}+\mathbf{d} = \mathbf{a}+\mathbf{c}$.
Midpoint of $AC$: $\frac{\mathbf{a}+\mathbf{c}}{2}$. Midpoint of $BD$: $\frac{\mathbf{b}+\mathbf{d}}{2}$. These are equal since $\mathbf{a}+\mathbf{c} = \mathbf{b}+\mathbf{d}$. Hence $M$ is the midpoint of both diagonals.
8.2 Vectors in 3D
All 2D vector concepts extend naturally to three dimensions by introducing a third unit vector $\mathbf{k} = \begin{pmatrix}0\\0\\1\end{pmatrix}$ along the $z$-axis. A general 3D vector is written $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$.
3D Magnitude and Distance
For $\mathbf{a} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$: $\quad|\mathbf{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$.
Distance between points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$:
$$|AB| = |\overrightarrow{AB}| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$
Example 8.2.1 — Magnitude in 3D
Find the magnitude of $\mathbf{v} = 2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}$.
$|\mathbf{v}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
Example 8.2.2 — Distance Between Two Points
Find the distance between $A(1, 2, -3)$ and $B(4, -1, 1)$.
$\overrightarrow{AB} = \begin{pmatrix}3\\-3\\4\end{pmatrix}$, $\quad |AB| = \sqrt{9 + 9 + 16} = \sqrt{34}$.
Example 8.2.3 — Unit Vector in 3D
Find the unit vector in the direction of $\mathbf{a} = \begin{pmatrix}1\\-2\\2\end{pmatrix}$.
$|\mathbf{a}| = \sqrt{1+4+4} = 3$, so $\hat{\mathbf{a}} = \dfrac{1}{3}\begin{pmatrix}1\\-2\\2\end{pmatrix} = \begin{pmatrix}1/3\\-2/3\\2/3\end{pmatrix}$.
Example 8.2.4 — Section Formula (Internal Division)
$P$ divides segment $AB$ internally in the ratio $2:3$, where $A$ has position vector $\begin{pmatrix}1\\0\\4\end{pmatrix}$ and $B$ has position vector $\begin{pmatrix}6\\5\\-1\end{pmatrix}$. Find the position vector of $P$.
$\overrightarrow{OP} = \mathbf{a} + \dfrac{2}{5}\overrightarrow{AB} = \begin{pmatrix}1\\0\\4\end{pmatrix} + \dfrac{2}{5}\begin{pmatrix}5\\5\\-5\end{pmatrix} = \begin{pmatrix}1\\0\\4\end{pmatrix} + \begin{pmatrix}2\\2\\-2\end{pmatrix} = \begin{pmatrix}3\\2\\2\end{pmatrix}$.
8.3 Scalar (Dot) Product
The scalar product (also called the dot product) takes two vectors and returns a scalar. It has both an algebraic definition that is easy to compute and a geometric definition that connects directly to the angle between the vectors.
Scalar Product — Two Equivalent Definitions
Algebraic: For $\mathbf{a} = \begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}$:
$$\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$$
Geometric: $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}|\,|\mathbf{b}|\cos\theta$, where $\theta$ is the angle between the vectors ($0 \leq \theta \leq \pi$).
Perpendicularity test: $\mathbf{a} \perp \mathbf{b} \iff \mathbf{a} \cdot \mathbf{b} = 0$ (for non-zero vectors).
Note: the scalar product is commutative ($\mathbf{a}\cdot\mathbf{b} = \mathbf{b}\cdot\mathbf{a}$) and distributive ($\mathbf{a}\cdot(\mathbf{b}+\mathbf{c}) = \mathbf{a}\cdot\mathbf{b} + \mathbf{a}\cdot\mathbf{c}$).
Example 8.3.1 — Computing the Scalar Product
Find $\mathbf{a} \cdot \mathbf{b}$ where $\mathbf{a} = \begin{pmatrix}2\\-1\\3\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}4\\5\\-2\end{pmatrix}$.
$\mathbf{a} \cdot \mathbf{b} = 2(4) + (-1)(5) + 3(-2) = 8 - 5 - 6 = -3$.
Example 8.3.2 — Perpendicular Vectors
Show that $\mathbf{p} = \begin{pmatrix}1\\2\\2\end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix}2\\1\\-\frac{4}{2}\end{pmatrix} = \begin{pmatrix}2\\1\\-2\end{pmatrix}$ are perpendicular.
$\mathbf{p}\cdot\mathbf{q} = 1(2) + 2(1) + 2(-2) = 2 + 2 - 4 = 0$. Since the dot product is zero, $\mathbf{p} \perp \mathbf{q}$.
Example 8.3.3 — Angle Between Two Vectors
Find the angle between $\mathbf{a} = \begin{pmatrix}3\\0\\4\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}1\\1\\0\end{pmatrix}$, giving your answer to one decimal place.
$\mathbf{a}\cdot\mathbf{b} = 3$, $\quad |\mathbf{a}| = \sqrt{9+0+16} = 5$, $\quad |\mathbf{b}| = \sqrt{1+1+0} = \sqrt{2}$.
$\cos\theta = \dfrac{3}{5\sqrt{2}} \Rightarrow \theta = \arccos\!\left(\dfrac{3}{5\sqrt{2}}\right) \approx 64.9°$.
Example 8.3.4 — Finding an Unknown Component
$\mathbf{a} = \begin{pmatrix}k\\3\\-1\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}2\\k\\4\end{pmatrix}$ are perpendicular. Find $k$.
$\mathbf{a}\cdot\mathbf{b} = 0 \Rightarrow 2k + 3k - 4 = 0 \Rightarrow 5k = 4 \Rightarrow k = \dfrac{4}{5}$.
Exam Tip
When using $\cos\theta = \frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{a}||\mathbf{b}|}$ to find the angle between two lines (not vectors), always take the acute angle: if $\cos\theta$ is negative, compute $180° - \theta$ instead. Direction vectors pointing in opposite senses represent the same line, so the angle between lines is always between $0°$ and $90°$.
8.4 Equations of Lines
Any straight line in space is completely determined by a point on the line and a direction. The vector equation captures this idea directly and is far more versatile than the Cartesian form in three dimensions.
Three Forms of the Equation of a Line
Vector form: $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$, where $\mathbf{a}$ is the position vector of any point on the line, $\mathbf{b}$ is the direction vector, and $\lambda \in \mathbb{R}$ is a scalar parameter.
Parametric form: $x = a_1 + \lambda b_1$, $y = a_2 + \lambda b_2$, $z = a_3 + \lambda b_3$.
Cartesian (symmetric) form: $\dfrac{x - a_1}{b_1} = \dfrac{y - a_2}{b_2} = \dfrac{z - a_3}{b_3}$ (provided all $b_i \neq 0$).
Figure 8.2 — The line through $A(1, 2)$ and $B(4, 4)$ in 2D. Direction vector $\overrightarrow{AB} = \begin{pmatrix}3\\2\end{pmatrix}$; vector equation $\mathbf{r} = \begin{pmatrix}1\\2\end{pmatrix} + \lambda\begin{pmatrix}3\\2\end{pmatrix}$. As $\lambda$ varies, the point traces the entire line.
Example 8.4.1 — Equation of a Line Through Two Points
Find the vector equation of the line through $A(2, -1, 3)$ and $B(5, 0, 1)$.
Direction vector: $\overrightarrow{AB} = \begin{pmatrix}5-2\\0-(-1)\\1-3\end{pmatrix} = \begin{pmatrix}3\\1\\-2\end{pmatrix}$.
Vector equation: $\mathbf{r} = \begin{pmatrix}2\\-1\\3\end{pmatrix} + \lambda\begin{pmatrix}3\\1\\-2\end{pmatrix}$ (or equivalently using $B$ as the base point).
Example 8.4.2 — Converting to Cartesian Form
Convert $\mathbf{r} = \begin{pmatrix}1\\4\\-2\end{pmatrix} + \lambda\begin{pmatrix}2\\-1\\3\end{pmatrix}$ to Cartesian form.
Parametric equations: $x = 1 + 2\lambda$, $y = 4 - \lambda$, $z = -2 + 3\lambda$. Solve each for $\lambda$:
$$\frac{x-1}{2} = \frac{y-4}{-1} = \frac{z+2}{3}$$
Example 8.4.3 — Parallel Lines
Determine whether $L_1: \mathbf{r} = \begin{pmatrix}1\\2\\0\end{pmatrix} + \lambda\begin{pmatrix}2\\-1\\4\end{pmatrix}$ and $L_2: \mathbf{r} = \begin{pmatrix}0\\5\\1\end{pmatrix} + \mu\begin{pmatrix}-4\\2\\-8\end{pmatrix}$ are parallel and distinct.
$\begin{pmatrix}-4\\2\\-8\end{pmatrix} = -2\begin{pmatrix}2\\-1\\4\end{pmatrix}$, so the direction vectors are scalar multiples: $L_1 \parallel L_2$.
To confirm they are distinct, check whether $\overrightarrow{AB} = \begin{pmatrix}0-1\\5-2\\1-0\end{pmatrix} = \begin{pmatrix}-1\\3\\1\end{pmatrix}$ is a multiple of $\begin{pmatrix}2\\-1\\4\end{pmatrix}$. Ratios $-1/2$, $-3$, $1/4$ are not equal, so the lines are distinct parallel lines.
Example 8.4.4 — Angle Between Two Lines
Find the acute angle between lines with direction vectors $\mathbf{d}_1 = \begin{pmatrix}1\\2\\-1\end{pmatrix}$ and $\mathbf{d}_2 = \begin{pmatrix}2\\0\\1\end{pmatrix}$.
$\mathbf{d}_1\cdot\mathbf{d}_2 = 2 + 0 - 1 = 1$, $\quad |\mathbf{d}_1| = \sqrt{1+4+1} = \sqrt{6}$, $\quad |\mathbf{d}_2| = \sqrt{4+0+1} = \sqrt{5}$.
$\cos\theta = \dfrac{|1|}{\sqrt{6}\cdot\sqrt{5}} = \dfrac{1}{\sqrt{30}}$ (modulus taken for acute angle). $\quad \theta = \arccos\!\left(\dfrac{1}{\sqrt{30}}\right) \approx 79.5°$.
8.5 Points, Lines and Distances
This section addresses three closely related questions: whether a point lies on a given line, whether two lines intersect, and how to find the shortest distance from a point to a line.
Testing Whether a Point Lies on a Line
To test whether the point $P$ (position vector $\mathbf{p}$) lies on $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$, write down the three component equations and check whether a single consistent value of $\lambda$ satisfies all three simultaneously.
Example 8.5.1 — Does a Point Lie on a Line?
Does $P = (7, 1, -3)$ lie on the line $\mathbf{r} = \begin{pmatrix}1\\3\\1\end{pmatrix} + \lambda\begin{pmatrix}2\\-1\\-2\end{pmatrix}$?
$x$-component: $7 = 1 + 2\lambda \Rightarrow \lambda = 3$.
$y$-component: $1 = 3 + 3(-1) = 0 \neq 1$. Inconsistent. Therefore $P$ does not lie on the line.
Example 8.5.2 — Intersection of Two Lines (or Skew)
Determine whether lines $L_1: \mathbf{r} = \begin{pmatrix}2\\1\\0\end{pmatrix}+\lambda\begin{pmatrix}1\\1\\1\end{pmatrix}$ and $L_2: \mathbf{r} = \begin{pmatrix}4\\2\\-1\end{pmatrix}+\mu\begin{pmatrix}-1\\0\\1\end{pmatrix}$ intersect.
Set equal component-by-component: $2+\lambda = 4-\mu$ (i), $1+\lambda = 2$ (ii), $\lambda = -1+\mu$ (iii).
From (ii): $\lambda = 1$. From (iii): $\mu = 2$. Check (i): LHS $= 3$, RHS $= 4 - 2 = 2$. Inconsistent. The lines are skew (non-intersecting and non-parallel in 3D).
Distance from a Point to a Line
To find the shortest distance from point $P$ (position vector $\mathbf{p}$) to line $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$:
- Write a general point $Q = \mathbf{a} + \lambda\mathbf{b}$ on the line.
- Form $\overrightarrow{QP} = \mathbf{p} - \mathbf{a} - \lambda\mathbf{b}$.
- Apply the perpendicularity condition $\overrightarrow{QP} \cdot \mathbf{b} = 0$ and solve for $\lambda$.
- Substitute $\lambda$ back to find $Q$, then compute $|QP|$.
Example 8.5.3 — Distance from a Point to a Line
Find the shortest distance from $P(5, 3, 2)$ to the line $\mathbf{r} = \begin{pmatrix}1\\1\\0\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\2\end{pmatrix}$.
Step 1 General point: $Q = (1+2\lambda,\; 1+\lambda,\; 2\lambda)$.
Step 2 $\overrightarrow{QP} = \begin{pmatrix}4-2\lambda\\2-\lambda\\2-2\lambda\end{pmatrix}$.
Step 3 Perpendicularity: $\overrightarrow{QP}\cdot\begin{pmatrix}2\\1\\2\end{pmatrix} = 0$:
$2(4-2\lambda) + (2-\lambda) + 2(2-2\lambda) = 8 - 4\lambda + 2 - \lambda + 4 - 4\lambda = 14 - 9\lambda = 0 \Rightarrow \lambda = \dfrac{14}{9}$.
Step 4 $\overrightarrow{QP} = \dfrac{1}{9}\begin{pmatrix}4\cdot9 - 2\cdot14\\2\cdot9 - 14\\2\cdot9 - 2\cdot14\end{pmatrix} = \dfrac{1}{9}\begin{pmatrix}8\\4\\-10\end{pmatrix}$.
Distance $= \dfrac{1}{9}\sqrt{64 + 16 + 100} = \dfrac{\sqrt{180}}{9} = \dfrac{6\sqrt{5}}{9} = \dfrac{2\sqrt{5}}{3}$.
Example 8.5.4 — Foot of the Perpendicular
Find the coordinates of the foot of the perpendicular from $A(3, 0, 5)$ to the line $\mathbf{r} = \begin{pmatrix}0\\2\\1\end{pmatrix} + \lambda\begin{pmatrix}1\\0\\2\end{pmatrix}$.
General point $F = (\lambda,\; 2,\; 1+2\lambda)$. $\overrightarrow{FA} = (3-\lambda,\; -2,\; 4-2\lambda)$.
Perpendicularity: $(3-\lambda)\cdot1 + (-2)\cdot0 + (4-2\lambda)\cdot2 = 0 \Rightarrow 3-\lambda + 8-4\lambda = 0 \Rightarrow 11 = 5\lambda \Rightarrow \lambda = \dfrac{11}{5}$.
Foot: $F = \left(\dfrac{11}{5},\; 2,\; 1 + \dfrac{22}{5}\right) = \left(\dfrac{11}{5},\; 2,\; \dfrac{27}{5}\right)$.
Practice Problems
Problem 1
Given $\mathbf{a} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}$ and $\mathbf{b} = -\mathbf{i} + 4\mathbf{j} + 2\mathbf{k}$, find: (a) $|\mathbf{a}|$; (b) $2\mathbf{a} - \mathbf{b}$; (c) a unit vector in the direction of $\mathbf{a} + \mathbf{b}$.
Show solution
(b) $2\mathbf{a} - \mathbf{b} = \begin{pmatrix}4-(-1)\\-2-4\\6-2\end{pmatrix} = \begin{pmatrix}5\\-6\\4\end{pmatrix}$.
(c) $\mathbf{a}+\mathbf{b} = \begin{pmatrix}1\\3\\5\end{pmatrix}$, $|\mathbf{a}+\mathbf{b}| = \sqrt{1+9+25} = \sqrt{35}$. Unit vector $= \dfrac{1}{\sqrt{35}}\begin{pmatrix}1\\3\\5\end{pmatrix}$.
Problem 2
$A$, $B$, $C$ have position vectors $\begin{pmatrix}1\\2\\-1\end{pmatrix}$, $\begin{pmatrix}3\\0\\1\end{pmatrix}$, $\begin{pmatrix}5\\-2\\3\end{pmatrix}$ respectively. Show that $A$, $B$, $C$ are collinear.
Show solution
Problem 3
Find the angle between $\mathbf{a} = \begin{pmatrix}2\\1\\-2\end{pmatrix}$ and $\mathbf{b} = \begin{pmatrix}1\\-2\\2\end{pmatrix}$ to the nearest degree.
Show solution
$|\mathbf{a}| = \sqrt{4+1+4} = 3$, $|\mathbf{b}| = \sqrt{1+4+4} = 3$.
$\cos\theta = \dfrac{-4}{9} \Rightarrow \theta = \arccos\!\left(-\dfrac{4}{9}\right) \approx 116°$.
Problem 4
Find the value(s) of $t$ such that $\mathbf{p} = \begin{pmatrix}t\\1\\2\end{pmatrix}$ and $\mathbf{q} = \begin{pmatrix}2\\t\\-1\end{pmatrix}$ are perpendicular.
Show solution
Problem 5
Write the vector equation of the line through $P(3, -2, 1)$ with direction $\mathbf{d} = 2\mathbf{i} + \mathbf{j} - 3\mathbf{k}$, and express it in Cartesian form.
Show solution
Cartesian form: $\dfrac{x-3}{2} = \dfrac{y+2}{1} = \dfrac{z-1}{-3}$.
Problem 6
Find the position vector of the intersection of $L_1: \mathbf{r} = \begin{pmatrix}1\\0\\2\end{pmatrix} + \lambda\begin{pmatrix}2\\1\\-1\end{pmatrix}$ and $L_2: \mathbf{r} = \begin{pmatrix}3\\2\\0\end{pmatrix} + \mu\begin{pmatrix}0\\-1\\1\end{pmatrix}$.
Show solution
From (i): $\lambda = 1$. From (iii): $\mu = 1$. Check (ii): $1 = 2-1 = 1$. Consistent.
Intersection point: $\begin{pmatrix}1+2\\0+1\\2-1\end{pmatrix} = \begin{pmatrix}3\\1\\1\end{pmatrix}$.
Problem 7
Find the shortest distance from $B(4, 2, 0)$ to the line $\mathbf{r} = \lambda\begin{pmatrix}1\\1\\0\end{pmatrix}$ (passing through the origin).
Show solution
Perpendicularity: $(4-\lambda) + (2-\lambda) = 0 \Rightarrow 6 = 2\lambda \Rightarrow \lambda = 3$.
$\overrightarrow{QB} = \begin{pmatrix}1\\-1\\0\end{pmatrix}$. Distance $= \sqrt{1+1} = \sqrt{2}$.
Problem 8
Points $A$ and $B$ have position vectors $3\mathbf{i} + 2\mathbf{j}$ and $7\mathbf{i} - 2\mathbf{j}$ respectively. Point $C$ divides $AB$ externally in the ratio $3:1$. Find the position vector of $C$.
Show solution
Problem 9
Show that lines $L_1: \mathbf{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + \lambda\begin{pmatrix}1\\-1\\2\end{pmatrix}$ and $L_2: \mathbf{r} = \begin{pmatrix}2\\1\\0\end{pmatrix} + \mu\begin{pmatrix}2\\1\\1\end{pmatrix}$ are skew.
Show solution
Try intersection: $1+\lambda = 2+2\mu$ (i), $2-\lambda = 1+\mu$ (ii), $3+2\lambda = \mu$ (iii).
From (ii): $\lambda = 1-\mu$. Into (iii): $3+2(1-\mu) = \mu \Rightarrow 5 = 3\mu \Rightarrow \mu = \frac{5}{3}$, $\lambda = -\frac{2}{3}$.
Check (i): $1 + (-\frac{2}{3}) = \frac{1}{3}$ but $2 + \frac{10}{3} = \frac{16}{3}$. Inconsistent. Lines are skew.
Problem 10
Vectors $\mathbf{a}$ and $\mathbf{b}$ satisfy $|\mathbf{a}| = 3$, $|\mathbf{b}| = 5$ and $\mathbf{a}\cdot\mathbf{b} = 12$. Find $|\mathbf{a} + \mathbf{b}|$ and the angle between $\mathbf{a}$ and $\mathbf{b}$.
Show solution
$\cos\theta = \dfrac{12}{3 \times 5} = \dfrac{12}{15} = 0.8 \Rightarrow \theta = \arccos(0.8) \approx 36.9°$.