Arrows you can draw and lists of numbers you can compute with are the same objects: add them tip to tail, stretch them by scalars, and combine them into linear combinations $c_1\mathbf{v}_1+\dots+c_k\mathbf{v}_k$.
A pilot needs to know where the aircraft will actually go when a crosswind is blowing. An engineer needs the combined effect of three cables pulling on a bracket. A phone's GPS reports your position as three numbers, and a hospital stores each patient's heart rate, blood pressure, temperature, oxygen level and age as a list of five numbers. In every case the object has several numbers that belong together, and the natural things to do with it are to add two of them and to scale one up or down.
That is exactly what a vector is. In the plane and in space a vector is an arrow with a length and a direction, and we add arrows by placing them tip to tail. Written in coordinates, the same vector is an ordered list of numbers, and adding arrows becomes adding numbers entry by entry. The list form has a great advantage: it works just as well with $5$ or $500$ entries, where no picture is possible. In this section we build the algebra of vectors in $\mathbb{R}^n$ and meet the idea at the heart of the whole subject — the linear combination.
Vectors in the Plane and in Space
Geometrically, a vector is a directed line segment: an arrow from an initial point $P$ (its tail) to a terminal point $Q$ (its tip), written $\overrightarrow{PQ}$. Only the length and the direction matter, so two arrows that are parallel, point the same way and have the same length represent the same vector, wherever they are drawn. A displacement of "3 km east and 4 km north" is the same displacement whether it starts in Muscat or in Nairobi.
Because position does not matter, we can always slide a vector so that its tail is at the origin $O$. Its tip then lands on a single point, and the coordinates of that point describe the vector completely. If $P=(p_1,p_2,p_3)$ and $Q=(q_1,q_2,q_3)$, the vector from $P$ to $Q$ has components
$$\overrightarrow{PQ}=\begin{bmatrix}q_1-p_1\\q_2-p_2\\q_3-p_3\end{bmatrix}\qquad(\text{tip minus tail}).$$
The arrow from the origin to a point $P$ is the position vector $\mathbf{p}=\overrightarrow{OP}$ of $P$; its components are just the coordinates of $P$. Points and vectors are written with the same numbers, but they play different roles: a point is a location, a vector is a displacement.
📘 Definition 4.1 — Vectors in $\mathbb{R}^n$
For a positive integer $n$, a vector in $\mathbb{R}^n$ is an ordered list of $n$ real numbers, written as a column
$$\mathbf{v}=\begin{bmatrix}v_1\\v_2\\\vdots\\v_n\end{bmatrix}.$$
The numbers $v_1,\dots,v_n$ are the components (or entries) of $\mathbf{v}$, and $\mathbb{R}^n$ is the set of all such vectors. Inline we also write $\mathbf{v}=(v_1,v_2,\dots,v_n)$. Two vectors are equal when they are in the same $\mathbb{R}^n$ and all corresponding components are equal. The zero vector $\mathbf{0}=(0,0,\dots,0)$ has every component $0$. Real numbers, used to scale vectors, are called scalars.
Vectors in $\mathbb{R}^2$ and $\mathbb{R}^3$ can be drawn as arrows. A vector in $\mathbb{R}^n$ for larger $n$ cannot be drawn, but it is just as useful. The table shows one (illustrative) patient's record as a vector in $\mathbb{R}^5$ — the order of the entries is fixed once and for all, so the third entry is always the temperature.
entry
1: heart rate (bpm)
2: systolic BP (mmHg)
3: temperature (°C)
4: oxygen (%)
5: age (years)
$\mathbf{x}$
$72$
$118$
$36.8$
$98$
$41$
✏️ Example 1 — Components from two points
Find the components of the vector $\overrightarrow{PQ}$ from $P(2,-1,3)$ to $Q(5,3,1)$. Which point $R$ is reached if the same displacement is applied starting at $S(0,0,4)$?
💡 Show solution▶
Tip minus tail $$\overrightarrow{PQ}=\begin{bmatrix}5-2\\3-(-1)\\1-3\end{bmatrix}=\begin{bmatrix}3\\4\\-2\end{bmatrix}.$$
Move it The same vector drawn from $S$ ends at $S+\overrightarrow{PQ}=(0+3,\ 0+4,\ 4-2)=(3,4,2)$, so $R=(3,4,2)$.
Check $P+\overrightarrow{PQ}=(2+3,\,-1+4,\,3-2)=(5,3,1)=Q$. ✓ The arrows $\overrightarrow{PQ}$ and $\overrightarrow{SR}$ are different arrows but the same vector.
Addition, Scalar Multiples and Subtraction
Two displacements made one after the other give a single total displacement: walk along $\mathbf{u}$, then along $\mathbf{v}$, and you have walked along $\mathbf{u}+\mathbf{v}$. In pictures this is the triangle law: put the tail of $\mathbf{v}$ at the tip of $\mathbf{u}$; the sum runs from the tail of $\mathbf{u}$ to the tip of $\mathbf{v}$. Doing it in the other order ($\mathbf{v}$ first) gives the other two sides of a parallelogram — the parallelogram law — and the sum is its diagonal. In coordinates the walk simply adds the east–west parts and the north–south parts separately.
📘 Definition 4.2 — Vector addition and scalar multiplication
Let $\mathbf{u}=(u_1,\dots,u_n)$ and $\mathbf{v}=(v_1,\dots,v_n)$ be vectors in $\mathbb{R}^n$ and let $c$ be a scalar. Then
$$\mathbf{u}+\mathbf{v}=\begin{bmatrix}u_1+v_1\\u_2+v_2\\\vdots\\u_n+v_n\end{bmatrix},\qquad c\,\mathbf{v}=\begin{bmatrix}cv_1\\cv_2\\\vdots\\cv_n\end{bmatrix}.$$
The negative of $\mathbf{v}$ is $-\mathbf{v}=(-1)\mathbf{v}$, and the difference is $\mathbf{u}-\mathbf{v}=\mathbf{u}+(-\mathbf{v})$. Only vectors in the same $\mathbb{R}^n$ can be added: $(1,2)+(1,2,3)$ is not defined.
Geometric meaning. The scalar multiple $c\mathbf{v}$ is parallel to $\mathbf{v}$: it is $|c|$ times as long, pointing the same way when $c\gt0$ and the opposite way when $c\lt0$; $0\mathbf{v}=\mathbf{0}$. Two nonzero vectors are called parallel when one is a scalar multiple of the other. The difference $\mathbf{u}-\mathbf{v}$ is the vector that must be added to $\mathbf{v}$ to get $\mathbf{u}$, so it is the arrow from the tip of $\mathbf{v}$ to the tip of $\mathbf{u}$ — the other diagonal of the parallelogram. This is the same rule as "tip minus tail": $\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}$.
Interactive 1 — Adding, Subtracting and Scaling Arrows
Drag the tips of u and v (they snap to half-units). Choose what to build. Play slides the copies tip to tail so you can see why the triangle and parallelogram laws give the same diagonal, how $\mathbf{u}-\mathbf{v}$ joins the two tips, and how $c\mathbf{u}$ stretches and flips as $c$ changes.
1.5
$\mathbf{u}$—
$\mathbf{v}$—
result—
📐 Theorem 4.1 — Algebraic properties of vectors in $\mathbb{R}^n$
For all vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ in $\mathbb{R}^n$ and all scalars $c,d$:
(e) distributive over vectors$c(\mathbf{u}+\mathbf{v})=c\mathbf{u}+c\mathbf{v}$
(f) distributive over scalars$(c+d)\mathbf{u}=c\mathbf{u}+d\mathbf{u}$
(g) associative for scalars$c(d\mathbf{u})=(cd)\mathbf{u}$
(h) identity$1\mathbf{u}=\mathbf{u}$
Each property follows from the matching property of real numbers, applied one component at a time. For example, the $i$-th component of $c(\mathbf{u}+\mathbf{v})$ is $c(u_i+v_i)=cu_i+cv_i$, which is the $i$-th component of $c\mathbf{u}+c\mathbf{v}$. The practical meaning of Theorem 4.1 is that you may rearrange, group and expand vector expressions exactly as in school algebra — and solve vector equations the same way.
✏️ Example 2 — Vector arithmetic
Let $\mathbf{u}=\begin{bmatrix}1\\-2\\4\end{bmatrix}$ and $\mathbf{v}=\begin{bmatrix}3\\0\\-1\end{bmatrix}$. Compute $\mathbf{u}+\mathbf{v}$, $\mathbf{u}-\mathbf{v}$ and $2\mathbf{u}-3\mathbf{v}$.
💡 Show solution▶
Sum and difference $$\mathbf{u}+\mathbf{v}=\begin{bmatrix}1+3\\-2+0\\4+(-1)\end{bmatrix}=\begin{bmatrix}4\\-2\\3\end{bmatrix},\qquad \mathbf{u}-\mathbf{v}=\begin{bmatrix}1-3\\-2-0\\4-(-1)\end{bmatrix}=\begin{bmatrix}-2\\-2\\5\end{bmatrix}.$$
Scale first $2\mathbf{u}=(2,-4,8)$ and $3\mathbf{v}=(9,0,-3)$, so $$2\mathbf{u}-3\mathbf{v}=\begin{bmatrix}2-9\\-4-0\\8-(-3)\end{bmatrix}=\begin{bmatrix}-7\\-4\\11\end{bmatrix}.$$
Check $(\mathbf{u}+\mathbf{v})+(\mathbf{u}-\mathbf{v})$ should be $2\mathbf{u}$: $(4-2,\,-2-2,\,3+5)=(2,-4,8)$. ✓
✏️ Example 3 — Solving a vector equation
With $\mathbf{u}$ and $\mathbf{v}$ as in Example 2, find the vector $\mathbf{x}$ in $\mathbb{R}^3$ such that $3\mathbf{x}-\mathbf{u}=2(\mathbf{v}+\mathbf{x})$.
💡 Show solution▶
Expand By Theorem 4.1(e), $2(\mathbf{v}+\mathbf{x})=2\mathbf{v}+2\mathbf{x}$, so $3\mathbf{x}-\mathbf{u}=2\mathbf{v}+2\mathbf{x}$.
Collect Add $\mathbf{u}-2\mathbf{x}$ to both sides: $\mathbf{x}=\mathbf{u}+2\mathbf{v}$ — exactly as with numbers.
Check $3\mathbf{x}-\mathbf{u}=(21-1,\,-6+2,\,6-4)=(20,-4,2)$ and $2(\mathbf{v}+\mathbf{x})=2(10,-2,1)=(20,-4,2)$. ✓
✏️ Example 4 — Completing a parallelogram
Three vertices of a parallelogram $ABCD$ (labelled in order around it) are $A(1,0)$, $B(4,1)$ and $C(5,3)$. Find the fourth vertex $D$ and the point $M$ where the diagonals meet.
💡 Show solution▶
Equal sides In a parallelogram $ABCD$ the opposite sides $\overrightarrow{AD}$ and $\overrightarrow{BC}$ are the same vector. $\overrightarrow{BC}=\mathbf{c}-\mathbf{b}=(1,2)$.
Fourth vertex $\mathbf{d}=\mathbf{a}+\overrightarrow{BC}=(1,0)+(1,2)=(2,2)$, so $D=(2,2)$. In general $\mathbf{d}=\mathbf{a}+\mathbf{c}-\mathbf{b}$.
Midpoint The midpoint of $AC$ has position vector $\tfrac12(\mathbf{a}+\mathbf{c})=\tfrac12(6,3)=(3,1.5)$; the midpoint of $BD$ is $\tfrac12(\mathbf{b}+\mathbf{d})=\tfrac12(6,3)$ too. So $M=(3,\tfrac32)$ — the diagonals bisect each other.
Check $\overrightarrow{AB}=(3,1)$ and $\overrightarrow{DC}=(5-2,\,3-2)=(3,1)$: the other pair of sides also match. ✓
Linear Combinations and Standard Unit Vectors
With addition and scaling we can build new vectors from old ones. This single construction will turn out to be the main character of the whole course.
📘 Definition 4.3 — Linear combination
A vector $\mathbf{w}$ in $\mathbb{R}^n$ is a linear combination of the vectors $\mathbf{v}_1,\mathbf{v}_2,\dots,\mathbf{v}_k$ in $\mathbb{R}^n$ if there are scalars $c_1,c_2,\dots,c_k$ such that
$$\mathbf{w}=c_1\mathbf{v}_1+c_2\mathbf{v}_2+\dots+c_k\mathbf{v}_k .$$
The scalars $c_1,\dots,c_k$ are the coefficients (or weights) of the combination.
The simplest building blocks are the standard unit vectors. In $\mathbb{R}^n$ they are
$$\mathbf{e}_1=\begin{bmatrix}1\\0\\\vdots\\0\end{bmatrix},\quad \mathbf{e}_2=\begin{bmatrix}0\\1\\\vdots\\0\end{bmatrix},\quad\dots,\quad \mathbf{e}_n=\begin{bmatrix}0\\0\\\vdots\\1\end{bmatrix},$$
and in $\mathbb{R}^3$ they are also called $\mathbf{i}=\mathbf{e}_1$, $\mathbf{j}=\mathbf{e}_2$, $\mathbf{k}=\mathbf{e}_3$ (in $\mathbb{R}^2$: $\mathbf{i}=(1,0)$, $\mathbf{j}=(0,1)$). Every vector is a linear combination of them, with its own components as the coefficients:
$$\mathbf{v}=v_1\mathbf{e}_1+v_2\mathbf{e}_2+\dots+v_n\mathbf{e}_n,\qquad\text{for example}\quad (3,-1,5)=3\mathbf{i}-\mathbf{j}+5\mathbf{k}.$$
For other building blocks the coefficients have to be found, and that means solving a linear system — which is why Chapter 1 comes first.
✏️ Example 5 — Is it a linear combination?
Let $\mathbf{a}_1=\begin{bmatrix}1\\2\\-1\end{bmatrix}$ and $\mathbf{a}_2=\begin{bmatrix}2\\1\\1\end{bmatrix}$. (a) Is $\mathbf{b}=\begin{bmatrix}4\\5\\-1\end{bmatrix}$ a linear combination of $\mathbf{a}_1$ and $\mathbf{a}_2$? (b) Is $\mathbf{b}^\prime=\begin{bmatrix}4\\5\\0\end{bmatrix}$?
💡 Show solution▶
Set up We need $c_1,c_2$ with $c_1\mathbf{a}_1+c_2\mathbf{a}_2=\mathbf{b}$. Comparing components gives a linear system: $$\begin{aligned}c_1+2c_2&=4\\2c_1+c_2&=5\\-c_1+c_2&=-1.\end{aligned}$$
Solve Row reduce the augmented matrix ($R_2\to R_2-2R_1$, $R_3\to R_3+R_1$, then $R_2\to-\tfrac13R_2$, $R_3\to R_3-3R_2$): $$\left[\begin{array}{cc|c}1&2&4\\2&1&5\\-1&1&-1\end{array}\right]\to\left[\begin{array}{cc|c}1&2&4\\0&-3&-3\\0&3&3\end{array}\right]\to\left[\begin{array}{cc|c}1&2&4\\0&1&1\\0&0&0\end{array}\right].$$ So $c_2=1$ and $c_1=4-2=2$.
(b) Answer The first two equations are unchanged, so they still force $c_1=2$, $c_2=1$. The third would need $-c_1+c_2=0$, but $-2+1=-1\ne0$. The system is inconsistent, so $\mathbf{b}^\prime$ is not a linear combination of $\mathbf{a}_1$ and $\mathbf{a}_2$. Geometrically, all combinations of $\mathbf{a}_1,\mathbf{a}_2$ fill a plane through the origin, and $\mathbf{b}^\prime$ lies off that plane.
Interactive 2 — Reach the Target with a Linear Combination
Choose the weights $c_1$ and $c_2$ so that $c_1\mathbf{a}_1+c_2\mathbf{a}_2$ (drawn tip to tail) lands on the target. The slanted grid shows every combination with whole or half weights. Play lets the computer steer to each target in turn. Then try the last pair, where $\mathbf{a}_2$ is parallel to $\mathbf{a}_1$: the combinations only fill a line, and most targets cannot be reached at all.
00
$c_1\mathbf{a}_1$—
$c_2\mathbf{a}_2$—
$c_1\mathbf{a}_1+c_2\mathbf{a}_2$—
target $\mathbf{b}$—
Vectors as Data, and the Vector Form of a Linear System
Once vectors are lists, the operations have everyday meanings. If $\mathbf{w}_1$ and $\mathbf{w}_2$ record the numbers of five products a shop sold in two weeks, then $\mathbf{w}_1+\mathbf{w}_2$ is the two-week total, $\mathbf{w}_2-\mathbf{w}_1$ is the week-to-week change, $\tfrac12(\mathbf{w}_1+\mathbf{w}_2)$ is the weekly average, and $1.1\,\mathbf{w}_1$ is a 10% increase in every product. Spreadsheets, statistics packages and machine-learning libraries all work this way: a row of data is a vector, and averaging records is a linear combination with weights $\tfrac1m$.
Linear combinations also give a new way to read a linear system. Group the coefficients of each unknown into a column:
$$\begin{aligned}x_1+2x_2&=5\\3x_1-x_2&=1\end{aligned}\qquad\Longleftrightarrow\qquad x_1\begin{bmatrix}1\\3\end{bmatrix}+x_2\begin{bmatrix}2\\-1\end{bmatrix}=\begin{bmatrix}5\\1\end{bmatrix}.$$
The left system asks for a point on two lines; the right vector equation asks how much of each column vector is needed to reach the right-hand side. They are the same question.
📐 Theorem 4.2 — Linear systems as linear combinations
Let $\mathbf{a}_1,\dots,\mathbf{a}_n$ be the columns of the coefficient matrix of a linear system with $m$ equations and $n$ unknowns, and let $\mathbf{b}$ in $\mathbb{R}^m$ be its right-hand side. Then $(x_1,\dots,x_n)$ is a solution of the system if and only if
$$x_1\mathbf{a}_1+x_2\mathbf{a}_2+\dots+x_n\mathbf{a}_n=\mathbf{b}.$$
In particular, the system is consistent if and only if $\mathbf{b}$ is a linear combination of the columns $\mathbf{a}_1,\dots,\mathbf{a}_n$.
✏️ Example 6 — A system in vector form
(a) Solve the vector equation $x_1\begin{bmatrix}1\\3\end{bmatrix}+x_2\begin{bmatrix}2\\-1\end{bmatrix}=\begin{bmatrix}5\\1\end{bmatrix}$ and interpret the answer. (b) Write the system $2x_1-x_2+4x_3=1$, $x_1+5x_3=-3$ in vector form.
💡 Show solution▶
(a) Solve $$\left[\begin{array}{cc|c}1&2&5\\3&-1&1\end{array}\right]\xrightarrow{R_2\to R_2-3R_1}\left[\begin{array}{cc|c}1&2&5\\0&-7&-14\end{array}\right],$$ so $x_2=2$ and $x_1=5-2(2)=1$.
(a) Interpret $1\cdot(1,3)+2\cdot(2,-1)=(5,1)$: one copy of the first column plus two copies of the second reach $\mathbf{b}$. Equivalently, the two lines $x_1+2x_2=5$ and $3x_1-x_2=1$ meet at $(1,2)$.
(b) Columns The unknown $x_2$ is missing from the second equation, so its coefficient there is $0$: $$x_1\begin{bmatrix}2\\1\end{bmatrix}+x_2\begin{bmatrix}-1\\0\end{bmatrix}+x_3\begin{bmatrix}4\\5\end{bmatrix}=\begin{bmatrix}1\\-3\end{bmatrix}.$$ Here three vectors in $\mathbb{R}^2$ are being combined, so we should expect many solutions — Chapter 5 explains exactly why.
⚠️ Common mistakes
1. Tail minus tip. $\overrightarrow{PQ}=\mathbf{q}-\mathbf{p}$, not $\mathbf{p}-\mathbf{q}$; the second is $\overrightarrow{QP}$, the same arrow reversed.
2. Adding a point to a point. $P+Q$ has no geometric meaning (it depends on where the origin is). Point $+$ vector $=$ point, point $-$ point $=$ vector, vector $\pm$ vector $=$ vector.
3. Mixing sizes. Vectors with different numbers of components cannot be added or compared.
4. Columns, not rows. In the vector form of a system, each vector collects the coefficients of one unknown from every equation (a column), not one equation (a row).
Exercises
I
Computing with vectors. Let $\mathbf{u}=(2,-1,3)$, $\mathbf{v}=(-1,4,0)$ and $\mathbf{w}=(0,2,-5)$.
1 $\mathbf{u}+\mathbf{v}$
2 $3\mathbf{w}$
3 $\mathbf{u}-\mathbf{v}$
4 $2\mathbf{u}-3\mathbf{v}+\mathbf{w}$
5 $-(\mathbf{u}-2\mathbf{w})$
6 The components of $\overrightarrow{PQ}$ for $P(3,-2)$ and $Q(-1,5)$.
7 The components of $\overrightarrow{PQ}$ for $P(1,0,-4)$ and $Q(-2,3,2)$.
8 $\overrightarrow{AB}=(2,-3,1)$ and $B=(5,0,-2)$. Find the point $A$.
9 Write $(4,-7,2)$ in terms of $\mathbf{i}$, $\mathbf{j}$, $\mathbf{k}$.
10 In $\mathbb{R}^5$ let $\mathbf{a}=(1,0,-2,3,1)$ and $\mathbf{b}=(2,1,1,0,-1)$. Compute $\mathbf{a}+2\mathbf{b}$ and $3\mathbf{a}-\mathbf{b}$.
Check: a linear-combination calculator for $c\,\mathbf{x}+d\,\mathbf{y}$ in $\mathbb{R}^3$
$c\,\mathbf{x}$—
$d\,\mathbf{y}$—
$c\,\mathbf{x}+d\,\mathbf{y}$—
Type the vectors as three numbers separated by commas (fractions such as 1/2 are fine).
8 $\mathbf{b}-\mathbf{a}=\overrightarrow{AB}$, so $\mathbf{a}=\mathbf{b}-\overrightarrow{AB}=(5-2,\,0+3,\,-2-1)=(3,3,-3)$. Check: $(5,0,-2)-(3,3,-3)=(2,-3,1)$. ✓
9 $(4,-7,2)=4\mathbf{i}-7\mathbf{j}+2\mathbf{k}$.
10 $\mathbf{a}+2\mathbf{b}=(1+4,\,0+2,\,-2+2,\,3+0,\,1-2)=(5,2,0,3,-1)$ and $3\mathbf{a}-\mathbf{b}=(3-2,\,0-1,\,-6-1,\,9-0,\,3+1)=(1,-1,-7,9,4)$.
II
Vector equations and linear combinations ($\mathbf{u},\mathbf{v},\mathbf{w}$ as in Exercise I).
1 Solve $2\mathbf{x}+\mathbf{u}=3\mathbf{v}$ for $\mathbf{x}$.
2 Solve $4\mathbf{x}-2\mathbf{v}=\mathbf{x}+\mathbf{w}$ for $\mathbf{x}$.
3 Find $a$ and $b$ with $a(1,2)+b(3,-1)=(5,3)$.
4 Find all $a$, $b$ with $a(2,1)+b(-4,-2)=(6,3)$.
5 Show that $(1,1)$ is not a linear combination of $(2,1)$ and $(-4,-2)$.
6 Is $(3,1,5)$ a linear combination of $\mathbf{a}_1=(1,0,1)$ and $\mathbf{a}_2=(1,1,2)$?
7 Is $(2,-1,1)$ a linear combination of $\mathbf{a}_1=(1,0,1)$ and $\mathbf{a}_2=(1,1,2)$? If so, find the coefficients.
8 Write $(5,-1,4)$ as a linear combination of $(1,1,0)$, $(0,1,1)$ and $(1,0,1)$.
9 Write the system $2x_1-x_2+x_3=4$, $x_1+3x_3=-1$, $-x_1+2x_2=0$ as a vector equation.
10 Write $x_1(1,2)+x_2(-1,3)=(0,5)$ as a linear system and solve it.
💡 Show solutions▶
1 $2\mathbf{x}=3\mathbf{v}-\mathbf{u}=(-3,12,0)-(2,-1,3)=(-5,13,-3)$, so $\mathbf{x}=\left(-\tfrac52,\tfrac{13}2,-\tfrac32\right)$.
2 $3\mathbf{x}=2\mathbf{v}+\mathbf{w}=(-2,8,0)+(0,2,-5)=(-2,10,-5)$, so $\mathbf{x}=\left(-\tfrac23,\tfrac{10}3,-\tfrac53\right)$.
3 $a+3b=5$ and $2a-b=3$. From the second, $b=2a-3$; then $a+6a-9=5$, so $a=2$, $b=1$. Check: $2(1,2)+(3,-1)=(5,3)$. ✓
4 Both equations say $2a-4b=6$ and $a-2b=3$ — the same equation. So $a=3+2b$ with $b$ free: infinitely many solutions, e.g. $(a,b)=(3,0)$, $(5,1)$, $(1,-1)$. This happens because $(-4,-2)=-2(2,1)$.
5 $a(2,1)+b(-4,-2)=(a-2b)(2,1)$ is always a multiple of $(2,1)$, so its second component is half its first. For $(1,1)$ this fails ($1\ne\tfrac12$), so no $a,b$ exist.
6 $c_1+c_2=3$, $c_2=1$, $c_1+2c_2=5$. The first two give $c_1=2$, $c_2=1$, but then $c_1+2c_2=4\ne5$. Inconsistent: no.
7 $c_1+c_2=2$, $c_2=-1$, $c_1+2c_2=1$. So $c_2=-1$, $c_1=3$, and $3-2=1$ ✓. Yes: $(2,-1,1)=3\mathbf{a}_1-\mathbf{a}_2$.
8 $c_1+c_3=5$, $c_1+c_2=-1$, $c_2+c_3=4$. Adding all three: $2(c_1+c_2+c_3)=8$, so $c_1+c_2+c_3=4$. Then $c_2=4-5=-1$, $c_3=4-(-1)=5$, $c_1=4-4=0$: $(5,-1,4)=0(1,1,0)-(0,1,1)+5(1,0,1)$. Check: $(0+0+5,\,0-1+0,\,0-1+5)=(5,-1,4)$. ✓
10 $x_1-x_2=0$, $2x_1+3x_2=5$. Then $x_1=x_2$ and $5x_1=5$, so $x_1=x_2=1$. Check: $(1,2)+(-1,3)=(0,5)$. ✓
III
Geometry and applications.
1 Use position vectors to find the midpoint of $P(2,-1,4)$ and $Q(6,3,-2)$.
2 $ABCD$ is a parallelogram with $A(1,2)$, $B(4,3)$, $C(6,7)$. Find $D$.
3 Find the point one third of the way from $P(0,3,-3)$ to $Q(6,0,3)$.
4Statics. Forces $\mathbf{F}_1=(3,4)$ N and $\mathbf{F}_2=(-5,1)$ N act on a ring. Find the third force $\mathbf{F}_3$ that keeps the ring in equilibrium (total force $\mathbf{0}$).
5Navigation. A boat heads due north at $4$ m/s relative to the water while the current flows due east at $1.5$ m/s. Using east and north as the axes, find the boat's velocity relative to the ground, its speed, and its position after $100$ s.
6Nutrition. Using approximate, illustrative values: per $100$ g, oats contain $(13,68,7,10)$, milk $(3,5,4,0)$ and banana $(1,23,0,3)$ grams of (protein, carbohydrate, fat, fibre). Find the nutrient vector of a breakfast of $50$ g oats, $200$ g milk and $120$ g banana.
7Drones. A drone makes three displacements $(120,50,30)$, $(-40,80,0)$ and $(0,-30,-30)$ (metres; east, north, up). Where does it end up, and what single displacement brings it home?
8 Prove with vectors that the diagonals of every parallelogram bisect each other.
4 $\mathbf{F}_1+\mathbf{F}_2+\mathbf{F}_3=\mathbf{0}$, so $\mathbf{F}_3=-(\mathbf{F}_1+\mathbf{F}_2)=-(-2,5)=(2,-5)$ N.
5 Velocity $=(0,4)+(1.5,0)=(1.5,4)$ m/s. Speed (Pythagoras) $=\sqrt{1.5^2+4^2}=\sqrt{18.25}\approx4.27$ m/s. After $100$ s the position is $100(1.5,4)=(150,400)$: $150$ m east and $400$ m north of the start.
6 $0.5(13,68,7,10)+2(3,5,4,0)+1.2(1,23,0,3)=(6.5,34,3.5,5)+(6,10,8,0)+(1.2,27.6,0,3.6)=(13.7,\,71.6,\,11.5,\,8.6)$ g of protein, carbohydrate, fat and fibre.
7 Sum $=(120-40+0,\ 50+80-30,\ 30+0-30)=(80,100,0)$ m: $80$ m east, $100$ m north, back at take-off height. The way home is $-(80,100,0)=(-80,-100,0)$ m.
8 Let $ABCD$ be a parallelogram, so $\overrightarrow{AD}=\overrightarrow{BC}$, i.e. $\mathbf{d}-\mathbf{a}=\mathbf{c}-\mathbf{b}$ and $\mathbf{d}=\mathbf{a}+\mathbf{c}-\mathbf{b}$. The midpoint of $BD$ is $\tfrac12(\mathbf{b}+\mathbf{d})=\tfrac12(\mathbf{b}+\mathbf{a}+\mathbf{c}-\mathbf{b})=\tfrac12(\mathbf{a}+\mathbf{c})$, which is the midpoint of $AC$. The diagonals share their midpoint, so each bisects the other. $\blacksquare$
IV
Thinking about vectors.
1 Prove Theorem 4.1(f), $(c+d)\mathbf{u}=c\mathbf{u}+d\mathbf{u}$, for vectors in $\mathbb{R}^n$.
2 True or false: if $c\mathbf{u}=\mathbf{0}$, then $c=0$ or $\mathbf{u}=\mathbf{0}$. Justify.
3 Show that every vector $(a,b)$ in $\mathbb{R}^2$ is a linear combination of $(1,1)$ and $(1,-1)$, and find the coefficients.
4 Describe geometrically the set of all linear combinations of $(1,2,3)$ and $(2,4,6)$. Is $(1,0,0)$ one of them?
💡 Show solutions▶
1 The $i$-th component of $(c+d)\mathbf{u}$ is $(c+d)u_i=cu_i+du_i$ (distributive law for real numbers), which is the $i$-th component of $c\mathbf{u}+d\mathbf{u}$. All components agree, so the vectors are equal. $\blacksquare$
2True If $c\ne0$, multiply both sides by $\tfrac1c$: $\mathbf{u}=\tfrac1c(c\mathbf{u})=\tfrac1c\mathbf{0}=\mathbf{0}$ (Theorem 4.1(g), (h)). So either $c=0$ or $\mathbf{u}=\mathbf{0}$.
3 Solve $x(1,1)+y(1,-1)=(a,b)$: $x+y=a$, $x-y=b$. Adding and subtracting, $x=\tfrac{a+b}{2}$, $y=\tfrac{a-b}{2}$. These exist for every $a,b$, so every vector is reached: $(a,b)=\tfrac{a+b}{2}(1,1)+\tfrac{a-b}{2}(1,-1)$.
4 $(2,4,6)=2(1,2,3)$, so $c_1(1,2,3)+c_2(2,4,6)=(c_1+2c_2)(1,2,3)$: the combinations are all multiples of $(1,2,3)$ — a line through the origin in the direction $(1,2,3)$. $(1,0,0)$ is not on it: $t(1,2,3)=(1,0,0)$ would need $t=1$ and $2t=0$ at once.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes
Research question. How much must a pilot turn the aircraft into a crosswind so that it still flies straight to its destination, and how much time does the wind cost?
Why it matters. An aircraft moves through the air, and the air itself moves over the ground. The ground velocity is the vector sum air velocity + wind velocity. Flight-planning software, drone autopilots and ships crossing a current all solve this triangle of vectors many times per second.
What to do.
Sample case: a light aircraft flies at $220$ km/h through the air and must travel $500$ km due north. A $40$ km/h wind blows from west to east. Use axes east $=(1,0)$, north $=(0,1)$, so the wind is $\mathbf{w}=(40,0)$.
Let the air velocity be $\mathbf{a}=(a_1,a_2)$ with $a_1^2+a_2^2=220^2$. The ground velocity $\mathbf{a}+\mathbf{w}$ must point due north, so its first component is $0$. Find $a_1$ and $a_2$.
Find the heading correction angle $\theta$ west of north from $\sin\theta=\tfrac{40}{220}$, the ground speed, and the flight time. Compare with the time in still air.
Repeat for a wind of $40$ km/h blowing from the north-west, $\mathbf{w}=(20\sqrt2,-20\sqrt2)$. Is the time cost larger or smaller?
Tools: Calculator or spreadsheet; graph paper to draw the vector triangle to scale.
Hand in: A half-page note with your vector triangle, the heading correction, ground speed and flight time for both winds, and one sentence on why a crosswind always costs time.
Going further. Look up the wind triangle on a pilot's flight computer (the "E6B") or the way river ferries in Bangladesh or on the Nile angle into the current. How does a drone autopilot measure the wind it cannot see?
Instructor note: Case 1: $\mathbf{a}=(-40,\sqrt{46800})\approx(-40,216.33)$, heading $\theta\approx10.48^\circ$ west of north, ground speed $\approx216.33$ km/h, time $\approx2.311$ h versus $2.273$ h in still air (about $2.3$ min longer). Case 2: $a_1=-20\sqrt2\approx-28.28$, $a_2=\sqrt{220^2-800}\approx218.17$, ground speed $a_2-20\sqrt2\approx189.89$ km/h, time $\approx2.633$ h — the headwind part costs far more than the crosswind part.
🧪 Self-check quiz
0/5 correct
Q1. The vector from $P(1,4)$ to $Q(3,1)$ is
Q2. $2(1,-1,0)-(3,1,2)=$
Q3. Which vector is a linear combination of $(1,2,0)$ and $(0,1,1)$?
Q4. Drawn with both tails at the origin, the vector $\mathbf{u}-\mathbf{v}$ is represented by the arrow
Q5. A linear system with coefficient columns $\mathbf{a}_1,\dots,\mathbf{a}_n$ and right-hand side $\mathbf{b}$ is consistent exactly when