---
title: "Vectors"
book: "High School Mathematics"
subject: math
language: en
chapter: 6
exercises: 10
source: https://one-course.com/books/math/2/en/chapter/6-vectors
---

# Chapter 6 — Vectors

A [vector](#def-g10-vectors-vector) encodes a displacement: a direction and a length, regardless of the starting point. [Vectors](#def-g10-vectors-vector) give clean proofs of geometric facts and, once [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) enter the picture, reduce them to small computations. They will be a key tool in [Chapter 7](https://one-course.com/books/math/2/en/chapter/7-equations-of-lines-and-linear-systems#ch-g10-lines) and, later, for the scalar product in [Chapter 16](https://one-course.com/books/math/2/en/chapter/16-the-scalar-product-in-the-plane#ch-g11-scal).

## 6.1 Translations and vectors

**Definition 6.1 (Vector).**

Given two points $A$ and $B$, the *vector* $\vect{AB}$ represents the displacement from $A$ to $B$: it has a *direction* (the direction of the line $(AB)$, together with the way of traveling along it, from $A$ towards $B$) and a *length* (the distance $AB$, also written $\norm{\vect{AB}}$). Two vectors are *equal* when they encode the same displacement: $\vect{AB} = \vect{CD}$ means that applying the two displacements to any point gives the same result.

**Proposition 6.2 (Equal vectors and parallelograms).**

$\vect{AB} = \vect{CD}$ exactly when $ABDC$ (in this order!) is a parallelogram, possibly a flattened one — equivalently, when $[AD]$ and $[BC]$ have the same [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint).

**Proof.** *Admitted at this level.* ∎

![Equal vectors AB = CD: same direction, same way, same length. The quadrilateral ABDC is a parallelogram.](https://one-course.com/images/onecourse/chapters/math-2/g10-vectors/fig-999585eecc93.svg)

*Equal [vectors](#def-g10-vectors-vector) $\vect{AB} = \vect{CD}$: same direction, same way, same length. The quadrilateral $ABDC$ is a parallelogram.*

**Notation 6.3.**

A [vector](#def-g10-vectors-vector) is often named by a single letter, $\vec u$ or $\vec v$, when the endpoints do not matter. The *zero [vector](#def-g10-vectors-vector)* $\vec 0 = \vect{AA}$ is the displacement that does not move anything. The *opposite* of $\vect{AB}$ is $\vect{BA}$: same length, opposite way. We write $\vect{BA} = -\vect{AB}$.

## 6.2 Adding vectors

**Definition 6.4 (Sum of two vectors).**

The *sum* $\vec u + \vec v$ is the displacement obtained by performing $\vec u$ then $\vec v$.

**Theorem 6.5 (Chasles’ relation).**

For any three points $A$, $B$, $C$:

$$
\vect{AB} + \vect{BC} = \vect{AC}.
$$

**Proof.** Going from $A$ to $B$, then from $B$ to $C$, is a displacement that takes $A$ to $C$: it is the displacement encoded by $\vect{AC}$. ∎

**Remark 6.6 (Parallelogram rule).**

To add two [vectors](#def-g10-vectors-vector) drawn from the same point, $\vect{AB} + \vect{AC}$, complete the parallelogram $ABDC$: the [sum](#def-g10-vectors-sum) is the diagonal $\vect{AD}$. Indeed $\vect{AC} = \vect{BD}$, so $\vect{AB} + \vect{AC} = \vect{AB} + \vect{BD} = \vect{AD}$ by Chasles’ relation.

![Two pictures of the sum: tip-to-tail (Chasles’ relation, left) and the parallelogram rule (right).](https://one-course.com/images/onecourse/chapters/math-2/g10-vectors/fig-3ca1613ca043.svg)

![Two pictures of the sum: tip-to-tail (Chasles’ relation, left) and the parallelogram rule (right).](https://one-course.com/images/onecourse/chapters/math-2/g10-vectors/fig-3828ee349c85.svg)

*Two pictures of the [sum](#def-g10-vectors-sum): tip-to-tail (Chasles’ relation, left) and the parallelogram rule (right).*

**Example 6.7 (Simplifying with Chasles).**

Simplify $\vect{MN} + \vect{NP} + \vect{PQ}$: chaining the displacements gives $\vect{MQ}$. Simplify $\vect{AB} - \vect{AC}$: rewrite the difference as a [sum](#def-g10-vectors-sum) with the opposite [vector](#def-g10-vectors-vector),

$$
\vect{AB} - \vect{AC} = \vect{AB} + \vect{CA}
= \vect{CA} + \vect{AB} = \vect{CB}.
$$

## 6.3 Multiplying a vector by a number

**Definition 6.8 (Scalar multiple).**

Let $\vec u$ be a nonzero [vector](#def-g10-vectors-vector) and $k$ a [real number](https://one-course.com/books/math/2/en/chapter/1-numbers-and-sets-of-numbers#def-g10-numbers-sets). The [vector](#def-g10-vectors-vector) $k\vec u$ has the same direction as $\vec u$, length $\abs{k} \times \norm{\vec u}$, and points the same way as $\vec u$ if $k > 0$, the opposite way if $k < 0$. For $k = 0$, $0\vec u = \vec 0$.

**Definition 6.9 (Collinearity).**

Two [vectors](#def-g10-vectors-vector) $\vec u$ and $\vec v$ are *collinear* when they have the same direction, i.e. when $\vec v = k \vec u$ for some real $k$ (or one of them is $\vec 0$).

**Remark 6.10.**

Collinearity is the [vector](#def-g10-vectors-vector) language for parallelism and alignment:

- the lines $(AB)$ and $(CD)$ are parallel exactly when $\vect{AB}$ and $\vect{CD}$ are [collinear](#def-g10-vectors-collinear) ;
- the points $A$ , $B$ , $C$ are aligned exactly when $\vect{AB}$ and $\vect{AC}$ are [collinear](#def-g10-vectors-collinear) .

## 6.4 Coordinates of vectors

**Definition 6.11 (Coordinates of a vector).**

In a [coordinate system](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system), the *[coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system)* of the [vector](#def-g10-vectors-vector) $\vect{AB}$ are the numbers describing the displacement:

$$
\vect{AB}\,(x_B - x_A,\ y_B - y_A).
$$

A [vector](#def-g10-vectors-vector) $\vec u\,(a, b)$ moves every point $a$ units horizontally and $b$ units vertically.

**Proposition 6.12 (Computing with coordinates).**

Let $\vec u\,(a, b)$ and $\vec v\,(c, d)$, and $k \in \R$.

1. $\vec u = \vec v$ exactly when $a = c$ and $b = d$ ;
2. $\vec u + \vec v$ has [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) $(a + c,\ b + d)$ ;
3. $k\vec u$ has [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) $(ka,\ kb)$ ;
4. $\norm{\vec u} = \sqrt{a^2 + b^2}$ ( [orthonormal system](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) ).

**Proof.** 1–3 restate, coordinate by coordinate, what the displacements do: for instance performing $\vec u$ then $\vec v$ moves a point horizontally by $a$ then $c$, hence by $a + c$ in total. Point 4 is the distance formula of [Theorem 5.4](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#thm-g10-coordgeom-distance) applied to a segment representing $\vec u$. ∎

**Theorem 6.13 (Collinearity criterion).**

Two [vectors](#def-g10-vectors-vector) $\vec u\,(a, b)$ and $\vec v\,(c, d)$ are [collinear](#def-g10-vectors-collinear) if and only if

$$
ad - bc = 0 .
$$

**Proof.** If $\vec v = k\vec u$, then $c = ka$ and $d = kb$, so $ad - bc = a(kb) - b(ka) = 0$. The same holds if $\vec u = k \vec v$ or if either [vector](#def-g10-vectors-vector) is zero.

Conversely, suppose $ad - bc = 0$ with $\vec u \neq \vec 0$, say $a \neq 0$ (the case $b \neq 0$ is similar). Set $k = \frac{c}{a}$; then $c = ka$, and $ad = bc = b(ka)$ gives, after dividing by $a \neq 0$, $d = kb$. So $\vec v = k\vec u$. ∎

**Example 6.14.**

Are $\vec u\,(3, -2)$ and $\vec v\,(-6, 4)$ [collinear](#def-g10-vectors-collinear)? Compute $ad - bc = 3 \times 4 - (-2)\times(-6) = 12 - 12 = 0$: yes, and indeed $\vec v = -2\vec u$. For $\vec u\,(3, -2)$ and $\vec w\,(1, 5)$: $3 \times 5 - (-2) \times 1 = 17 \neq 0$: not [collinear](#def-g10-vectors-collinear).

**Method 6.15 (Alignment and parallelism).**

To prove that three points $A$, $B$, $C$ are aligned:

1. compute the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of $\vect{AB}$ and $\vect{AC}$ ;
2. check the criterion $ad - bc = 0$ of [Theorem 6.13](#thm-g10-vectors-det) ;
3. conclude: the [vectors](#def-g10-vectors-vector) are [collinear](#def-g10-vectors-collinear) and share the point $A$ , so the three points are aligned.

The same computation with $\vect{AB}$ and $\vect{CD}$ proves that the lines $(AB)$ and $(CD)$ are parallel.

**Example 6.16.**

Let $A(1, 2)$, $B(3, 5)$ and $C(7, 11)$. Then $\vect{AB}\,(2, 3)$ and $\vect{AC}\,(6, 9)$, and $2 \times 9 - 3 \times 6 = 18 - 18 = 0$: the points are aligned (in fact $\vect{AC} = 3\vect{AB}$).

![Aligned points: AC is a scalar multiple of AB.](https://one-course.com/images/onecourse/chapters/math-2/g10-vectors/fig-1e83277b8c55.svg)

*Aligned points: $\vect{AC}$ is a scalar multiple of $\vect{AB}$.*

## 6.5 Exercises

**Exercise 6.1 ★.**

Simplify using Chasles’ relation:

$$
\vect{AB} + \vect{BD}, \qquad
\vect{KL} + \vect{LM} + \vect{MK}, \qquad
\vect{AC} - \vect{BC}, \qquad
\vect{AB} + \vect{CA}.
$$

**Solution of Exercise 6.1.**

$\vect{AB} + \vect{BD} = \vect{AD}$ (Chasles).

$\vect{KL} + \vect{LM} + \vect{MK} = \vect{KK} = \vec 0$.

$\vect{AC} - \vect{BC} = \vect{AC} + \vect{CB} = \vect{AB}$.

$\vect{AB} + \vect{CA} = \vect{CA} + \vect{AB} = \vect{CB}$.

**Exercise 6.2 ★.**

Let $A(2, 1)$, $B(5, 3)$, $C(-1, 4)$. Compute the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of $\vect{AB}$, $\vect{BC}$, $\vect{AC}$, and check on the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) that $\vect{AB} + \vect{BC} = \vect{AC}$.

**Solution of Exercise 6.2.**

$\vect{AB}\,(5-2,\ 3-1) = (3, 2)$; $\vect{BC}\,(-1-5,\ 4-3) = (-6, 1)$; $\vect{AC}\,(-1-2,\ 4-1) = (-3, 3)$. Check: $(3, 2) + (-6, 1) = (-3, 3)$, coordinate by coordinate.

**Exercise 6.3 ★.**

Let $\vec u\,(2, -3)$ and $\vec v\,(-1, 4)$. Give the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of $\vec u + \vec v$, $3\vec u$, $2\vec u - \vec v$, and compute $\norm{\vec u}$.

**Solution of Exercise 6.3.**

$\vec u + \vec v = (2 + (-1),\ -3 + 4) = (1, 1)$.

$3\vec u = (6, -9)$.

$2\vec u - \vec v = (4 - (-1),\ -6 - 4) = (5, -10)$.

$\norm{\vec u} = \sqrt{2^2 + (-3)^2} = \sqrt{13}$.

**Exercise 6.4 ★.**

In each case, say whether $\vec u$ and $\vec v$ are [collinear](#def-g10-vectors-collinear):

$$
\text{(a) } \vec u\,(4, 6),\ \vec v\,(6, 9);
\qquad
\text{(b) } \vec u\,(2, -5),\ \vec v\,(-4, 10);
\qquad
\text{(c) } \vec u\,(3, 1),\ \vec v\,(1, 3).
$$

**Solution of Exercise 6.4.**

(a) $4 \times 9 - 6 \times 6 = 36 - 36 = 0$: [collinear](#def-g10-vectors-collinear) ($\vec v = \frac32 \vec u$).

(b) $2 \times 10 - (-5) \times (-4) = 20 - 20 = 0$: [collinear](#def-g10-vectors-collinear) ($\vec v = -2\vec u$).

(c) $3 \times 3 - 1 \times 1 = 8 \neq 0$: not [collinear](#def-g10-vectors-collinear).

**Exercise 6.5 ★.**

Let $A(0, 2)$, $B(4, 0)$, $C(6, 3)$. Find the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of the point $D$ such that $ABCD$ is a parallelogram, i.e. such that $\vect{AB} = \vect{DC}$.

**Solution of Exercise 6.5.**

$\vect{AB}\,(4, -2)$, and $\vect{DC}\,(6 - x_D,\ 3 - y_D)$. The condition $\vect{AB} = \vect{DC}$ gives $6 - x_D = 4$ and $3 - y_D = -2$, so $D(2, 5)$.

**Exercise 6.6 ★★.**

Let $A(-2, 1)$, $B(1, 3)$, $C(4, -1)$.

1. Compute the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of the point $M$ such that $\vect{AM} = \vect{AB} + \vect{AC}$ .
2. What is the quadrilateral $ABMC$ ? Justify.

**Solution of Exercise 6.6.**

*1.* $\vect{AB}\,(3, 2)$ and $\vect{AC}\,(6, -2)$, so $\vect{AB} + \vect{AC} = (9, 0)$ and $M = (-2 + 9,\ 1 + 0) = (7, 1)$.

*2.* By the parallelogram rule, $\vect{AM} = \vect{AB} + \vect{AC}$ means that $ABMC$ is a parallelogram ($M$ is the fourth vertex opposite $A$). One can check: $\vect{AB}\,(3,2)$ and $\vect{CM}\,(7-4,\ 1-(-1)) =
(3, 2)$ are equal.

**Exercise 6.7 ★★.**

Let $A(1, -1)$, $B(4, 1)$, $C(10, 5)$. Are the points $A$, $B$, $C$ aligned? Same question for $A(0, 3)$, $B(2, 2)$, $C(8, -1)$.

**Solution of Exercise 6.7.**

First triple: $\vect{AB}\,(3, 2)$, $\vect{AC}\,(9, 6)$; $3 \times 6 - 2 \times 9 = 0$: aligned ($\vect{AC} = 3\vect{AB}$).

Second triple: $\vect{AB}\,(2, -1)$, $\vect{AC}\,(8, -4)$; $2 \times (-4) - (-1) \times 8 = -8 + 8 = 0$: aligned as well ($\vect{AC} = 4\vect{AB}$).

**Exercise 6.8 ★★.**

Let $A(1, 2)$, $B(5, 4)$, $C(6, 1)$, $D(2, -1)$. Show that $(AB)$ and $(CD)$ are parallel, then that $ABCD$ is a parallelogram. Which verification implies the other?

**Solution of Exercise 6.8.**

$\vect{AB}\,(4, 2)$ and $\vect{DC}\,(6-2,\ 1-(-1)) = (4, 2)$: the two [vectors](#def-g10-vectors-vector) are *equal*, so $ABCD$ is a parallelogram, and in particular $(AB) \parallel (CD)$. The parallelogram property is the stronger one: it implies the parallelism (collinearity of the [vectors](#def-g10-vectors-vector)), not conversely.

**Exercise 6.9 ★★.**

Let $ABC$ be a triangle, and let $I$ be the point defined by $\vect{AI} = \frac23 \vect{AB}$. With $A(0,0)$, $B(6, 3)$, $C(2, 5)$ (so that computations stay simple):

1. compute the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of $I$ ;
2. let $J$ be such that $\vect{CJ} = \frac23\vect{CB}$ ; compute the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of $J$ ;
3. show that $(IJ)$ is parallel to $(AC)$ .

**Solution of Exercise 6.9.**

*1.* $\vect{AB}\,(6, 3)$, so $\vect{AI} = \frac23\vect{AB} = (4, 2)$ and $I(4, 2)$.

*2.* $\vect{CB}\,(6-2,\ 3-5) = (4, -2)$, so $\vect{CJ} = \frac23\vect{CB} = \left(\frac83, -\frac43\right)$ and $J\left(2 + \frac83,\ 5 - \frac43\right) = \left(\frac{14}{3},
\frac{11}{3}\right)$.

*3.* $\vect{IJ}\,\left(\frac{14}{3} - 4,\ \frac{11}{3} - 2\right) =
\left(\frac23, \frac53\right)$ and $\vect{AC}\,(2, 5)$. Collinearity criterion:

$$
\frac23 \times 5 - \frac53 \times 2 = \frac{10}{3} - \frac{10}{3} = 0 :
$$

the [vectors](#def-g10-vectors-vector) are [collinear](#def-g10-vectors-collinear), so $(IJ)$ is parallel to $(AC)$. (Both $I$ and $J$ sit one third of the way from the side towards $B$: the small triangle $IBJ$ is a reduction of $ABC$.)

**Exercise 6.10 ★★★.**

Let $ABCD$ be any quadrilateral, and let $I$, $J$, $K$, $L$ be the [midpoints](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[AB]$, $[BC]$, $[CD]$, $[DA]$. Using [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) $A(x_A, y_A)$, etc., show that $\vect{IJ} = \vect{LK}$, and conclude that the [midpoints](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of the sides of *any* quadrilateral form a parallelogram.

**Solution of Exercise 6.10.**

The [midpoints](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) are

$$
I\left(\tfrac{x_A + x_B}{2}, \tfrac{y_A + y_B}{2}\right), \quad
J\left(\tfrac{x_B + x_C}{2}, \tfrac{y_B + y_C}{2}\right), \quad
K\left(\tfrac{x_C + x_D}{2}, \tfrac{y_C + y_D}{2}\right), \quad
L\left(\tfrac{x_D + x_A}{2}, \tfrac{y_D + y_A}{2}\right).
$$

First [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of the two [vectors](#def-g10-vectors-vector):

$$
x_J - x_I = \frac{x_B + x_C}{2} - \frac{x_A + x_B}{2}
= \frac{x_C - x_A}{2},
\qquad
x_K - x_L = \frac{x_C + x_D}{2} - \frac{x_D + x_A}{2}
= \frac{x_C - x_A}{2},
$$

and the same computation works for the second [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system). So $\vect{IJ} = \vect{LK}$ (both equal $\frac12\vect{AC}$), and $IJKL$ is a parallelogram, whatever the quadrilateral $ABCD$ looks like.

## 6.6 Problem: The algebra of arrows

**Problem 6.1.**

Weekend problem — Chasles gymnastics, the centroid re-proved in two lines, and forces in equilibrium: what vectors are really for

Grade 8 proved that the medians of a triangle meet at two thirds of their length — with an ingenious parallelogram hidden inside the triangle (the centroid weekend problem of the Middle School volume). [Vectors](#def-g10-vectors-vector) reprove it in two lines, and hand out the concurrency for free. That is what this new algebra is for: theorems become computations with arrows. This problem trains the computation (Chasles above all), delivers the two-line proof, extends Varignon ([Exercise 6.10](#exo-g10-vectors-10)), and ends where [vectors](#def-g10-vectors-vector) were born: forces and velocities.

**Part I — Chasles gymnastics.**

1. Simplify using Chasles’ relation ( [Theorem 6.5](#thm-g10-vectors-chasles) ): $\vect{AB} + \vect{BC} + \vect{CD}$ ; $\vect{MA} - \vect{MB}$ ; $\vect{AB} + \vect{CD} + \vect{BC} + \vect{DA}$ .
2. The origin trick: show that for *any* point $O$ , $\vect{AB} = \vect{OB} - \vect{OA}$ .
3. Prove the two characterizations of the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) $I$ of $[AB]$: $$\vect{IA} + \vect{IB} = \vec 0  \qquad\text{and}\qquad  \vect{OI} = \tfrac12\left(\vect{OA} + \vect{OB}\right)  \ \text{for any } O .$$
4. Using $\vect{AB} = \vect{DC}$ , re-find the fourth vertex $D$ of the parallelogram $ABCD$ with $A(-1, 2)$ , $B(3, 4)$ , $C(6, 0)$ — and compare with the midpoint-trick answer of [Problem 5.1](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#pb-g10-coordgeom-1) .
5. Adding a fixed [vector](#def-g10-vectors-vector) $\vec u$ to every point of the plane performs which transformation — the one discovered with two mirrors in the two-mirrors weekend problem of the Middle School volume and two half-turns in the half-turns weekend problem of the Middle School volume? Compute the [image](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) of $(1, 2)$ under $\vec u = (3, -1)$ .

**Part II — The centroid, third proof.** Define the point $G$ of a triangle $ABC$ by the elegant condition

$$
\vect{GA} + \vect{GB} + \vect{GC} = \vec 0 .
$$

6. Using the origin trick, show that this condition determines $G$ uniquely: $\vect{OG} = \frac13\left(\vect{OA} + \vect{OB} +  \vect{OC}\right)$ for any $O$ — so the [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) of $G$ are the *averages* of the vertices’ (the centroid weekend problem of the Middle School volume saw it numerically).
7. Prove $\vect{AG} = \frac13\left(\vect{AB} +  \vect{AC}\right)$ , and, writing $K$ for the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[BC]$ , $\vect{AK} = \frac12\left(\vect{AB} +  \vect{AC}\right)$ . Conclude: $\vect{AG} = \frac23\,\vect{AK}$ — the two-thirds theorem, re-proved.
8. Why do the medians from $B$ and from $C$ pass through the *same* $G$ with no further computation? Compare the three available proofs of this theorem — grade 8’s hidden parallelogram, [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) , [vectors](#def-g10-vectors-vector) — in one sentence each.
9. Numerical check: $A(1, 1)$ , $B(5, 2)$ , $C(3, 6)$ . Compute $G$ , then $K$ , then verify $\vect{AG} = \frac23 \vect{AK}$ with the collinearity criterion ( [Theorem 6.13](#thm-g10-vectors-det) ).
10. Physics reads $G$ as the balance point of three equal masses at the vertices. Double the mass at $A$ (masses $2, 1, 1$ ): the balance point becomes $\vect{OG'} = \frac{2\vect{OA} + \vect{OB} +  \vect{OC}}{4}$ . Compute $G'$ for the triangle of question 9 and compare its position with $G$ .

**Part III — Collinearity at work.**

11. Are $\vec u(3, -2)$ and $\vec v(-6, 4)$ [collinear](#def-g10-vectors-collinear) ? And $\vec w(2, 5)$ , $\vec z(4, 9)$ ? Decide with the determinant criterion.
12. Are the points $A(1, 2)$ , $B(3, 5)$ , $C(7, 11)$ aligned?
13. Beyond Varignon ( [Exercise 6.10](#exo-g10-vectors-10) ): in any quadrilateral $ABCD$ , the *bimedians* join the [midpoints](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of opposite sides ( $[IK]$ and $[JL]$ ). Using position [vectors](#def-g10-vectors-vector) ( $\vect{OI} = \frac12(\vect{OA} +  \vect{OB})$ , etc.), show that both bimedians have the same [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) , of position [vector](#def-g10-vectors-vector) $\frac14\left(\vect{OA} + \vect{OB} + \vect{OC} +  \vect{OD}\right)$ : the centroid of the quadrilateral.
14. Thales in one line: $M$ and $N$ are the points defined by $\vect{AM} = \frac13\vect{AB}$ and $\vect{AN} = \frac13\vect{AC}$ . Compute $\vect{MN}$ in terms of $\vect{BC}$ and conclude parallelism and ratio.
15. Generalize question 14 to an arbitrary ratio $k$ , and state in one sentence how the [vector](#def-g10-vectors-vector) calculus absorbs the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) theorem (the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) theorem) and Thales (Thales’ theorem) at once.

**Part IV — Forces and velocities.**

16. A boat is pushed by a current of $3$ km/h due east while rowing at $4$ km/h due north. Give the resultant velocity [vector](#def-g10-vectors-vector) , its magnitude, and describe the actual course. (A very old triple appears.)
17. Three forces act on a ring: $\vec u(2, 1)$ , $\vec v(-3, 2)$ , $\vec w(1, -3)$ . Compute the resultant. What does the ring do? And why is the [sum](#def-g10-vectors-sum) of the arrows along any *closed* polygon always $\vec 0$ (Chasles)?
18. Two forces $\vec u(4, -1)$ and $\vec v(-1, 3)$ act on a hook. What third force holds the hook in equilibrium?
19. A swimmer crosses a river heading straight across at $2$ m/s; the current runs at $1.5$ m/s. Give the resultant velocity and its magnitude. The river is $50$ m wide: how long does the crossing take, and how far downstream does the swimmer land?
20. Finale: one object, three costumes — displacement arrow, coordinate pair, force or velocity. List the three classical theorems this problem re-proved with [vector](#def-g10-vectors-vector) algebra, and name the one geometric quantity the [vectors](#def-g10-vectors-vector) of this chapter still cannot measure — the tool that fixes it arrives with the scalar product, in grade 11.

**Solution of Problem 6.1.**

**1.** $\vect{AB} + \vect{BC} + \vect{CD} = \vect{AD}$; $\vect{MA} - \vect{MB} = \vect{MA} + \vect{BM} =
\vect{BA}$; $\vect{AB} + \vect{BC} + \vect{CD} +
\vect{DA} = \vect{AA} = \vec 0$ (reordering the third [sum](#def-g10-vectors-sum)).

**2.** $\vect{OB} - \vect{OA} = \vect{AO} + \vect{OB} =
\vect{AB}$ by Chasles.

**3.** $I$ is the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[AB]$ exactly when $\vect{IA} = -\vect{IB}$, i.e. $\vect{IA} + \vect{IB} = \vec 0$. Then, for any $O$: $\vec 0 = \vect{IO} + \vect{OA} + \vect{IO} + \vect{OB}$, so $2\,\vect{OI} = \vect{OA} + \vect{OB}$ — and conversely.

**4.** $\vect{AB} = (4, 2)$, and $D$ satisfies $\vect{DC} = \vect{AB}$: $D = C - (4, 2) = (2, -2)$ — the same $D$ as [Problem 5.1](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#pb-g10-coordgeom-1), one line faster.

**5.** The translation of [vector](#def-g10-vectors-vector) $\vec u$ — exactly the transformation produced by two parallel mirrors or two half-turns. [Image](https://one-course.com/books/math/2/en/chapter/3-functions#def-g10-functions-function) of $(1, 2)$: $(4, 1)$.

**6.** [Expanding](https://one-course.com/books/math/2/en/chapter/2-algebra-equations-and-inequalities#def-g10-algebra-expand) each $\vect{GX}$ as $\vect{GO} + \vect{OX}$: $3\,\vect{GO} + \vect{OA} + \vect{OB} + \vect{OC} = \vec 0$, so $\vect{OG} = \frac13(\vect{OA} + \vect{OB} + \vect{OC})$: one and only one such point, whose [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) are the averages of the three vertices’.

**7.** $\vect{AB} + \vect{AC} =
(\vect{AG} + \vect{GB}) + (\vect{AG} + \vect{GC})
= 2\vect{AG} + (\vect{GB} + \vect{GC})
= 2\vect{AG} - \vect{GA} = 3\vect{AG}$, whence $\vect{AG} = \frac13(\vect{AB} + \vect{AC})$. The [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) formula (question 3, seen from $A$) gives $\vect{AK} = \frac12(\vect{AB} + \vect{AC})$. Comparing: $\vect{AG} = \frac23\vect{AK}$ — $G$ lies on the median $[AK]$, two thirds of the way from $A$.

**8.** The defining condition treats $A$, $B$, $C$ identically, so the same computation from $B$ or from $C$ places the *same* $G$ at two thirds of those medians: concurrency costs nothing. The parallelogram proof was ingenious but ad hoc; [coordinates](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#def-g10-coordgeom-system) compute but obscure; [vectors](#def-g10-vectors-vector) make the symmetry visible and the proof two lines.

**9.** $G\left(\frac{1+5+3}{3}, \frac{1+2+6}{3}\right) =
(3, 3)$; $K(4, 4)$; $\vect{AG} = (2, 2)$ and $\vect{AK} = (3, 3)$: determinant $2 \times 3 - 2 \times 3 = 0$, [collinear](#def-g10-vectors-collinear), and indeed $\vect{AG} = \frac23\vect{AK}$.

**10.** $G'\left(\frac{2 + 5 + 3}{4}, \frac{2 + 2 +
6}{4}\right) = (2.5,\ 2.5)$: pulled from $G(3,3)$ towards the doubled vertex $A(1,1)$, as a balance point should be.

**11.** $3 \times 4 - (-2)(-6) = 12 - 12 = 0$: [collinear](#def-g10-vectors-collinear) ($\vec v = -2\vec u$). $2 \times 9 - 5 \times 4 = -2 \neq 0$: not [collinear](#def-g10-vectors-collinear).

**12.** $\vect{AB} = (2, 3)$, $\vect{AC} = (6, 9)$: determinant $2 \times 9 - 3 \times 6 = 0$: aligned.

**13.** $\vect{OI} = \frac12(\vect{OA} + \vect{OB})$ and $\vect{OK} = \frac12(\vect{OC} + \vect{OD})$, so the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) of $[IK]$ has position [vector](#def-g10-vectors-vector) $\frac12(\vect{OI} + \vect{OK}) = \frac14(\vect{OA} +
\vect{OB} + \vect{OC} + \vect{OD})$ — and the computation for $[JL]$ gives the very same [vector](#def-g10-vectors-vector). The two bimedians (and, by Varignon, the parallelogram $IJKL$’s diagonals) all share this center: the quadrilateral’s centroid.

**14.** $\vect{MN} = \vect{AN} - \vect{AM} =
\frac13\vect{AC} - \frac13\vect{AB} = \frac13\vect{BC}$: so $(MN) \parallel (BC)$ and $MN = \frac13 BC$.

**15.** With $\vect{AM} = k\,\vect{AB}$ and $\vect{AN} = k\,\vect{AC}$: $\vect{MN} = k\,\vect{BC}$ — one line containing the [midpoint](https://one-course.com/books/math/2/en/chapter/5-coordinate-geometry#prop-g10-coordgeom-midpoint) theorem ($k = \frac12$) and Thales (any $k$): the ratio passes through the subtraction untouched.

**16.** Resultant $(3, 4)$, magnitude $\sqrt{9 + 16} = 5$ km/h: the boat actually travels north-east-by-north along the $3$–$4$–$5$ triangle’s hypotenuse.

**17.** $\vec u + \vec v + \vec w = (0, 0)$: the ring stays put — equilibrium. Along a closed polygon $A_1 A_2 \dots A_n A_1$, Chasles telescopes the [sum](#def-g10-vectors-sum) to $\vect{A_1 A_1} = \vec 0$: going around brings you back.

**18.** $\vec w = -(\vec u + \vec v) = -(3, 2) =
(-3, -2)$.

**19.** Resultant $(1.5,\ 2)$, magnitude $\sqrt{2.25 + 4} = 2.5$ m/s. Crossing: $50$ m at $2$ m/s across: $25$ s; drift: $1.5 \times 25 = 37.5$ m downstream.

**20.** Re-proved this weekend: the centroid’s two-thirds theorem (question 7), the bimedian/Varignon center (question 13), Thales–midpoint (questions 14–15). What arrows still cannot do: measure *angles* (and hence lengths in arbitrary directions) — the scalar product of grade 11 is precisely the missing instrument.