High School Mathematics · Grades 10–12

6Vectors

A vector encodes a displacement: a direction and a length, regardless of the starting point. Vectors give clean proofs of geometric facts and, once coordinates enter the picture, reduce them to small computations. They will be a key tool in Chapter 7 and, later, for the scalar product in Chapter 16.

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.

Proof. Admitted at this level. ∎

Equal vectors AB = CD: same direction, same way, same length. The quadrilateral ABDC is a parallelogram. — Equal vectors $\vect{AB} = \vect{CD}$ : same direction, same way, same length. The quadrilateral $ABDC$ is a parallelogram.

Notation 6.3

A vector is often named by a single letter, $\vec u$ or $\vec v$ , when the endpoints do not matter. The zero 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 drawn from the same point, $\vect{AB} + \vect{AC}$ , complete the parallelogram $ABDC$ : the 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).

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 with the opposite 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 and $k$ a real number. The 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 $\vec u$ 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 language for parallelism and alignment:

the lines $(AB)$ and $(CD)$ are parallel exactly when $\vect{AB}$ and $\vect{CD}$ are collinear;
the points $A$ , $B$ , $C$ are aligned exactly when $\vect{AB}$ and $\vect{AC}$ are collinear.

6.4 Coordinates of vectors

Definition 6.11 (Coordinates of a vector)

In a coordinate system, the coordinates of the vector $\vect{AB}$ are the numbers describing the displacement:

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

A 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 $k \in \R$ .

$\vec u = \vec v$ exactly when $a = c$ and $b = d$ ;
$\vec u + \vec v$ has coordinates $(a + c,\ b + d)$ ;
$k\vec u$ has coordinates $(ka,\ kb)$ ;
$\norm{\vec u} = \sqrt{a^2 + b^2}$ (orthonormal 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 applied to a segment representing $\vec u$ . ∎

Theorem 6.13 (Collinearity criterion)

Two vectors $\vec u\,(a, b)$ and $\vec v\,(c, d)$ are 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 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? 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.

Method 6.15 (Alignment and parallelism)

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

compute the coordinates of $\vect{AB}$ and $\vect{AC}$ ;
check the criterion $ad - bc = 0$ of Theorem 6.13;
conclude: the vectors are 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. — 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

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 of $\vect{AB}$ , $\vect{BC}$ , $\vect{AC}$ , and check on the coordinates that $\vect{AB} + \vect{BC} = \vect{AC}$ .

Solution

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 of $\vec u + \vec v$ , $3\vec u$ , $2\vec u - \vec v$ , and compute $\norm{\vec u}$ .

Solution

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:

\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

Solution of Exercise 6.4.

(a) $4 \times 9 - 6 \times 6 = 36 - 36 = 0$ : collinear ( $\vec v = \frac32 \vec u$ ).

(b) $2 \times 10 - (-5) \times (-4) = 20 - 20 = 0$ : collinear ( $\vec v = -2\vec u$ ).

Exercise 6.5 ★

Let $A(0, 2)$ , $B(4, 0)$ , $C(6, 3)$ . Find the coordinates of the point $D$ such that $ABCD$ is a parallelogram, i.e. such that $\vect{AB} = \vect{DC}$ .

Solution

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

Compute the coordinates of the point $M$ such that $\vect{AM} = \vect{AB} + \vect{AC}$ .
What is the quadrilateral $ABMC$ ? Justify.

Solution

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

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

Solution of Exercise 6.8.

$\vect{AB}\,(4, 2)$ and $\vect{DC}\,(6-2,\ 1-(-1)) = (4, 2)$ : the two vectors 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), 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):

compute the coordinates of $I$ ;
let $J$ be such that $\vect{CJ} = \frac23\vect{CB}$ ; compute the coordinates of $J$ ;
show that $(IJ)$ is parallel to $(AC)$ .

Solution

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 are 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 of $[AB]$ , $[BC]$ , $[CD]$ , $[DA]$ . Using coordinates $A(x_A, y_A)$ , etc., show that $\vect{IJ} = \vect{LK}$ , and conclude that the midpoints of the sides of any quadrilateral form a parallelogram.

Solution

Solution of Exercise 6.10.

The midpoints 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 of the two vectors:

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. 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 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), and ends where vectors were born: forces and velocities.

Part I — Chasles gymnastics.

Simplify using Chasles’ relation (Theorem 6.5): $\vect{AB} + \vect{BC} + \vect{CD}$ ; $\vect{MA} - \vect{MB}$ ; $\vect{AB} + \vect{CD} + \vect{BC} + \vect{DA}$ .
The origin trick: show that for any point $O$ , $\vect{AB} = \vect{OB} - \vect{OA}$ .
Prove the two characterizations of the midpoint $I$ of $[AB]$ :

$\vect{IA} + \vect{IB} = \vec 0 \qquad\text{and}\qquad \vect{OI} = \tfrac12\left(\vect{OA} + \vect{OB}\right) \ \text{for any } O .$
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.
Adding a fixed 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 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 .

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 of $G$ are the averages of the vertices’ (the centroid weekend problem of the Middle School volume saw it numerically).
Prove $\vect{AG} = \frac13\left(\vect{AB} + \vect{AC}\right)$ , and, writing $K$ for the midpoint of $[BC]$ , $\vect{AK} = \frac12\left(\vect{AB} + \vect{AC}\right)$ . Conclude: $\vect{AG} = \frac23\,\vect{AK}$ — the two-thirds theorem, re-proved.
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, vectors — in one sentence each.
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).
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.

Are $\vec u(3, -2)$ and $\vec v(-6, 4)$ collinear? And $\vec w(2, 5)$ , $\vec z(4, 9)$ ? Decide with the determinant criterion.
Are the points $A(1, 2)$ , $B(3, 5)$ , $C(7, 11)$ aligned?
Beyond Varignon (Exercise 6.10): in any quadrilateral $ABCD$ , the bimedians join the midpoints of opposite sides ( $[IK]$ and $[JL]$ ). Using position vectors ( $\vect{OI} = \frac12(\vect{OA} + \vect{OB})$ , etc.), show that both bimedians have the same midpoint, of position vector $\frac14\left(\vect{OA} + \vect{OB} + \vect{OC} + \vect{OD}\right)$ : the centroid of the quadrilateral.
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.
Generalize question 14 to an arbitrary ratio $k$ , and state in one sentence how the vector calculus absorbs the midpoint theorem (the midpoint theorem) and Thales (Thales’ theorem) at once.

Part IV — Forces and velocities.

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, its magnitude, and describe the actual course. (A very old triple appears.)
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 of the arrows along any closed polygon always $\vec 0$ (Chasles)?
Two forces $\vec u(4, -1)$ and $\vec v(-1, 3)$ act on a hook. What third force holds the hook in equilibrium?
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?
Finale: one object, three costumes — displacement arrow, coordinate pair, force or velocity. List the three classical theorems this problem re-proved with vector algebra, and name the one geometric quantity the vectors of this chapter still cannot measure — the tool that fixes it arrives with the scalar product, in grade 11.

Solution

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).

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

3. $I$ is the 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, one line faster.

5. The translation of vector $\vec u$ — exactly the transformation produced by two parallel mirrors or two half-turns. Image of $(1, 2)$ : $(4, 1)$ .

6. Expanding 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 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 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 compute but obscure; vectors 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, 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 ( $\vec v = -2\vec u$ ). $2 \times 9 - 5 \times 4 = -2 \neq 0$ : not 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 of $[IK]$ has position 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. 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 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 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.