Vectors in the coordinate plane | Grade 10

A map app tells you "walk 3 blocks east and 2 blocks north". Those two numbers are everything you need to know about the move: how far along the street and how far up the avenue. Placing vectors on a coordinate grid turns this idea into a tool. Instead of drawing arrows we can compute with pairs of numbers, which is faster and also works for problems too complicated to draw.

Coordinates of a vector

In the plane with axes $Ox$ and $Oy$ , let $\vec{i}$ be the vector of length $1$ along $Ox$ and $\vec{j}$ the vector of length $1$ along $Oy$ . These are the unit vectors.

The vector

\vec{a}=3\vec{i}+2\vec{j}

Two vectors are equal exactly when their coordinates are equal: $(x, y)=(x', y')$ means $x=x'$ and $y=y'$ . The zero vector is $(0, 0)$ .

From points to a vector

If $A(x_A, y_A)$ and $B(x_B, y_B)$ , then to go from $A$ to $B$ you move $x_B-x_A$ horizontally and $y_B-y_A$ vertically: $\overrightarrow{AB}=(x_B-x_A,\ y_B-y_A).$

Operations in coordinates

So you add, subtract and scale vectors one coordinate at a time.

Length of a vector and distance

By Pythagoras, the vector $\vec{a}=(x, y)$ has length $|\vec{a}|=\sqrt{x^2+y^2}.$ Since $AB$ is the length of $\overrightarrow{AB}$ , the distance between two points is $AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}.$

Midpoint and centroid

Using $\overrightarrow{OM}=\frac12(\overrightarrow{OA}+\overrightarrow{OB})$ and $\overrightarrow{OG}=\frac13(\overrightarrow{OA}+\overrightarrow{OB}+\overrightarrow{OC})$ with $O$ as the origin, we get coordinates by averaging.

Collinear vectors and collinear points

Two vectors $\vec{a}=(x_1, y_1)$ and $\vec{b}=(x_2, y_2)$ are collinear when one is a multiple of the other. In coordinates this means $x_1y_2-x_2y_1=0.$ (Write it as a product test instead of a ratio $\frac{x_1}{x_2}=\frac{y_1}{y_2}$ , which fails when a coordinate is $0$ .) Three points $A$ , $B$ , $C$ are collinear when $\overrightarrow{AB}$ and $\overrightarrow{AC}$ are collinear.

A missing vertex of a parallelogram

$ABCD$ is a parallelogram exactly when $\overrightarrow{AB}=\overrightarrow{DC}$ , which gives $D=A+C-B$ in coordinates.

Parallelogram

ABCD

Key terms

English	Tiếng Việt
unit vectors	các vectơ đơn vị
coordinates of a vector	toạ độ của vectơ
equal vectors	hai vectơ bằng nhau
length of a vector	độ dài vectơ
distance between two points	khoảng cách giữa hai điểm
midpoint	trung điểm
centroid	trọng tâm
collinear	cùng phương
parallelogram	hình bình hành

Summary

$\vec{a}=x\vec{i}+y\vec{j}$ is written $\vec{a}=(x, y)$ ; $\overrightarrow{AB}=(x_B-x_A,\ y_B-y_A)$ .
Add, subtract and scale one coordinate at a time.
$|\vec{a}|=\sqrt{x^2+y^2}$ and $AB=\sqrt{(x_B-x_A)^2+(y_B-y_A)^2}$ .
Midpoint: average of the coordinates; centroid: average of three.
$\vec{a}$ , $\vec{b}$ collinear $\iff x_1y_2-x_2y_1=0$ ; parallelogram $ABCD$ : $D=A+C-B$ .

Practice

Let $\vec{a}=(3, -2)$ and $\vec{b}=(-1, 5)$ . Find $2\vec{a}+\vec{b}$ and $\vec{a}-3\vec{b}$ .
Find $\overrightarrow{AB}$ and the distance $AB$ for $A(-2, 3)$ and $B(4, -1)$ .
Find the midpoint of $A(-3, 5)$ and $B(7, 1)$ .
Find the centroid of the triangle with vertices $A(1, 2)$ , $B(4, -1)$ , $C(-2, 8)$ .
Show that $A(1, 1)$ , $B(2, 3)$ , $C(4, 7)$ are collinear.
Find $m$ so that $\vec{a}=(m, 3)$ and $\vec{b}=(4, 6)$ are collinear.
$ABCD$ is a parallelogram with $A(-1, 0)$ , $B(2, 1)$ , $C(3, 4)$ . Find $D$ .
Write $\vec{c}=(5, 1)$ as $x\vec{a}+y\vec{b}$ with $\vec{a}=(1, 1)$ and $\vec{b}=(1, -1)$ .
Challenge. Find the point $M$ on the $x$ -axis that is collinear with $A(1, 2)$ and $B(3, -2)$ .
Challenge. Show that the triangle with $A(1, 1)$ , $B(4, 5)$ , $C(7, 1)$ is isosceles and find its perimeter.

Show solutions

$2\vec{a}+\vec{b}=(6-1,\ -4+5)=(5, 1)$ . $\vec{a}-3\vec{b}=(3+3,\ -2-15)=(6, -17)$ .
$\overrightarrow{AB}=(4+2,\ -1-3)=(6, -4)$ and $AB=\sqrt{36+16}=\sqrt{52}=2\sqrt{13}$ .
$\left(\dfrac{-3+7}{2},\ \dfrac{5+1}{2}\right)=(2, 3)$ .
$G\left(\dfrac{1+4-2}{3},\ \dfrac{2-1+8}{3}\right)=(1, 3)$ .
$\overrightarrow{AB}=(1, 2)$ and $\overrightarrow{AC}=(3, 6)$ . $1\cdot 6-3\cdot 2=0$ , so they are collinear (and $\overrightarrow{AC}=3\overrightarrow{AB}$ ).
$m\cdot 6-4\cdot 3=0$ , so $6m=12$ and $m=2$ .
$D=A+C-B=(-1+3-2,\ 0+4-1)=(0, 3)$ .
$x+y=5$ and $x-y=1$ . Adding gives $2x=6$ , $x=3$ , then $y=2$ . So $\vec{c}=3\vec{a}+2\vec{b}$ .
$M(t, 0)$ . $\overrightarrow{AM}=(t-1, -2)$ and $\overrightarrow{AB}=(2, -4)$ . Collinear: $(t-1)(-4)-2\cdot(-2)=0$ , so $-4t+4+4=0$ and $t=2$ . $M(2, 0)$ .
$AB=\sqrt{9+16}=5$ , $BC=\sqrt{9+16}=5$ , $AC=6$ . Two equal sides, so it is isosceles. Perimeter $5+5+6=16$ .