Vector Coordinates

When we place vectors in a coordinate plane, we can represent them by pairs of numbers — their coordinates — which make calculations with vectors purely algebraic.

Coordinate Representation

📐Definition — Coordinates of a Vector

Let $A = (x_1, y_1)$ and $B = (x_2, y_2)$ be two points in the plane. The coordinates of the vector $\overrightarrow{AB}$ are:

$\boxed{\overrightarrow{AB} = (x_2 - x_1,\; y_2 - y_1)}$

If $\vec{a} = (a_x, a_y)$ , then $a_x$ is the $x$ -component and $a_y$ is the $y$ -component of $\vec{a}$ .

The coordinates of a vector measure the net displacement from tail to head along each axis. Translating the vector (keeping the same direction and magnitude) leaves its coordinates unchanged, which is why equal vectors have identical coordinates.

Special cases:

Zero vector: $\vec{0} = (0, 0)$ .
Unit vector along the $x$ -axis: $\vec{e}_x = (1, 0)$ .
Unit vector along the $y$ -axis: $\vec{e}_y = (0, 1)$ .

Standard Position

Every vector $\vec{a} = (a_x, a_y)$ can be placed in standard position by locating its tail at the origin $O = (0, 0)$ , so that its head is at the point $(a_x, a_y)$ . This gives a unique representative for each vector.

Magnitude Formula

⚡Theorem — Magnitude of a Vector

If $\vec{a} = (a_x, a_y)$ , then:

$\boxed{|\vec{a}| = \sqrt{a_x^2 + a_y^2}}$

Proof. By the distance formula, $|\overrightarrow{AB}| = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{a_x^2 + a_y^2}$ . $\blacktriangleleft$

Equal Vectors and Coordinates

⚡Theorem — Equal Vectors Have Equal Coordinates

Two vectors $\vec{a} = (a_x, a_y)$ and $\vec{b} = (b_x, b_y)$ are equal if and only if their corresponding coordinates are equal:

$\vec{a} = \vec{b} \iff a_x = b_x \text{ and } a_y = b_y$

Worked Examples

✎Example — Example 1 — Coordinates and Magnitude

Given $A = (3, -1)$ and $B = (-2, 4)$ , find the coordinates and magnitude of $\overrightarrow{AB}$ .

Solution. Applying the coordinate formula:

$\overrightarrow{AB} = (-2 - 3,\; 4 - (-1)) = (-5,\; 5)$

Magnitude:

$|\overrightarrow{AB}| = \sqrt{(-5)^2 + 5^2} = \sqrt{25 + 25} = \sqrt{50} = 5\sqrt{2}$

Answer: $\overrightarrow{AB} = (-5, 5)$ , $|\overrightarrow{AB}| = 5\sqrt{2}$ . $\blacktriangleleft$

✎Example — Example 2 — Finding an Endpoint

The vector $\vec{a} = (4, -3)$ starts at the point $A = (1, 5)$ . Find the endpoint $B$ .

Solution. By the coordinate definition, $\overrightarrow{AB} = (x_B - x_A,\; y_B - y_A)$ , so:

$x_B - 1 = 4 \implies x_B = 5$ $y_B - 5 = -3 \implies y_B = 2$

Answer: $B = (5, 2)$ . $\blacktriangleleft$

Exercises

✏Exercise

Given points $P = (-3, 2)$ and $Q = (5, -1)$ :

(a) Find the coordinates of $\overrightarrow{PQ}$ and $\overrightarrow{QP}$ . (b) Compute $|\overrightarrow{PQ}|$ . (c) Verify that $\overrightarrow{PQ}$ and $\overrightarrow{QP}$ are opposite vectors.

✏Exercise

A vector $\vec{b} = (-6, 8)$ ends at the point $B = (2, 3)$ . Find the initial point $A$ .

Hint: Use $\overrightarrow{AB} = \vec{b}$ to set up equations for $x_A$ and $y_A$ .