"Walk $5$ km" is not enough to find your way; you need "walk $5$ km northeast." Quantities like displacement, velocity and force have both a size and a direction. Your work with angles will help describe the direction.
Key idea
A vector has a magnitude (length) and a direction, and is drawn as an arrow. In component form, $\mathbf{v} = \langle a, b \rangle$ means "move $a$ across and $b$ up." The vector from $P(x_1, y_1)$ to $Q(x_2, y_2)$ is $$\overrightarrow{PQ} = \langle x_2 - x_1,\ y_2 - y_1 \rangle.$$ Arrows with the same length and direction are the same vector, wherever they start.
Why it works
A vector records a move, not a place. "Three right and four up" is the same instruction whether you start at your door or across town; only the starting point differs. That is why a vector can slide anywhere without changing, and why subtraction (end minus start) gives its components.
Worked examples
Worked example: From two points
Find the vector from $P(1, 2)$ to $Q(4, -1)$.
- End minus start: $\langle 4 - 1,\ -1 - 2 \rangle$.
- $= \langle 3, -3 \rangle$: three right and three down.
Answer: $\langle 3, -3 \rangle$
Worked example: Where does it end?
The vector $\langle -2, 5 \rangle$ starts at $(3, 1)$. Where does it end?
- Make the move from the starting point: $x$: $3 + (-2) = 1$, $y$: $1 + 5 = 6$.
Answer: $(1, 6)$
Worked example: Same vector?
Do the arrows from $(0, 0)$ to $(2, 3)$ and from $(4, 1)$ to $(6, 4)$ represent the same vector?
- First: $\langle 2, 3 \rangle$. Second: $\langle 6 - 4, 4 - 1 \rangle = \langle 2, 3 \rangle$.
- Same move, so the same vector, drawn in two places.
Answer: Yes
Common mistake
Treating the components $\langle 3, 4 \rangle$ as the point $(3, 4)$. They match only when the arrow starts at the origin. If the arrow starts at $(1, 1)$, it ends at $(4, 5)$. Think "move," not "location." The angle brackets are a reminder of that difference.
Vectors are written in bold ($\mathbf{v}$) or with an arrow ($\vec{v}$). Another common form uses the unit vectors $\mathbf{i} = \langle 1, 0 \rangle$ and $\mathbf{j} = \langle 0, 1 \rangle$: $\langle 3, -4 \rangle = 3\mathbf{i} - 4\mathbf{j}$.