Since a vector is a move, combining vectors means combining moves. With components, the arithmetic happens one coordinate at a time.
Key idea
$\langle a, b \rangle + \langle c, d \rangle = \langle a + c,\ b + d \rangle$: place the vectors tip to tail. $k\langle a, b \rangle = \langle ka, kb \rangle$: this stretches the length by $|k|$, and a negative $k$ reverses the direction. $\mathbf{u} - \mathbf{v} = \mathbf{u} + (-\mathbf{v})$.
Why it works
Doing move $\mathbf{u}$ and then move $\mathbf{v}$ takes you as far across as both moves together, and as far up as both together, so the components add. Drawn as arrows, start the second where the first ends; the sum goes from the very start to the very end. Doing the same move twice, $2\mathbf{u}$, doubles every component.
Worked examples
Worked example: Combining
Let $\mathbf{u} = \langle 2, -1 \rangle$ and $\mathbf{v} = \langle -3, 4 \rangle$. Find $\mathbf{u} + \mathbf{v}$ and $2\mathbf{u} - \mathbf{v}$.
- $$\mathbf{u} + \mathbf{v} = \langle 2 - 3,\ -1 + 4 \rangle = \langle -1, 3 \rangle.$$
- $2\mathbf{u} = \langle 4, -2 \rangle$, so $$2\mathbf{u} - \mathbf{v} = \langle 4 - (-3),\ -2 - 4 \rangle = \langle 7, -6 \rangle.$$
Answer: $\langle -1, 3 \rangle$ and $\langle 7, -6 \rangle$
Worked example: Lengths don't simply add
For the same vectors, compare $|\mathbf{u} + \mathbf{v}|$ with $|\mathbf{u}| + |\mathbf{v}|$.
- $$|\mathbf{u} + \mathbf{v}| = \sqrt{1 + 9} = \sqrt{10} \approx 3.16.$$
- $|\mathbf{u}| + |\mathbf{v}| = \sqrt{5} + 5 \approx 7.24$.
- The sum is shorter because the two moves partly point in different directions.
Answer: $\sqrt{10} < \sqrt{5} + 5$
Worked example: Solving for a vector
Find $\mathbf{w}$ if $3\mathbf{w} + \langle 1, 2 \rangle = \langle 7, -4 \rangle$.
- Subtract: $3\mathbf{w} = \langle 6, -6 \rangle$.
- Divide by $3$.
Answer: $\mathbf{w} = \langle 2, -2 \rangle$
Common mistake
Subtracting in the wrong order. $\mathbf{u} - \mathbf{v}$ and $\mathbf{v} - \mathbf{u}$ have the same length but point in opposite directions. Check with a sketch: $\mathbf{u} - \mathbf{v}$ is the arrow from the tip of $\mathbf{v}$ to the tip of $\mathbf{u}$ (when both start at the same point), because $\mathbf{v} + (\mathbf{u} - \mathbf{v}) = \mathbf{u}$.