Componentwise:
$$\mathbf{a} + \mathbf{b} = (a_1+b_1)\mathbf{i} + (a_2+b_2)\mathbf{j} + (a_3+b_3)\mathbf{k}$$
$$\mathbf{a} - \mathbf{b} = (a_1-b_1)\mathbf{i} + (a_2-b_2)\mathbf{j} + (a_3-b_3)\mathbf{k}$$
Geometric interpretation:
- $\mathbf{a} + \mathbf{b}$: triangle rule or parallelogram rule (diagonal)
- $\mathbf{a} - \mathbf{b}$: the vector from the tip of $\mathbf{b}$ to the tip of $\mathbf{a}$ (when both originate from the same point)
Properties:
- Commutative: $\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}$
- Associative: $(\mathbf{a}+\mathbf{b})+\mathbf{c} = \mathbf{a}+(\mathbf{b}+\mathbf{c})$
- Identity: $\mathbf{a} + \mathbf{0} = \mathbf{a}$
- Inverse: $\mathbf{a} + (-\mathbf{a}) = \mathbf{0}$
$$\lambda\mathbf{a} = (\lambda a_1)\mathbf{i} + (\lambda a_2)\mathbf{j} + (\lambda a_3)\mathbf{k}$$
Properties:
- $\lambda(\mathbf{a}+\mathbf{b}) = \lambda\mathbf{a} + \lambda\mathbf{b}$ (distributive)
- $(\lambda + \mu)\mathbf{a} = \lambda\mathbf{a} + \mu\mathbf{a}$
- $(\lambda\mu)\mathbf{a} = \lambda(\mu\mathbf{a})$
A linear combination of vectors $\mathbf{v}_1, \ldots, \mathbf{v}_n$ is:
$$\lambda_1\mathbf{v}_1 + \lambda_2\mathbf{v}_2 + \cdots + \lambda_n\mathbf{v}_n, \quad \lambda_i \in \mathbb{R}$$
The span of ${\mathbf{v}_1, \ldots, \mathbf{v}_n}$ is the set of all linear combinations.
Example: Express $\mathbf{c} = 5\mathbf{i} - \mathbf{j} + 8\mathbf{k}$ as a linear combination of $\mathbf{a} = \mathbf{i}+\mathbf{j}+\mathbf{k}$ and $\mathbf{b} = 2\mathbf{i}-\mathbf{j}+3\mathbf{k}$.
Solve $\lambda\mathbf{a} + \mu\mathbf{b} = \mathbf{c}$:
$$\lambda + 2\mu = 5, \quad \lambda - \mu = -1, \quad \lambda + 3\mu = 8$$
From equations 1 and 2: $3\mu = 6 \Rightarrow \mu = 2$, $\lambda = 1$. Check eq. 3: $1+6=7 \neq 8$. So not a linear combination of $\mathbf{a}$ and $\mathbf{b}$ alone.
Linear combinations are used to prove geometric results.
Example: $ABCD$ is a parallelogram with $\overrightarrow{AB} = \mathbf{p}$ and $\overrightarrow{AD} = \mathbf{q}$. Show the diagonals bisect each other.
Midpoint of $AC$: position vector of $A$ $+ \tfrac{1}{2}\overrightarrow{AC} = \mathbf{a} + \tfrac{1}{2}(\mathbf{p}+\mathbf{q})$.
Midpoint of $BD$: $\mathbf{b} + \tfrac{1}{2}\overrightarrow{BD} = (\mathbf{a}+\mathbf{p}) + \tfrac{1}{2}(-\mathbf{p}+\mathbf{q}) = \mathbf{a} + \tfrac{1}{2}(\mathbf{p}+\mathbf{q})$. Same point. $\square$
KEY TAKEAWAY: Vector operations obey algebraic laws identical to those of real numbers, except there is no vector-vector multiplication (only dot and cross products). This makes vector algebra powerful and intuitive.
EXAM TIP: In vector geometry proofs, express all relevant vectors in terms of a small set of basis vectors ($\mathbf{p}$, $\mathbf{q}$ etc.) and show the required equality algebraically.
COMMON MISTAKE: Subtracting vectors by subtracting magnitudes: $|\mathbf{a} - \mathbf{b}| \neq |\mathbf{a}| - |\mathbf{b}|$ in general.
