Vector Operations in 3D

Let $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ and $\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j} + b_3\mathbf{k}$. All three operations act on components in parallel:

1. Addition $\mathbf{a} + \mathbf{b} = (a_1 + b_1)\mathbf{i} + (a_2 + b_2)\mathbf{j} + (a_3 + b_3)\mathbf{k}$, commutative and associative.
2. Subtraction $\mathbf{a} - \mathbf{b} = (a_1 - b_1)\mathbf{i} + (a_2 - b_2)\mathbf{j} + (a_3 - b_3)\mathbf{k}$, equivalent to $\mathbf{a} + (-\mathbf{b})$.
3. Scalar multiplication $\lambda\mathbf{a} = \lambda a_1\mathbf{i} + \lambda a_2\mathbf{j} + \lambda a_3\mathbf{k}$, distributes over addition: $\lambda(\mathbf{a} + \mathbf{b}) = \lambda\mathbf{a} + \lambda\mathbf{b}$.

Worked through the hook: Drones at $A(2, 1, 3)$ and $B(4, 2, 6)$. Position vectors:

• $\vec{OA} = 2\mathbf{i} + \mathbf{j} + 3\mathbf{k}$
• $\vec{OB} = 4\mathbf{i} + 2\mathbf{j} + 6\mathbf{k}$
• Test: $\vec{OB} = 2 \cdot \vec{OA}$? Check each component: $4 = 2 \cdot 2$ ✓, $2 = 2 \cdot 1$ ✓, $6 = 2 \cdot 3$ ✓. Yes.
• So $\vec{OA}$ and $\vec{OB}$ are parallel with scalar $\lambda = 2$, drone $B$ is exactly twice as far from the pad as drone $A$ along the same direction.

Add / subtract / scale: act on each of the three components in parallel · $|\lambda\mathbf{a}| = |\lambda|\,|\mathbf{a}|$, absolute value on the scalar · Parallel test: $\mathbf{b} = \lambda\mathbf{a}$ for some scalar $\lambda \neq 0$ · Drone example: $\vec{OB} = 2\vec{OA}$ so they're parallel ($\lambda = 2$)

Pause, copy componentwise $+/-/\times$ rules, $|\lambda\mathbf{a}| = |\lambda||\mathbf{a}|$, and the parallel test $\mathbf{b} = \lambda\mathbf{a}$ into your book.

We just saw that vector addition, subtraction and scalar multiplication act componentwise, with the parallel test $\mathbf{b} = \lambda\mathbf{a}$ for some $\lambda \neq 0$. That raises a question: how do the triangle and parallelogram laws represent these operations geometrically? This card answers it → head-to-tail gives the triangle law; tail-to-tail with the diagonal gives the parallelogram law; $\mathbf{a}-\mathbf{b}$ is the other diagonal.

The algebraic rules from card 05 each have a geometric mirror, and that mirror is unchanged from 2D, only lifted into 3D space:

• Triangle law: place $\mathbf{a}$ then $\mathbf{b}$ head-to-tail; the sum $\mathbf{a} + \mathbf{b}$ is the arrow from the start of $\mathbf{a}$ to the head of $\mathbf{b}$.
• Parallelogram law: place $\mathbf{a}$ and $\mathbf{b}$ tail-to-tail at the same point; complete the parallelogram. Then $\mathbf{a} + \mathbf{b}$ is the diagonal from that common point, and $\mathbf{a} - \mathbf{b}$ is the other diagonal (from the head of $\mathbf{b}$ to the head of $\mathbf{a}$).
• Zero vector $\mathbf{0}$: the unique vector of magnitude $0$. It is the additive identity ($\mathbf{a} + \mathbf{0} = \mathbf{a}$) and equals $\mathbf{a} - \mathbf{a}$ for any $\mathbf{a}$. It has no defined direction, so we never say "$\mathbf{0}$ is parallel to $\mathbf{a}$".

Parallel test in components:

• $\mathbf{b} = \lambda\mathbf{a}$ means $b_1 = \lambda a_1$, $b_2 = \lambda a_2$, $b_3 = \lambda a_3$. Compute the ratios $b_1/a_1$, $b_2/a_2$, $b_3/a_3$ (where defined), if all three are equal to some common $\lambda$, the vectors are parallel.

Triangle law: head-to-tail, sum = arrow from start of first to head of second · Parallelogram law: tail-to-tail at common point, sum = diagonal · $\mathbf{a} - \mathbf{b}$ = other diagonal (head of $\mathbf{b}$ to head of $\mathbf{a}$) · Parallel test: all three component ratios equal a common $\lambda$ · Zero components must match: $0 \to 0$ in the parallel partner

Pause, copy the triangle law (head-to-tail, sum from start to end), the parallelogram law (tail-to-tail, sum = diagonal), the other-diagonal rule for $\mathbf{a}-\mathbf{b}$, and the zero-component parallel condition into your book.