Vector Operations
Textbook Reference
| Primary source | OpenStax Calculus Volume 3, Section 2.1: “Vectors in the Plane” |
| Direct link | https://openstax.org/books/calculus-volume-3/pages/2-1-vectors-in-the-plane |
| Textbook used in class | Stewart, Calculus, Section 12.2: “Vectors” (Examples 4, 5, 6) |
Opening Scenario
Two forces act on a point: one $3$ N to the right and one $4$ N upward. The net force is found by adding the two force vectors. Vector addition and scalar multiplication are the algebra of direction-aware quantities.
Quick Reference
Vector addition: $\mathbf{a} + \mathbf{b} = \langle a_1 + b_1,\; a_2 + b_2,\; a_3 + b_3\rangle$.
Scalar multiplication: $c\,\mathbf{a} = \langle ca_1,\; ca_2,\; ca_3\rangle$.
Vector subtraction: $\mathbf{a} - \mathbf{b} = \mathbf{a} + (-1)\mathbf{b} = \langle a_1 - b_1,\; a_2 - b_2,\; a_3 - b_3\rangle$.
Geometric rules:
- Addition: triangle law (place $\mathbf{b}$ at the tip of $\mathbf{a}$; the sum runs from the tail of $\mathbf{a}$ to the tip of $\mathbf{b}$).
- Scalar multiplication by $c > 0$: same direction, $|c|$ times as long.
- Scalar multiplication by $c < 0$: reversed direction, $|c|$ times as long.
Properties:
- $\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}$ (commutative)
- $(\mathbf{a} + \mathbf{b}) + \mathbf{c} = \mathbf{a} + (\mathbf{b} + \mathbf{c})$ (associative)
- $c(\mathbf{a} + \mathbf{b}) = c\mathbf{a} + c\mathbf{b}$
Key Concepts
1. Computing Sums and Scalar Multiples
Example 1. Let $\mathbf{a} = \langle 2, -1, 3\rangle$ and $\mathbf{b} = \langle -1, 4, 0\rangle$. Find $\mathbf{a} + \mathbf{b}$, $3\mathbf{a}$, and $2\mathbf{a} - \mathbf{b}$. (Stewart 12.2, Example 4.)
\[ \mathbf{a} + \mathbf{b} = \langle 1, 3, 3\rangle, \quad 3\mathbf{a} = \langle 6, -3, 9\rangle, \quad 2\mathbf{a} - \mathbf{b} = \langle 5, -6, 6\rangle. \]
2. Geometric Interpretation of Subtraction
$\mathbf{b} - \mathbf{a}$ is the vector from the tip of $\mathbf{a}$ to the tip of $\mathbf{b}$ (when both are placed at the same starting point). This is the displacement you add to $\mathbf{a}$ to reach $\mathbf{b}$.
Example 2. Given $A = (1, 0, 2)$ and $B = (3, 4, -1)$, express $\overrightarrow{AB}$ as $\mathbf{b} - \mathbf{a}$ where $\mathbf{a}$ and $\mathbf{b}$ are the position vectors of $A$ and $B$.
$\overrightarrow{AB} = \langle 3-1, 4-0, -1-2\rangle = \langle 2, 4, -3\rangle = \mathbf{b} - \mathbf{a}$.
scaling a vector changes its direction. Multiplying by a positive scalar preserves the direction and stretches the magnitude. Multiplying by a negative scalar reverses the direction. Only a scalar of $0$ produces the zero vector. Multiplying by $-1$ gives the vector pointing in the exact opposite direction.
Common Errors Summary
| Error | Correction |
|---|---|
| Adding vectors component-by-component but using wrong components | Keep each component in its own slot: $a_1 + b_1$, $a_2 + b_2$, $a_3 + b_3$ |
| Confusing $c\mathbf{a}$ (scalar mult) with $\mathbf{c}\cdot\mathbf{a}$ (dot product) | Scalar multiplication produces a vector; the dot product produces a scalar |
Common Misconceptions
multiplying a vector by a positive scalar changes its direction.
This is the multiplicative-not-additive error. Scalar multiplication by any positive constant stretches or compresses the magnitude while leaving the direction unchanged. The direction reverses only when the scalar is negative. For example, $3\langle 1, 2, 0\rangle = \langle 3, 6, 0\rangle$ points in the same direction as $\langle 1, 2, 0\rangle$ with three times the length. Only $-3\langle 1, 2, 0\rangle = \langle -3, -6, 0\rangle$ points in the opposite direction.
Leveled Practice
Problem 1. For $\mathbf{u} = \langle 3, -2, 1\rangle$ and $\mathbf{v} = \langle -1, 0, 4\rangle$, compute $\mathbf{u} - 2\mathbf{v}$ and $|\mathbf{u} - 2\mathbf{v}|$.
Show answer
$\mathbf{u} - 2\mathbf{v} = \langle 3-(-2), -2-0, 1-8\rangle = \langle 5, -2, -7\rangle$.
$|\mathbf{u} - 2\mathbf{v}| = \sqrt{25 + 4 + 49} = \sqrt{78}$.