Concept of a Vector

A vector is an ordered pair of points $(A, B)$ in the plane, written $\overrightarrow{AB}$.

• Point $A$ is called the initial point (tail) and point $B$ is called the terminal point (head).
• The magnitude (or length) of $\overrightarrow{AB}$ is the distance between $A$ and $B$:

$|\overrightarrow{AB}| = AB$

• A vector is often denoted by a single bold or arrowed letter: $\vec{a}$, $\vec{b}$, $\overrightarrow{AB}$.

The zero vector $\vec{0}$ is the vector whose initial and terminal points coincide: $A = B$. Its magnitude is zero: $|\vec{0}| = 0$.

The zero vector has no determined direction; by convention it is considered parallel (collinear) to every vector.

Two vectors $\vec{a}$ and $\vec{b}$ are equal ($\vec{a} = \vec{b}$) if and only if they have the same magnitude and the same direction.

Equivalently, $\overrightarrow{AB} = \overrightarrow{CD}$ if and only if $ABDC$ is a parallelogram (or $A = C$ and $B = D$).

The vector opposite to $\vec{a}$ is denoted $-\vec{a}$. It has the same magnitude as $\vec{a}$ but points in the opposite direction:

$|-\vec{a}| = |\vec{a}|, \qquad -\overrightarrow{AB} = \overrightarrow{BA}$

Two vectors are collinear if they lie on parallel lines (or on the same line). In other words, $\vec{a}$ and $\vec{b}$ are collinear if they have either the same direction or opposite directions.

The zero vector $\vec{0}$ is collinear with every vector.

Points $A(1, 2)$, $B(4, 5)$, $C(3, 0)$, $D(6, 3)$ are given. Show that $\overrightarrow{AB} = \overrightarrow{CD}$.

Solution. Compute the displacements:

$\overrightarrow{AB}: \quad \Delta x = 4 - 1 = 3,\quad \Delta y = 5 - 2 = 3$

$\overrightarrow{CD}: \quad \Delta x = 6 - 3 = 3,\quad \Delta y = 3 - 0 = 3$

Both vectors have displacement $(3, 3)$, so they have the same magnitude $\sqrt{3^2 + 3^2} = 3\sqrt{2}$ and the same direction. Therefore $\overrightarrow{AB} = \overrightarrow{CD}$. $\blacktriangleleft$

Let $\vec{a} = \overrightarrow{PQ}$ where $P = (0, 0)$ and $Q = (2, -3)$.

(a) Write the opposite vector $-\vec{a}$. (b) Is $\overrightarrow{PQ}$ collinear with $\overrightarrow{RS}$ where $R = (1, 1)$ and $S = (3, -2)$?

Solution.

(a) $-\vec{a} = \overrightarrow{QP}$, which starts at $Q = (2, -3)$ and ends at $P = (0, 0)$. Its displacement is $(-2, 3)$ — opposite in direction to $(2, -3)$. $|\!-\vec{a}| = \sqrt{4+9} = \sqrt{13} = |\vec{a}|$. $\checkmark$

(b) $\overrightarrow{RS}$ has displacement $(3-1,\, -2-1) = (2, -3)$. This equals the displacement of $\overrightarrow{PQ}$, so the vectors are equal (and in particular collinear). $\blacktriangleleft$

Points $M(2, 1)$, $N(5, 4)$, $K(-1, 3)$, $L(2, 6)$ are given.

(a) Are $\overrightarrow{MN}$ and $\overrightarrow{KL}$ equal? (b) Write the vector opposite to $\overrightarrow{MN}$. (c) What is $|\overrightarrow{MN}|$?

Three vectors $\vec{a}$, $\vec{b}$, $\vec{c}$ are given. $\vec{a}$ has magnitude $5$ and points north. $\vec{b}$ has magnitude $5$ and points south. $\vec{c}$ has magnitude $3$ and points north.

(a) Which pairs of vectors are collinear? (b) Which pairs are equal? (c) Write the vector $-\vec{b}$ in terms of $\vec{a}$, $\vec{b}$, $\vec{c}$.

Hint: Two vectors are equal only if both magnitude and direction match.