Addition and Subtraction of Vectors

To add $\vec{a}$ and $\vec{b}$ using the triangle rule, place the tail of $\vec{b}$ at the head of $\vec{a}$. The sum $\vec{a} + \vec{b}$ is the vector from the tail of $\vec{a}$ to the head of $\vec{b}$:

$\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$

To add $\vec{a}$ and $\vec{b}$ using the parallelogram rule, place both vectors at the same starting point. Complete the parallelogram with $\vec{a}$ and $\vec{b}$ as adjacent sides. The sum $\vec{a} + \vec{b}$ is the diagonal of the parallelogram from the common starting point.

For any vectors $\vec{a}$, $\vec{b}$, $\vec{c}$:

1. Commutative law: $\vec{a} + \vec{b} = \vec{b} + \vec{a}$
2. Associative law: $(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$
3. Identity element: $\vec{a} + \vec{0} = \vec{a}$
4. Additive inverse: $\vec{a} + (-\vec{a}) = \vec{0}$

Proof of commutativity. Let $\vec{a} = \overrightarrow{AB}$ and $\vec{b} = \overrightarrow{BC}$. In the parallelogram $ABCD$, $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AC}$. Also $\overrightarrow{AD} = \overrightarrow{BC} = \vec{b}$ and $\overrightarrow{DC} = \overrightarrow{AB} = \vec{a}$, so $\vec{b} + \vec{a} = \overrightarrow{AD} + \overrightarrow{DC} = \overrightarrow{AC}$. Hence $\vec{a} + \vec{b} = \vec{b} + \vec{a}$. $\blacktriangleleft$

If $\vec{a} = (a_1, a_2)$ and $\vec{b} = (b_1, b_2)$, then:

$\boxed{\vec{a} + \vec{b} = (a_1 + b_1,\; a_2 + b_2)}$

Proof. Place $\vec{a}$ from $O = (0,0)$ to $A = (a_1, a_2)$, and $\vec{b}$ from $A$ to $B = (a_1 + b_1,\, a_2 + b_2)$. Then $\vec{a} + \vec{b} = \overrightarrow{OB} = (a_1 + b_1,\, a_2 + b_2)$. $\blacktriangleleft$

The difference $\vec{a} - \vec{b}$ is defined as:

$\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$

In coordinates:

$\boxed{(a_1, a_2) - (b_1, b_2) = (a_1 - b_1,\; a_2 - b_2)}$

Given $\vec{a} = (3, -2)$ and $\vec{b} = (-1, 5)$, find: (a) $\vec{a} + \vec{b}$, (b) $\vec{a} - \vec{b}$, (c) $|\vec{a} + \vec{b}|$.

Solution.

(a) $\vec{a} + \vec{b} = (3 + (-1),\; -2 + 5) = (2, 3)$

(b) $\vec{a} - \vec{b} = (3 - (-1),\; -2 - 5) = (4, -7)$

(c) $|\vec{a} + \vec{b}| = |(2, 3)| = \sqrt{4 + 9} = \sqrt{13}$

Answer: $(2, 3)$; $(4, -7)$; $\sqrt{13}$. $\blacktriangleleft$

In parallelogram $ABCD$, let $\vec{a} = \overrightarrow{AB}$ and $\vec{b} = \overrightarrow{AD}$. Express the diagonals $\overrightarrow{AC}$ and $\overrightarrow{BD}$ in terms of $\vec{a}$ and $\vec{b}$.

Solution. By the triangle rule:

$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} = \vec{a} + \vec{b}$

(since $\overrightarrow{BC} = \overrightarrow{AD} = \vec{b}$).

For $\overrightarrow{BD}$: from $B$, going to $D$ means first going back along $\vec{a}$ and then along $\vec{b}$:

$\overrightarrow{BD} = \overrightarrow{BA} + \overrightarrow{AD} = -\vec{a} + \vec{b} = \vec{b} - \vec{a}$

Answer: $\overrightarrow{AC} = \vec{a} + \vec{b}$, $\overrightarrow{BD} = \vec{b} - \vec{a}$. $\blacktriangleleft$

Let $\vec{p} = (2, 7)$, $\vec{q} = (-4, 3)$, $\vec{r} = (1, -5)$.

(a) Compute $\vec{p} + \vec{q} + \vec{r}$. (b) Compute $\vec{p} - \vec{q}$ and $|\vec{p} - \vec{q}|$. (c) Find $\vec{x}$ such that $\vec{p} + \vec{x} = \vec{r}$.

Points $O = (0, 0)$, $A = (4, 1)$, $B = (-1, 3)$ are given.

(a) Express $\overrightarrow{AB}$ as a difference of position vectors from $O$. (b) Find the position vector of the point $C$ such that $\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{OB}$.

Hint: Use the relation $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ for part (a).