Multiplication of a Vector by a Scalar

Let $\vec{a}$ be a vector and $k$ a real number. The scalar multiple $k\vec{a}$ is a vector defined as follows:

• $|k\vec{a}| = |k| \cdot |\vec{a}|$ (magnitude is scaled by $|k|$)
• If $k > 0$: $k\vec{a}$ has the same direction as $\vec{a}$
• If $k < 0$: $k\vec{a}$ has the opposite direction to $\vec{a}$
• If $k = 0$: $k\vec{a} = \vec{0}$ regardless of $\vec{a}$

In particular, $(-1)\vec{a} = -\vec{a}$ (the opposite vector).

If $\vec{a} = (a_1, a_2)$ and $k \in \mathbb{R}$, then:

$\boxed{k\vec{a} = (ka_1,\; ka_2)}$

Proof. In standard position, $\vec{a}$ goes from $O = (0,0)$ to $A = (a_1, a_2)$. The point $kA = (ka_1, ka_2)$ is on the ray $OA$ (same side if $k > 0$, opposite if $k < 0$) at distance $|k| \cdot |OA| = |k| \cdot |\vec{a}|$ from $O$. So $k\vec{a} = \overrightarrow{O(kA)} = (ka_1, ka_2)$. $\blacktriangleleft$

For vectors $\vec{a}$, $\vec{b}$ and scalars $k$, $m$:

1. Distributive over vectors: $k(\vec{a} + \vec{b}) = k\vec{a} + k\vec{b}$
2. Distributive over scalars: $(k + m)\vec{a} = k\vec{a} + m\vec{a}$
3. Associative: $(km)\vec{a} = k(m\vec{a})$
4. Identity: $1 \cdot \vec{a} = \vec{a}$
5. Zero scalar: $0 \cdot \vec{a} = \vec{0}$
6. Zero vector: $k \cdot \vec{0} = \vec{0}$

Let $\vec{a} \neq \vec{0}$. A vector $\vec{b}$ is collinear with $\vec{a}$ if and only if there exists a real number $k$ such that:

$\boxed{\vec{b} = k\vec{a}}$

Proof ($\Rightarrow$). If $\vec{b} \parallel \vec{a}$ and $\vec{b} \neq \vec{0}$, choose $k = |\vec{b}|/|\vec{a}|$ if $\vec{b}$ has the same direction as $\vec{a}$, and $k = -|\vec{b}|/|\vec{a}|$ if opposite. Then $|k\vec{a}| = |k| \cdot |\vec{a}| = |\vec{b}|$ and the direction matches, so $\vec{b} = k\vec{a}$. If $\vec{b} = \vec{0}$, take $k = 0$.

Proof ($\Leftarrow$). If $\vec{b} = k\vec{a}$, then by definition $k\vec{a}$ is parallel to $\vec{a}$, so $\vec{b} \parallel \vec{a}$. $\blacktriangleleft$

Let $A$ and $B$ be two points, with position vectors $\overrightarrow{OA}$ and $\overrightarrow{OB}$ from the origin $O$. The point $C$ that divides segment $AB$ in the ratio $m : n$ from $A$ (i.e., $AC : CB = m : n$) has position vector:

$\boxed{\overrightarrow{OC} = \overrightarrow{OA} + \dfrac{m}{m+n}\,\overrightarrow{AB} = \dfrac{n\,\overrightarrow{OA} + m\,\overrightarrow{OB}}{m + n}}$

Special case — Midpoint ($m = n$):

$\overrightarrow{OM} = \dfrac{\overrightarrow{OA} + \overrightarrow{OB}}{2}, \quad \text{i.e.,} \quad M = \left(\dfrac{x_A + x_B}{2},\; \dfrac{y_A + y_B}{2}\right)$

Proof. Since $AC : CB = m : n$, we have $\overrightarrow{AC} = \dfrac{m}{m+n}\overrightarrow{AB}$. Then: $\overrightarrow{OC} = \overrightarrow{OA} + \overrightarrow{AC} = \overrightarrow{OA} + \dfrac{m}{m+n}\overrightarrow{AB}$

Substituting $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$ and simplifying gives the second form. $\blacktriangleleft$

Given $\vec{a} = (2, -3)$.

(a) Find $3\vec{a}$ and $-\dfrac{1}{2}\vec{a}$. (b) Show that $\vec{b} = (-6, 9)$ is collinear with $\vec{a}$, and find $k$ such that $\vec{b} = k\vec{a}$.

Solution.

(a) $3\vec{a} = (6, -9)$. $\quad -\dfrac{1}{2}\vec{a} = (-1,\; \dfrac{3}{2})$.

(b) We need $(-6, 9) = k(2, -3)$, so $2k = -6 \Rightarrow k = -3$ and $-3k = 9 \Rightarrow k = -3$. Both equations give $k = -3$, so $\vec{b} = -3\vec{a}$ and the vectors are collinear (opposite direction). $\blacktriangleleft$

Points $A = (1, 4)$ and $B = (7, -2)$ are given.

(a) Find the midpoint $M$ of $AB$. (b) Find the point $C$ that divides $AB$ in the ratio $2 : 1$ from $A$.

Solution.

(a) $M = \left(\dfrac{1+7}{2},\; \dfrac{4+(-2)}{2}\right) = (4,\; 1)$.

(b) Using the section formula with $m = 2$, $n = 1$:

$\overrightarrow{OC} = \dfrac{1 \cdot \overrightarrow{OA} + 2 \cdot \overrightarrow{OB}}{2 + 1} = \dfrac{(1, 4) + 2(7, -2)}{3} = \dfrac{(1 + 14,\; 4 - 4)}{3} = \dfrac{(15, 0)}{3} = (5, 0)$

Answer: $M = (4, 1)$; $C = (5, 0)$. $\blacktriangleleft$

Given $\vec{u} = (-3, 6)$ and $\vec{v} = (1, -2)$.

(a) Compute $2\vec{u} - 3\vec{v}$. (b) Find the scalar $k$ such that $\vec{u} = k\vec{v}$ (if it exists). (c) What does your answer to (b) tell you about the directions of $\vec{u}$ and $\vec{v}$?

Points $P = (-2, 3)$ and $Q = (10, -1)$ are given.

(a) Find the midpoint of $PQ$. (b) Find the point $R$ that divides $PQ$ in the ratio $1 : 3$ from $P$. (c) Find the point $S$ that divides $PQ$ in the ratio $3 : 1$ from $P$.

Hint: For (b) use the formula with $m = 1$, $n = 3$.