Scalar Multiplication of Vectors

Doubling your velocity, or reversing a force, means multiplying a vector by a number. This scalar multiplication stretches, shrinks, or flips a vector without turning it sideways. Combined with addition, it lets you build vectors out of simpler pieces, test whether points lie on a line, and prove facts about geometric figures with a few lines of algebra.

Key ideas

What k times a vector means

For a real number $k$ (a scalar) and a vector $\vec{v}$, the vector $k\vec{v}$ has:

• magnitude $\lvert k\rvert\,\lvert\vec{v}\rvert$, so it’s $\lvert k\rvert$ times as long;
• the same direction as $\vec{v}$ if $k \gt 0$, and the opposite direction if $k \lt 0$.

Also $0\vec{v} = \vec{0}$ and $(-1)\vec{v} = -\vec{v}$. For example, if $\vec{v}$ is $6$ km/h north, then $3\vec{v}$ is $18$ km/h north and $-2\vec{v}$ is $12$ km/h south.

Scalar multiples of $\vec{v}$ are all parallel to $\vec{v}$. A negative scalar reverses the direction.

Collinear (parallel) vectors

Two non-zero vectors $\vec{u}$ and $\vec{v}$ are collinear (parallel) exactly when one is a scalar multiple of the other:

$$\vec{u} = k\vec{v} \quad \text{for some scalar } k$$

This gives a test for collinear points: $A$, $B$ and $C$ lie on one line if $\overrightarrow{AB} = k\overrightarrow{AC}$ for some $k$. The vectors are parallel and share the point $A$, so they lie along the same line.

Unit vectors

A unit vector has magnitude $1$. To get the unit vector in the direction of a non-zero vector $\vec{v}$, divide by its magnitude:

$$\frac{1}{\lvert\vec{v}\rvert}\,\vec{v}$$

It points the same way as $\vec{v}$ (since $\dfrac{1}{\lvert\vec{v}\rvert} \gt 0$) and has length $\dfrac{1}{\lvert\vec{v}\rvert} \cdot \lvert\vec{v}\rvert = 1$. To make a vector of any length $L$ in that direction, multiply the unit vector by $L$.

Properties of scalar multiplication

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

Property	Statement
Distributive (over vector addition)	$k(\vec{a} + \vec{b}) = k\vec{a} + k\vec{b}$
Distributive (over scalar addition)	$(k + m)\vec{a} = k\vec{a} + m\vec{a}$
Associative	$k(m\vec{a}) = (km)\vec{a}$
Identity	$1\vec{a} = \vec{a}$

These mean you can expand brackets and collect “like terms” in vector expressions exactly as in algebra. The first one has a nice picture: stretching a whole tip-to-tail triangle by $k$ makes a similar triangle whose sides are $k\vec{a}$, $k\vec{b}$ and $k(\vec{a} + \vec{b})$.

Linear combinations

An expression like $3\vec{a} - 2\vec{b}$ is a linear combination of $\vec{a}$ and $\vec{b}$: a sum of scalar multiples. If $\vec{a}$ and $\vec{b}$ are not collinear, then every vector in their plane can be written as $m\vec{a} + n\vec{b}$ in exactly one way. So if

$$m\vec{a} + n\vec{b} = p\vec{a} + q\vec{b}$$

with $\vec{a}$ and $\vec{b}$ non-collinear, you can conclude $m = p$ and $n = q$. This is how vector proofs often finish.

Proving things with vectors

To prove a geometric fact with vectors:

1. Label the figure and choose one or two vectors (often two sides) to work with.
2. Write the vectors you care about in terms of those, using tip-to-tail paths around the figure.
3. Simplify, and interpret the result: $\vec{u} = k\vec{v}$ means parallel and $\lvert k\rvert$ times as long; equal vectors mean the same point or the same segment.

Worked examples

Example 1: Describing scalar multiples

$\vec{v}$ is a velocity of $6$ km/h at N $30^\circ$ E. Describe $3\vec{v}$, $-2\vec{v}$, and the unit vector in the direction of $\vec{v}$.

Solution.

$3\vec{v}$: $3 \times 6 = 18$ km/h, same direction: $18$ km/h at N $30^\circ$ E.

$-2\vec{v}$: $2 \times 6 = 12$ km/h, opposite direction: $12$ km/h at S $30^\circ$ W.

Unit vector: $\dfrac{1}{6}\vec{v}$, which has magnitude $1$ (km/h) at N $30^\circ$ E.

Example 2: Simplifying and solving

• (a) Simplify $3(2\vec{a} - \vec{b}) - 2(\vec{a} + 4\vec{b})$.
• (b) Solve $2\vec{x} + 3\vec{a} = 5\vec{b} - \vec{x}$ for $\vec{x}$.

Solution.

(a) Expand with the distributive property, then collect like terms:

$$\begin{aligned} 3(2\vec{a} - \vec{b}) - 2(\vec{a} + 4\vec{b}) &= 6\vec{a} - 3\vec{b} - 2\vec{a} - 8\vec{b} \\ &= 4\vec{a} - 11\vec{b} \end{aligned}$$

(b) Treat it like a linear equation:

$$\begin{aligned} 2\vec{x} + \vec{x} &= 5\vec{b} - 3\vec{a} \\ 3\vec{x} &= 5\vec{b} - 3\vec{a} \\ \vec{x} &= \tfrac{5}{3}\vec{b} - \vec{a} \end{aligned}$$

Check: $2\vec{x} + 3\vec{a} = \tfrac{10}{3}\vec{b} - 2\vec{a} + 3\vec{a} = \tfrac{10}{3}\vec{b} + \vec{a}$, and $5\vec{b} - \vec{x} = 5\vec{b} - \tfrac{5}{3}\vec{b} + \vec{a} = \tfrac{10}{3}\vec{b} + \vec{a}$. ✓

Example 3: Linear combinations in a parallelogram

$OABC$ is a parallelogram (vertices in order) with $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OC} = \vec{c}$. $M$ is the midpoint of $AB$. Write $\overrightarrow{OB}$, $\overrightarrow{OM}$ and $\overrightarrow{CM}$ in terms of $\vec{a}$ and $\vec{c}$.

Solution. In a parallelogram, $\overrightarrow{AB} = \overrightarrow{OC} = \vec{c}$.

$$\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \vec{a} + \vec{c}$$

$M$ is halfway along $AB$, so $\overrightarrow{AM} = \tfrac{1}{2}\vec{c}$:

$$\overrightarrow{OM} = \overrightarrow{OA} + \overrightarrow{AM} = \vec{a} + \tfrac{1}{2}\vec{c}$$

To get from $C$ to $M$, go back to $O$ and then to $M$:

$$\overrightarrow{CM} = \overrightarrow{CO} + \overrightarrow{OM} = -\vec{c} + \vec{a} + \tfrac{1}{2}\vec{c} = \vec{a} - \tfrac{1}{2}\vec{c}$$

Example 4: Proving the midsegment theorem

In $\triangle ABC$, $M$ is the midpoint of $AB$ and $N$ is the midpoint of $AC$. Prove that $MN$ is parallel to $BC$ and half its length.

The segment joining the midpoints of two sides is parallel to the third side.

Solution. Since $M$ and $N$ are midpoints, $\overrightarrow{MA} = \tfrac{1}{2}\overrightarrow{BA}$ and $\overrightarrow{AN} = \tfrac{1}{2}\overrightarrow{AC}$. Go from $M$ to $N$ through $A$:

$$\begin{aligned} \overrightarrow{MN} &= \overrightarrow{MA} + \overrightarrow{AN} \\ &= \tfrac{1}{2}\overrightarrow{BA} + \tfrac{1}{2}\overrightarrow{AC} \\ &= \tfrac{1}{2}\left(\overrightarrow{BA} + \overrightarrow{AC}\right) && \text{distributive property} \\ &= \tfrac{1}{2}\overrightarrow{BC} && \text{tip to tail} \end{aligned}$$

$\overrightarrow{MN}$ is a positive scalar multiple of $\overrightarrow{BC}$, so $MN$ is parallel to $BC$ (and points the same way), and $\lvert\overrightarrow{MN}\rvert = \tfrac{1}{2}\lvert\overrightarrow{BC}\rvert$. $\blacksquare$

Common mistakes

Forgetting that a negative scalar flips the direction. $-3\vec{v}$ is three times as long as $\vec{v}$ and points the opposite way. Its magnitude is $3\lvert\vec{v}\rvert$, not $-3\lvert\vec{v}\rvert$.

Writing $\lvert k\vec{v}\rvert = k\lvert\vec{v}\rvert$ when $k$ is negative. The correct rule is $\lvert k\vec{v}\rvert = \lvert k\rvert\,\lvert\vec{v}\rvert$. Magnitudes are never negative.

Dividing by a vector. There’s no such thing as $\dfrac{\vec{a}}{\vec{b}}$. In Example 2(b), you divide both sides by the scalar $3$, which is fine.

Calling vectors collinear because they “look close”. Collinear means one is an exact scalar multiple of the other. For points, you also need a shared point: $\overrightarrow{AB} = k\overrightarrow{CD}$ only shows the segments are parallel, not that $A$, $B$, $C$ and $D$ are on one line.

Going around the figure the wrong way. In a proof, every step must be tip to tail. $\overrightarrow{MA} + \overrightarrow{AN}$ is fine; $\overrightarrow{MA} + \overrightarrow{NA}$ is not a path from $M$ to $N$. If you need $\overrightarrow{AM}$ but know $\overrightarrow{MA}$, use $\overrightarrow{AM} = -\overrightarrow{MA}$.

Comparing coefficients when the vectors are collinear. $m\vec{a} + n\vec{b} = p\vec{a} + q\vec{b}$ implies $m = p$ and $n = q$ only when $\vec{a}$ and $\vec{b}$ are not collinear. Say so in your solution.

Practice

1. (Warm-up) $\vec{v}$ is $4$ N east. Describe $2.5\vec{v}$, $-\vec{v}$ and $-0.5\vec{v}$.

Solution

$2.5\vec{v}$: $2.5 \times 4 = 10$ N east.

$-\vec{v}$: $4$ N west.

$-0.5\vec{v}$: $0.5 \times 4 = 2$ N west.

2. (Warm-up) Simplify.

• (a) $4\vec{a} + 3\vec{b} - \vec{a} + 2\vec{b}$
• (b) $2(\vec{a} - 3\vec{b}) + 5(\vec{b} + \vec{a})$

Solution

(a) $(4 - 1)\vec{a} + (3 + 2)\vec{b} = 3\vec{a} + 5\vec{b}$.

(b) $2\vec{a} - 6\vec{b} + 5\vec{b} + 5\vec{a} = 7\vec{a} - \vec{b}$.

3. (Core) Solve for $\vec{x}$: $4\vec{x} - \vec{a} = 2(\vec{x} + 3\vec{b})$.

Solution

$$\begin{aligned} 4\vec{x} - \vec{a} &= 2\vec{x} + 6\vec{b} \\ 2\vec{x} &= \vec{a} + 6\vec{b} \\ \vec{x} &= \tfrac{1}{2}\vec{a} + 3\vec{b} \end{aligned}$$

Check: $4\vec{x} - \vec{a} = 2\vec{a} + 12\vec{b} - \vec{a} = \vec{a} + 12\vec{b}$, and $2(\vec{x} + 3\vec{b}) = 2\left(\tfrac{1}{2}\vec{a} + 6\vec{b}\right) = \vec{a} + 12\vec{b}$. ✓

4. (Core) A force $\vec{F}$ is $25$ N on a bearing of $300^\circ$.

• (a) Describe the unit vector in the direction of $\vec{F}$, and write it as a multiple of $\vec{F}$.
• (b) Write a force of $10$ N in the direction opposite to $\vec{F}$ as a multiple of $\vec{F}$, and give its bearing.

Solution

(a) $\dfrac{1}{25}\vec{F}$: magnitude $1$ on a bearing of $300^\circ$.

(b) It has $\dfrac{10}{25} = \dfrac{2}{5}$ of the magnitude and the opposite direction, so it’s $-\dfrac{2}{5}\vec{F}$. Its bearing is $300^\circ - 180^\circ = 120^\circ$.

5. (Core) $\vec{a}$ and $\vec{b}$ are not collinear, and $\overrightarrow{AB} = 2\vec{a} - 3\vec{b}$ and $\overrightarrow{BC} = 4\vec{a} - 6\vec{b}$. Show that $A$, $B$ and $C$ are collinear, and find the ratio $AB : BC$.

Solution

$\overrightarrow{BC} = 4\vec{a} - 6\vec{b} = 2(2\vec{a} - 3\vec{b}) = 2\overrightarrow{AB}$.

So $\overrightarrow{AB}$ and $\overrightarrow{BC}$ are parallel, and they share the point $B$, so $A$, $B$ and $C$ lie on one line. Since $\lvert\overrightarrow{BC}\rvert = 2\lvert\overrightarrow{AB}\rvert$, the ratio is $AB : BC = 1 : 2$.

6. (Core) In $\triangle OAB$, $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OB} = \vec{b}$. Point $P$ is on $AB$ with $AP : PB = 1 : 2$. Write $\overrightarrow{OP}$ in terms of $\vec{a}$ and $\vec{b}$.

Solution

$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \vec{b} - \vec{a}$. $P$ is one third of the way from $A$ to $B$, so $\overrightarrow{AP} = \tfrac{1}{3}(\vec{b} - \vec{a})$.

$$\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \vec{a} + \tfrac{1}{3}\vec{b} - \tfrac{1}{3}\vec{a} = \tfrac{2}{3}\vec{a} + \tfrac{1}{3}\vec{b}$$

7. (Core) $ABCDEF$ is a regular hexagon (vertices in order) with $\overrightarrow{AB} = \vec{u}$ and $\overrightarrow{BC} = \vec{v}$. Write each vector in terms of $\vec{u}$ and $\vec{v}$.

• (a) $\overrightarrow{AD}$ (Hint: the long diagonal $AD$ is parallel to $BC$ and twice as long.)
• (b) $\overrightarrow{CD}$
• (c) $\overrightarrow{DE}$

Solution

(a) $\overrightarrow{AD}$ points the same way as $\overrightarrow{BC}$ and is twice as long, so $\overrightarrow{AD} = 2\vec{v}$.

(b) $\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CD}$, so $2\vec{v} = \vec{u} + \vec{v} + \overrightarrow{CD}$, which gives $\overrightarrow{CD} = \vec{v} - \vec{u}$.

(c) $DE$ is the side opposite $AB$; in a regular hexagon opposite sides are parallel and equal, and $\overrightarrow{DE}$ points the opposite way to $\overrightarrow{AB}$. So $\overrightarrow{DE} = -\vec{u}$.

8. (Challenge) Use vectors to prove that the diagonals of a parallelogram bisect each other. (Hint: in parallelogram $OABC$, let $\overrightarrow{OA} = \vec{a}$ and $\overrightarrow{OC} = \vec{c}$, and find the midpoints of both diagonals.)

Solution

The diagonals are $OB$ and $AC$. As in Example 3, $\overrightarrow{OB} = \vec{a} + \vec{c}$.

Let $M$ be the midpoint of $OB$: $\overrightarrow{OM} = \tfrac{1}{2}\overrightarrow{OB} = \tfrac{1}{2}\vec{a} + \tfrac{1}{2}\vec{c}$.

Let $N$ be the midpoint of $AC$. Since $\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \vec{c} - \vec{a}$,

$$\overrightarrow{ON} = \overrightarrow{OA} + \tfrac{1}{2}\overrightarrow{AC} = \vec{a} + \tfrac{1}{2}\vec{c} - \tfrac{1}{2}\vec{a} = \tfrac{1}{2}\vec{a} + \tfrac{1}{2}\vec{c}$$

$\overrightarrow{OM} = \overrightarrow{ON}$, so $M$ and $N$ are the same point. The midpoint of each diagonal is on the other diagonal, so the diagonals bisect each other. $\blacksquare$

9. (Challenge) $\vec{a}$ and $\vec{b}$ are not collinear. Find the scalars $m$ and $n$ such that

$$m(\vec{a} + 2\vec{b}) + n(3\vec{a} - \vec{b}) = 5\vec{a} + 3\vec{b}$$

Solution

Expand and collect: $(m + 3n)\vec{a} + (2m - n)\vec{b} = 5\vec{a} + 3\vec{b}$.

Because $\vec{a}$ and $\vec{b}$ are not collinear, the coefficients must match:

$$m + 3n = 5 \qquad\text{and}\qquad 2m - n = 3$$

From the second, $n = 2m - 3$. Substitute: $m + 3(2m - 3) = 5$, so $7m = 14$ and $m = 2$. Then $n = 1$.

Check: $2(\vec{a} + 2\vec{b}) + (3\vec{a} - \vec{b}) = 5\vec{a} + 3\vec{b}$. ✓