Multiplying a vector by a number | Grade 10

A cyclist rides $3$ km east. A second cyclist rides twice as far in the same direction. A third rides the same distance as the first but back to the west. All three journeys are made from the same displacement $\vec{a}$ : the second is $2\vec{a}$ and the third is $-\vec{a}$ . Multiplying a vector by a number stretches it, shrinks it or turns it around. It is the second basic operation on vectors, and it gives us a clean way to say "these points lie on one line" and "this point is one third of the way along".

The product of a number and a vector

The vectors

\vec{a}

2\vec{a}

and

-\frac{1}{2}\vec{a}

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

Properties

For all vectors $\vec{a}$ , $\vec{b}$ and numbers $k$ , $l$ :

Property	Formula
Distributes over vector sums	$k(\vec{a} + \vec{b}) = k\vec{a} + k\vec{b}$
Distributes over number sums	$(k + l)\vec{a} = k\vec{a} + l\vec{a}$
Associative	$k(l\vec{a}) = (kl)\vec{a}$
Identity	$1\vec{a} = \vec{a}$ , $(-1)\vec{a} = -\vec{a}$

These are the same rules you use with letters in algebra, so you may expand and collect terms in the usual way, treating vectors like letters. Also $k\vec{a} = \vec{0}$ only when $k = 0$ or $\vec{a} = \vec{0}$ .

Example Lengths and simplification

a) Suppose $|\vec{a}| = 5$ . Find $|3\vec{a}|$ and $|-2\vec{a}|$ , and compare their directions with $\vec{a}$ . b) Simplify $2(\vec{a} - 3\vec{b}) + 3(\vec{a} + \vec{b})$ .

Solution. a) $|3\vec{a}| = |3| \cdot 5 = 15$ , and $3 > 0$ , so $3\vec{a}$ has the same direction as $\vec{a}$ . $|-2\vec{a}| = |-2| \cdot 5 = 10$ , and $-2 < 0$ , so $-2\vec{a}$ has the opposite direction to $\vec{a}$ .

b) Expand each bracket: $2\vec{a} - 6\vec{b} + 3\vec{a} + 3\vec{b}$ . Collect like terms: $(2 + 3)\vec{a} + (-6 + 3)\vec{b} = 5\vec{a} - 3\vec{b}$ .

Collinear vectors

If $\vec{b} = k\vec{a}$ , then $\vec{b}$ is a stretched or reversed copy of $\vec{a}$ , so the two are collinear. The converse is also true.

Applying this to the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ gives a test for points on one line.

Example Are the points collinear?

Let $\vec{u}$ and $\vec{v}$ be two non-collinear vectors. Points $A$ , $B$ , $C$ satisfy $\overrightarrow{AB} = \vec{u} + 2\vec{v}$ and $\overrightarrow{AC} = 3\vec{u} + m\vec{v}$ .

a) For $m = 6$ , show that $A$ , $B$ , $C$ are collinear. b) Find all $m$ for which they are collinear.

Solution. a) For $m = 6$ , $\overrightarrow{AC} = 3\vec{u} + 6\vec{v} = 3(\vec{u} + 2\vec{v}) = 3\overrightarrow{AB}$ . So $\overrightarrow{AC} = 3\overrightarrow{AB}$ and the points are collinear.

b) The points are collinear if and only if $\overrightarrow{AC} = k\overrightarrow{AB}$ for some $k$ , that is $3\vec{u} + m\vec{v} = k\vec{u} + 2k\vec{v}$ . Since $\vec{u}$ and $\vec{v}$ are not collinear, the coefficients must match: $k = 3$ and $m = 2k = 6$ . So $m = 6$ is the only value.

Midpoint and centroid formulas

In the previous lesson we met $\overrightarrow{MA} + \overrightarrow{MB} = \vec{0}$ for the midpoint $M$ of $AB$ . Insert an arbitrary point $O$ : $\overrightarrow{MA} = \overrightarrow{MO} + \overrightarrow{OA}$ and $\overrightarrow{MB} = \overrightarrow{MO} + \overrightarrow{OB}$ , so $\overrightarrow{OA} + \overrightarrow{OB} = 2\overrightarrow{OM}.$ The same trick with the centroid gives $\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC} = 3\overrightarrow{OG}$ . Dividing by $2$ and $3$ :

Writing a vector along two directions

If $\vec{a}$ and $\vec{b}$ are not collinear, every vector $\vec{w}$ in the plane can be written in exactly one way as $\vec{w} = x\vec{a} + y\vec{b}.$ Think of walking $x$ steps in direction $\vec{a}$ and then $y$ steps in direction $\vec{b}$ .

\vec{w} = 2\vec{a} + \vec{b}

Example Decomposing along two sides of a triangle

In triangle $ABC$ let $\vec{b} = \overrightarrow{AB}$ and $\vec{c} = \overrightarrow{AC}$ . Let $M$ be the midpoint of $BC$ and $G$ the centroid. Express $\overrightarrow{AM}$ , $\overrightarrow{AG}$ and $\overrightarrow{BG}$ in terms of $\vec{b}$ and $\vec{c}$ .

Solution.

Take $A$ as the reference point in the midpoint formula: $\overrightarrow{AM} = \dfrac{1}{2}(\overrightarrow{AB} + \overrightarrow{AC}) = \dfrac{1}{2}(\vec{b} + \vec{c})$ .
The centroid formula with $O = A$ : $\overrightarrow{AG} = \dfrac{1}{3}(\overrightarrow{AA} + \overrightarrow{AB} + \overrightarrow{AC}) = \dfrac{1}{3}(\vec{b} + \vec{c})$ .
By the triangle rule, $\overrightarrow{BG} = \overrightarrow{BA} + \overrightarrow{AG} = -\vec{b} + \dfrac{1}{3}(\vec{b} + \vec{c}) = -\dfrac{2}{3}\vec{b} + \dfrac{1}{3}\vec{c}$ .

As a check, $\overrightarrow{AG} = \dfrac{2}{3}\overrightarrow{AM}$ , which is the known property of the centroid.

Solving geometry problems with vectors

The method is always the same:

Choose two non-collinear vectors as a basis, usually two sides from one vertex.
Write every vector in the problem as $x\vec{a} + y\vec{b}$ , using the triangle rule, midpoints and the given ratios.
Compare coefficients, or look for a relation $\overrightarrow{AB} = k\overrightarrow{AC}$ .

Parallelogram

ABCD

with

M

BC

and

N

AC

Example Three points on a line

$ABCD$ is a parallelogram, $M$ is the midpoint of $BC$ , and $N$ is the point of the diagonal $AC$ with $\overrightarrow{AN} = \dfrac{2}{3}\overrightarrow{AC}$ . Prove that $D$ , $N$ , $M$ are collinear.

Solution.

Let $\vec{b} = \overrightarrow{AB}$ and $\vec{d} = \overrightarrow{AD}$ . These are not collinear. In the parallelogram, $\overrightarrow{AC} = \vec{b} + \vec{d}$ and $\overrightarrow{DC} = \vec{b}$ .
$\overrightarrow{DM} = \overrightarrow{DC} + \overrightarrow{CM} = \vec{b} - \dfrac{1}{2}\vec{d}$ (because $\overrightarrow{CM} = -\dfrac{1}{2}\overrightarrow{BC} = -\dfrac{1}{2}\vec{d}$ ).
$\overrightarrow{DN} = \overrightarrow{AN} - \overrightarrow{AD} = \dfrac{2}{3}(\vec{b} + \vec{d}) - \vec{d} = \dfrac{2}{3}\vec{b} - \dfrac{1}{3}\vec{d}$ .
Compare: $\overrightarrow{DN} = \dfrac{2}{3}\left(\vec{b} - \dfrac{1}{2}\vec{d}\right) = \dfrac{2}{3}\overrightarrow{DM}$ . So the three points are collinear, and $N$ is two thirds of the way from $D$ to $M$ .

Key terms

English	Tiếng Việt
product of a number and a vector	tích của một số với một vectơ
scalar	số thực (vô hướng)
collinear vectors	các vectơ cùng phương
three collinear points	ba điểm thẳng hàng
midpoint	trung điểm
centroid	trọng tâm
decomposition of a vector	phân tích một vectơ
non-collinear vectors	các vectơ không cùng phương

Summary

$k\vec{a}$ has length $|k||\vec{a}|$ , the direction of $\vec{a}$ if $k > 0$ and the opposite direction if $k < 0$ .
The usual algebra rules hold: $k(\vec{a} + \vec{b}) = k\vec{a} + k\vec{b}$ , $(k + l)\vec{a} = k\vec{a} + l\vec{a}$ , $k(l\vec{a}) = (kl)\vec{a}$ .
For $\vec{a} \neq \vec{0}$ : $\vec{b}$ is collinear with $\vec{a}$ iff $\vec{b} = k\vec{a}$ . Points $A$ , $B$ , $C$ are collinear iff $\overrightarrow{AB} = k\overrightarrow{AC}$ .
For any point $O$ : $\overrightarrow{OM} = \frac{1}{2}(\overrightarrow{OA} + \overrightarrow{OB})$ for a midpoint and $\overrightarrow{OG} = \frac{1}{3}(\overrightarrow{OA} + \overrightarrow{OB} + \overrightarrow{OC})$ for a centroid.
Every vector can be written uniquely as $x\vec{a} + y\vec{b}$ when $\vec{a}$ , $\vec{b}$ are non-collinear.

Practice

Suppose $|\vec{a}| = 6$ . Find $|2\vec{a}|$ , $\left|-\dfrac{1}{2}\vec{a}\right|$ and $|0\vec{a}|$ , and say which of $2\vec{a}$ , $-\frac{1}{2}\vec{a}$ point in the same direction as $\vec{a}$ .
Simplify $3(\vec{a} - 2\vec{b}) - 2(\vec{a} - \vec{b})$ .
In triangle $ABC$ let $M$ , $N$ be the midpoints of $AB$ , $AC$ . Prove that $\overrightarrow{MN} = \dfrac{1}{2}\overrightarrow{BC}$ .
Vectors $\vec{u}$ , $\vec{v}$ are non-collinear, $\overrightarrow{AB} = 2\vec{u} - \vec{v}$ and $\overrightarrow{AC} = -6\vec{u} + 3\vec{v}$ . Show that $A$ , $B$ , $C$ are collinear, and find $k$ with $\overrightarrow{AC} = k\overrightarrow{AB}$ .
Let $\overrightarrow{AB} = \vec{u} - 2\vec{v}$ and $\overrightarrow{AC} = 3\vec{u} + m\vec{v}$ , with $\vec{u}$ , $\vec{v}$ non-collinear. Find $m$ so that $A$ , $B$ , $C$ are collinear.
$ABCD$ is a parallelogram and $O$ is any point. Prove that $\overrightarrow{OA} + \overrightarrow{OC} = \overrightarrow{OB} + \overrightarrow{OD}$ .
In triangle $ABC$ the point $D$ lies on $BC$ with $BD = \dfrac{1}{4}BC$ . Express $\overrightarrow{AD}$ in terms of $\overrightarrow{AB}$ and $\overrightarrow{AC}$ .
$G$ is the centroid of triangle $ABC$ , $\vec{b} = \overrightarrow{AB}$ and $\vec{c} = \overrightarrow{AC}$ . Express $\overrightarrow{CG}$ in terms of $\vec{b}$ and $\vec{c}$ .
(Challenge) In triangle $ABC$ let $D$ be on $BC$ with $\overrightarrow{BD} = 2\overrightarrow{DC}$ , $E$ the midpoint of $AD$ , and $F$ the point of $AC$ with $\overrightarrow{AF} = \dfrac{2}{5}\overrightarrow{AC}$ . Prove that $B$ , $E$ , $F$ are collinear.
(Challenge) Let $\vec{a}$ , $\vec{b}$ be non-collinear, $\vec{u} = \vec{a} + 2\vec{b}$ , $\vec{v} = \vec{a} - \vec{b}$ and $\vec{w} = 5\vec{a} + \vec{b}$ . Find $x$ , $y$ with $\vec{w} = x\vec{u} + y\vec{v}$ .

Show solutions

$|2\vec{a}| = 2 \cdot 6 = 12$ , $\left|-\frac{1}{2}\vec{a}\right| = \frac{1}{2} \cdot 6 = 3$ and $|0\vec{a}| = 0$ . Only $2\vec{a}$ has the same direction as $\vec{a}$ ( $2 > 0$ ); $-\frac{1}{2}\vec{a}$ is opposite.
$3\vec{a} - 6\vec{b} - 2\vec{a} + 2\vec{b} = \vec{a} - 4\vec{b}$ .
Take $A$ as the reference point: $\overrightarrow{AM} = \frac{1}{2}\overrightarrow{AB}$ and $\overrightarrow{AN} = \frac{1}{2}\overrightarrow{AC}$ . Then $\overrightarrow{MN} = \overrightarrow{AN} - \overrightarrow{AM} = \frac{1}{2}(\overrightarrow{AC} - \overrightarrow{AB}) = \frac{1}{2}\overrightarrow{BC}$ .
$\overrightarrow{AC} = -6\vec{u} + 3\vec{v} = -3(2\vec{u} - \vec{v}) = -3\overrightarrow{AB}$ . So $\overrightarrow{AC} = k\overrightarrow{AB}$ with $k = -3$ , and the points are collinear ( $C$ is on the other side of $A$ from $B$ ).
We need $\overrightarrow{AC} = k\overrightarrow{AB}$ : $3\vec{u} + m\vec{v} = k\vec{u} - 2k\vec{v}$ . Then $k = 3$ and $m = -2k = -6$ .
Let $I$ be the centre of the parallelogram, the midpoint of both $AC$ and $BD$ . Then $\overrightarrow{OA} + \overrightarrow{OC} = 2\overrightarrow{OI}$ and $\overrightarrow{OB} + \overrightarrow{OD} = 2\overrightarrow{OI}$ , so they are equal.
$\overrightarrow{BD} = \frac{1}{4}\overrightarrow{BC} = \frac{1}{4}(\overrightarrow{AC} - \overrightarrow{AB})$ . So $\overrightarrow{AD} = \overrightarrow{AB} + \overrightarrow{BD} = \overrightarrow{AB} + \frac{1}{4}\overrightarrow{AC} - \frac{1}{4}\overrightarrow{AB} = \frac{3}{4}\overrightarrow{AB} + \frac{1}{4}\overrightarrow{AC}$ .
$\overrightarrow{CG} = \overrightarrow{CA} + \overrightarrow{AG} = -\vec{c} + \frac{1}{3}(\vec{b} + \vec{c}) = \frac{1}{3}\vec{b} - \frac{2}{3}\vec{c}$ .
Let $\vec{b} = \overrightarrow{AB}$ , $\vec{c} = \overrightarrow{AC}$ . Since $BD = \frac{2}{3}BC$ , $\overrightarrow{AD} = \vec{b} + \frac{2}{3}(\vec{c} - \vec{b}) = \frac{1}{3}\vec{b} + \frac{2}{3}\vec{c}$ , and $\overrightarrow{AE} = \frac{1}{2}\overrightarrow{AD} = \frac{1}{6}\vec{b} + \frac{1}{3}\vec{c}$ . Then $\overrightarrow{BE} = \overrightarrow{AE} - \vec{b} = -\frac{5}{6}\vec{b} + \frac{1}{3}\vec{c}$ and $\overrightarrow{BF} = \overrightarrow{AF} - \vec{b} = -\vec{b} + \frac{2}{5}\vec{c}$ . Now $\frac{6}{5}\overrightarrow{BE} = -\vec{b} + \frac{2}{5}\vec{c} = \overrightarrow{BF}$ , so $B$ , $E$ , $F$ are collinear.
$x\vec{u} + y\vec{v} = (x + y)\vec{a} + (2x - y)\vec{b}$ . Comparing with $5\vec{a} + \vec{b}$ (the vectors $\vec{a}$ , $\vec{b}$ are non-collinear): $x + y = 5$ and $2x - y = 1$ . Adding gives $3x = 6$ , so $x = 2$ and $y = 3$ . Check: $2(\vec{a} + 2\vec{b}) + 3(\vec{a} - \vec{b}) = 5\vec{a} + \vec{b}$ .