Skip to content
M
hscscience Ext 1 · Y12
0/100daily goal
0
0
0 due
0
L1 · 0 XP
KJ
Your weak spots
Insights load after your first practice round.
Module 6 · L6 of 20 ~35 min ⚡ +95 XP available

Scalar Multiplication

Multiplying a vector by a number stretches or shrinks it — and if the number is negative, the arrow flips. This operation is the gateway to parallel vectors, linear combinations, and ultimately the entire algebraic theory of vectors. Understand $k\vec{a}$ and you'll unlock vector proofs in geometry.

Today's hook — A vector $\vec{a}$ points north-east. Without calculating, what do you predict $3\vec{a}$ looks like? What about $-2\vec{a}$? Jot your guess about direction and length, then check after card 05.
0/5QUESTS
01
Recall — your gut answer first
+5 XP warm-up

A vector $\vec{a}$ points north-east with magnitude 5. Before learning the rule — predict: what is the direction of $3\vec{a}$? What is the direction of $-2\vec{a}$? How long is each? Write your reasoning.

auto-saved
02
The core idea
+5 XP to read

Multiplying a vector by a scalar (a real number) scales the magnitude and may reverse the direction:

If $k$ is a scalar and $\vec{a}$ is a vector:

  • $|k\vec{a}| = |k| \cdot |\vec{a}|$ — the magnitude is scaled by $|k|$
  • If $k > 0$: same direction as $\vec{a}$
  • If $k < 0$: opposite direction to $\vec{a}$
  • If $k = 0$: the zero vector $\vec{0}$
a 2a −a k=1 (same length) k=2 (twice as long) k=−1 (reversed)
$|k\vec{a}| = |k|\,|\vec{a}|$
Positive scalar
$k > 0$: vector stretches or shrinks but keeps the same direction. $2\vec{a}$ is twice as long as $\vec{a}$, pointing the same way.
Negative scalar
$k < 0$: direction reverses AND magnitude scales by $|k|$. So $-3\vec{a}$ is three times as long as $\vec{a}$ but points the opposite way.
Parallel vectors
Two vectors are parallel if and only if one is a scalar multiple of the other: $\vec{b} = k\vec{a}$ for some scalar $k$. This is the test for parallelism.
03
What you'll master
Know

Key facts

  • $|k\vec{a}| = |k| \cdot |\vec{a}|$ — magnitude scales by $|k|$
  • $k > 0$: same direction; $k < 0$: opposite direction; $k = 0$: zero vector
  • $\vec{b} \parallel \vec{a}$ if and only if $\vec{b} = k\vec{a}$ for some scalar $k$
Understand

Concepts

  • Why the direction flips when a negative scalar is applied
  • The distributive and associative laws for scalar multiplication
  • How scalar multiples are used to test and express parallelism
Can do

Skills

  • Compute $k\vec{a}$ in component form for any scalar $k$
  • Find the magnitude of $k\vec{a}$ without computing components
  • Determine whether two vectors are parallel using scalar multiples
04
Key terms
ScalarA real number with only magnitude (no direction). Examples: 3, $-2$, $\frac{1}{2}$. Contrast with a vector, which has both magnitude and direction.
Scalar multiple $k\vec{a}$The vector obtained by multiplying each component of $\vec{a}$ by the scalar $k$. Its magnitude is $|k||\vec{a}|$.
Parallel vectorsTwo non-zero vectors are parallel if $\vec{b} = k\vec{a}$ for some non-zero scalar $k$. They point in the same or opposite directions.
Zero vector $\vec{0}$The result of multiplying any vector by 0. It has zero magnitude and no defined direction.
Distributive law$(k+m)\vec{a} = k\vec{a} + m\vec{a}$ and $k(\vec{a}+\vec{b}) = k\vec{a} + k\vec{b}$. Scalar multiplication distributes over vector addition.
Associative law$k(m\vec{a}) = (km)\vec{a}$. Two successive scalar multiplications equal one multiplication by the product of the scalars.
05
Scalar multiplication
core concept

Given a vector $\vec{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$ and a scalar $k$, the scalar multiple is:

$$k\vec{a} = \begin{pmatrix} ka_1 \\ ka_2 \end{pmatrix}$$

Multiply each component by $k$. The magnitude of the result is:

$$|k\vec{a}| = |k|\,|\vec{a}|$$

The direction depends on the sign of $k$:

  • $k > 0$: $k\vec{a}$ points in the same direction as $\vec{a}$
  • $k < 0$: $k\vec{a}$ points in the opposite direction to $\vec{a}$
  • $k = 0$: $k\vec{a} = \vec{0}$ (zero vector)
Checking your hook prediction. If $\vec{a}$ points north-east with magnitude 5, then $3\vec{a}$ also points north-east but has magnitude $3 \times 5 = 15$. And $-2\vec{a}$ points south-west (opposite) with magnitude $2 \times 5 = 10$. Did your prediction match?

Given a vector $\vec{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$ and a scalar $k$, the scalar multiple is:

Pause — copy the scalar multiplication rule: $k\begin{pmatrix}a_1\\a_2\end{pmatrix}=\begin{pmatrix}ka_1\\ka_2\end{pmatrix}$, with the magnitude result $|k\vec{a}|=|k|\,|\vec{a}|$ into your book.

Quick check: If $\vec{a} = \begin{pmatrix} 2 \\ -3 \end{pmatrix}$, what is $-4\vec{a}$?

06
Parallel vectors and algebraic laws
core concept

We just saw that $k\vec{a}=\begin{pmatrix}ka_1\\ka_2\end{pmatrix}$; positive $k$ preserves direction, negative $k$ reverses it, and $|k|$ scales the magnitude. That raises a question: two vectors are parallel if and only if one is a scalar multiple of the other — how do you test this algebraically, and what does it mean when $k$ is negative? This card answers it → $\vec{b}\parallel\vec{a}$ iff $\vec{b}=k\vec{a}$ for some $k\neq0$; $k<0$ means the vectors point in opposite directions (anti-parallel).

Parallel vectors: Two non-zero vectors $\vec{a}$ and $\vec{b}$ are parallel if and only if $\vec{b} = k\vec{a}$ for some non-zero scalar $k$.

  • If $k > 0$, they point in the same direction.
  • If $k < 0$, they point in opposite directions.

Key algebraic laws (hold for all vectors $\vec{a}$, $\vec{b}$ and scalars $k$, $m$):

$$k(\vec{a}+\vec{b}) = k\vec{a} + k\vec{b} \qquad (k+m)\vec{a} = k\vec{a} + m\vec{a} \qquad k(m\vec{a}) = (km)\vec{a}$$

These laws make it possible to expand and simplify vector expressions, just as in algebra.

Testing parallelism in components. To check if $\vec{b} = \begin{pmatrix} b_1 \\ b_2 \end{pmatrix}$ is parallel to $\vec{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$, check whether $\dfrac{b_1}{a_1} = \dfrac{b_2}{a_2}$ (i.e., the ratios of corresponding components are equal). If so, $k = \dfrac{b_1}{a_1}$ is the scalar.

Parallel vectors: Two non-zero vectors $\vec{a}$ and $\vec{b}$ are parallel if and only if $\vec{b} = k\vec{a}$ for some non-zero scalar $k$.

Pause — copy the parallel-vector test: $\vec{b}\parallel\vec{a}\iff\vec{b}=k\vec{a}$ for some $k\neq0$; $k>0$ same direction; $k<0$ opposite direction into your book.

Did you get this? True or false: the vectors $\begin{pmatrix} 3 \\ 6 \end{pmatrix}$ and $\begin{pmatrix} -2 \\ -4 \end{pmatrix}$ are parallel.

PROBLEM 1 · COMPUTE SCALAR MULTIPLE

Given $\vec{u} = \begin{pmatrix} -3 \\ 4 \end{pmatrix}$, find $2\vec{u}$ and $-\frac{1}{2}\vec{u}$, and state the magnitude of each.

1
$2\vec{u} = \begin{pmatrix} 2(-3) \\ 2(4) \end{pmatrix} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}$
Multiply each component by 2. Direction unchanged since $k = 2 > 0$.
PROBLEM 2 · PARALLEL VECTORS

Determine whether $\vec{v} = \begin{pmatrix} 6 \\ -9 \end{pmatrix}$ is parallel to $\vec{w} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}$. If so, find the scalar $k$ such that $\vec{v} = k\vec{w}$.

1
Check ratios: $\dfrac{6}{-4} = -\dfrac{3}{2}$ and $\dfrac{-9}{6} = -\dfrac{3}{2}$. Ratios equal, so $\vec{v} \parallel \vec{w}$.
Equal component ratios confirm a scalar relationship exists.
PROBLEM 3 · SIMPLIFY EXPRESSION

Simplify $3\vec{a} - 2(\vec{a} + \vec{b}) + 4\vec{b}$ using the laws of scalar multiplication.

1
Expand: $3\vec{a} - 2(\vec{a} + \vec{b}) + 4\vec{b} = 3\vec{a} - 2\vec{a} - 2\vec{b} + 4\vec{b}$
Apply the distributive law: $-2(\vec{a}+\vec{b}) = -2\vec{a} - 2\vec{b}$.

Fill the gap: If $|\vec{a}| = 7$, then $|-3\vec{a}| = $ .

Trap 01
Forgetting to negate both components
When $k < 0$, every component changes sign. A common error is negating the first component but not the second. Write $-2\begin{pmatrix} 3 \\ -4 \end{pmatrix} = \begin{pmatrix} -6 \\ 8 \end{pmatrix}$, NOT $\begin{pmatrix} -6 \\ -8 \end{pmatrix}$. Expand the multiplication fully.
Trap 02
Using $|k\vec{a}| = k|\vec{a}|$ without the absolute value
Magnitude is always non-negative. For a negative scalar, $|-3\vec{a}| = |-3| \times |\vec{a}| = 3|\vec{a}|$, not $-3|\vec{a}|$. Magnitude of a vector can never be negative.
Trap 03
Claiming all vectors with one matching component are parallel
Both component ratios must be equal. $\begin{pmatrix} 2 \\ 4 \end{pmatrix}$ and $\begin{pmatrix} 2 \\ 6 \end{pmatrix}$ have the same first component but are NOT parallel. Check $\frac{2}{2} = 1 \neq \frac{4}{6}$.

Did you get this? True or false: $\begin{pmatrix} 4 \\ 10 \end{pmatrix}$ and $\begin{pmatrix} 6 \\ 15 \end{pmatrix}$ are parallel.

Work mode · how are you completing this lesson?
1

Given $\vec{v} = \begin{pmatrix} 5 \\ -12 \end{pmatrix}$, find $3\vec{v}$, $-\vec{v}$, and $\frac{1}{5}\vec{v}$. State the magnitude of each.

2

Determine whether $\vec{p} = \begin{pmatrix} 8 \\ -6 \end{pmatrix}$ and $\vec{q} = \begin{pmatrix} -12 \\ 9 \end{pmatrix}$ are parallel. If so, find the scalar $k$.

3

Simplify $5\vec{a} + 3(\vec{a} - 2\vec{b}) - \vec{b}$.

4

If $|\vec{a}| = 4$, find the magnitudes of $5\vec{a}$, $-\frac{3}{4}\vec{a}$, and $0\vec{a}$.

5

Find the value of $m$ such that $m\vec{a} + \vec{b} = \vec{0}$, given $\vec{a} = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ and $\vec{b} = \begin{pmatrix} -6 \\ -9 \end{pmatrix}$.

Odd one out: Three of these are true for any vector $\vec{a}$ and scalars $k$, $m$. Which one is NOT always true?

11
Revisit your thinking

Earlier you predicted the directions and magnitudes of $3\vec{a}$ and $-2\vec{a}$ when $\vec{a}$ points north-east with magnitude 5.

The answers: $3\vec{a}$ points north-east (same direction) with magnitude $3 \times 5 = 15$. $-2\vec{a}$ points south-west (opposite direction) with magnitude $2 \times 5 = 10$. The key insight is that direction is determined by the sign of the scalar and magnitude is scaled by the absolute value. These two facts unlock every scalar multiplication problem.

auto-saved
01
Multiple choice
+5 XP per correct · +25 XP all-correct

Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.

02
Short answer
ApplyBand 32 marks

Q1. Given $\vec{a} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}$, find $-2\vec{a}$ and $|\vec{-2\vec{a}}|$. (2 marks)

auto-saved
ApplyBand 42 marks

Q2. Show that $\vec{u} = \begin{pmatrix} 10 \\ -15 \end{pmatrix}$ and $\vec{v} = \begin{pmatrix} -4 \\ 6 \end{pmatrix}$ are parallel, and find the scalar $k$ such that $\vec{u} = k\vec{v}$. (2 marks)

auto-saved
AnalyseBand 53 marks

Q3. In parallelogram $OABC$, $\vec{OA} = \vec{a}$ and $\vec{OC} = \vec{c}$. Point $M$ is the midpoint of $OB$. Using scalar multiplication and vector addition, show that $\vec{OM} = \frac{1}{2}(\vec{a} + \vec{c})$. (3 marks)

auto-saved
Comprehensive answers (click to reveal)

Activity answers:

1. $3\vec{v} = \begin{pmatrix} 15 \\ -36 \end{pmatrix}$, magnitude 39; $-\vec{v} = \begin{pmatrix} -5 \\ 12 \end{pmatrix}$, magnitude 13; $\frac{1}{5}\vec{v} = \begin{pmatrix} 1 \\ -\frac{12}{5} \end{pmatrix}$, magnitude $\frac{13}{5}$

2. $\frac{8}{-12} = -\frac{2}{3}$ and $\frac{-6}{9} = -\frac{2}{3}$. Equal, so parallel. $k = -\frac{2}{3}$.

3. $5\vec{a}+3\vec{a}-6\vec{b}-\vec{b} = 8\vec{a}-7\vec{b}$

4. $|5\vec{a}| = 20$; $\left|-\frac{3}{4}\vec{a}\right| = 3$; $|0\vec{a}| = 0$

5. $m\begin{pmatrix} 2 \\ 3 \end{pmatrix} = \begin{pmatrix} 6 \\ 9 \end{pmatrix}$, so $m = 3$.


Q1 (2 marks): $-2\vec{a} = -2\begin{pmatrix} -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 6 \\ -10 \end{pmatrix}$ [1].   $|{-2\vec{a}}| = \sqrt{36+100} = \sqrt{136} = 2\sqrt{34}$ [1]. (Or: $|-2||\vec{a}| = 2\sqrt{9+25} = 2\sqrt{34}$.)

Q2 (2 marks): $\dfrac{10}{-4} = -\dfrac{5}{2}$ and $\dfrac{-15}{6} = -\dfrac{5}{2}$ — ratios equal so vectors are parallel [1]. $k = -\dfrac{5}{2}$, i.e. $\vec{u} = -\dfrac{5}{2}\vec{v}$ [1].

Q3 (3 marks): In parallelogram $OABC$, $\vec{AB} = \vec{OC} = \vec{c}$ (opposite sides equal and parallel) [1].   $\vec{OB} = \vec{OA} + \vec{AB} = \vec{a} + \vec{c}$ [1].   $M$ is midpoint of $OB$, so $\vec{OM} = \dfrac{1}{2}\vec{OB} = \dfrac{1}{2}(\vec{a}+\vec{c})$ [1].

01
Boss battle · The Scalar Stretcher
earn bronze · silver · gold

Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.

⚔ Enter the arena
02
Science Jump · platform challenge

Climb platforms by answering scalar multiplication questions. Lighter alternative to the boss.

Mark lesson as complete

Tick when you've finished the practice and review.

🎓
Want help with Scalar Multiplication?

Work through this topic 1-on-1 with an experienced HSC tutor.

Book a free session →