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.
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.
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 $\vec{0}$
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 $k$
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
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
Given a vector $\vec{a} = \begin{pmatrix} a_1 \\ a_2 \end{pmatrix}$ and a scalar $k$, the scalar multiple is:
Multiply each component by $k$. The magnitude of the result is:
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)
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 $-4\vec{a}$?
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 $k$, $m$):
These laws make it possible to expand and simplify vector expressions, just as in algebra.
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.
Worked examples · 3 in a row, reveal as you go
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.
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 $\vec{v} = k\vec{w}$.
Simplify $3\vec{a} - 2(\vec{a} + \vec{b}) + 4\vec{b}$ using the laws of scalar multiplication.
Fill the gap: If $|\vec{a}| = 7$, then $|-3\vec{a}| = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: $\begin{pmatrix} 4 \\ 10 \end{pmatrix}$ and $\begin{pmatrix} 6 \\ 15 \end{pmatrix}$ are parallel.
Activities · practice with the ideas
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.
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 $k$.
Simplify $5\vec{a} + 3(\vec{a} - 2\vec{b}) - \vec{b}$.
If $|\vec{a}| = 4$, find the magnitudes of $5\vec{a}$, $-\frac{3}{4}\vec{a}$, and $0\vec{a}$.
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?
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.
Pick your answer, then rate your confidence — that tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Given $\vec{a} = \begin{pmatrix} -3 \\ 5 \end{pmatrix}$, find $-2\vec{a}$ and $|\vec{-2\vec{a}}|$. (2 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)
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)
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].
Five timed questions. Beat the boss to bank a tier — gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering scalar multiplication questions. Lighter alternative to the boss.
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