Vector arithmetic: addition, subtraction and scalar multiplication

Vector arithmetic combines directed changes. Addition joins movements, subtraction compares them, and scalar multiplication changes their size and possibly their direction.

For example, moving by

$$\mathbf a=\begin{pmatrix}3\\-1\end{pmatrix}$$

and then by

$$\mathbf b=\begin{pmatrix}-2\\4\end{pmatrix}$$

produces the overall movement

$$\mathbf a+\mathbf b =\begin{pmatrix}3+(-2)\\-1+4\end{pmatrix} =\begin{pmatrix}1\\3\end{pmatrix}.$$

The calculation is componentwise because horizontal changes combine with horizontal changes, while vertical changes combine with vertical changes.

Prerequisites

You should be able to:

• represent vectors in column form and in $\mathbf i$, $\mathbf j$, $\mathbf k$ notation;
• distinguish vectors from scalars and points;
• calculate confidently with negatives and fractions;
• solve linear equations and simple simultaneous equations.

Review vectors in two and three dimensions or algebraic fluency if needed.

Adding vectors

Add corresponding components:

$$\boxed{ \begin{pmatrix}a_1\\a_2\end{pmatrix} +\begin{pmatrix}b_1\\b_2\end{pmatrix} =\begin{pmatrix}a_1+b_1\\a_2+b_2\end{pmatrix}}$$

and, in three dimensions,

$$\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix} +\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix} =\begin{pmatrix}a_1+b_1\\a_2+b_2\\a_3+b_3\end{pmatrix}.$$

Only vectors with the same number of components can be added. A two dimensional vector and a three dimensional vector belong to different spaces.

Worked example 1: addition in two forms

Let

$$\mathbf p=\begin{pmatrix}-5\\7\end{pmatrix}, \qquad \mathbf q=\begin{pmatrix}8\\-3\end{pmatrix}.$$

Then

$$\begin{aligned} \mathbf p+\mathbf q &=\begin{pmatrix}-5+8\\7+(-3)\end{pmatrix}\\ &=\boxed{\begin{pmatrix}3\\4\end{pmatrix}}. \end{aligned}$$

Equivalently, $\mathbf p=-5\mathbf i+7\mathbf j$ and $\mathbf q=8\mathbf i-3\mathbf j$, so

$$\mathbf p+\mathbf q =(-5+8)\mathbf i+(7-3)\mathbf j =\boxed{3\mathbf i+4\mathbf j}.$$

The two answers are different notations for the same vector.

Why addition is head to tail

Suppose a route goes from $A$ to $B$, then from $B$ to $C$. The combined displacement goes directly from $A$ to $C$:

$$\boxed{\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}}.$$

This is the triangle law of vector addition. Translate the second arrow without rotating or resizing it so that its tail meets the head of the first. The resultant joins the first tail to the final head.

The parallelogram law gives the same sum. Draw $\mathbf a$ and $\mathbf b$ from one point and complete a parallelogram. Its diagonal from that point is $\mathbf a+\mathbf b$.

Order does not affect the resultant:

$$\boxed{\mathbf a+\mathbf b=\mathbf b+\mathbf a}.$$

The head to tail routes differ, but they are opposite sides of the same parallelogram and finish at the same place.

Worked example 2: simplify a route

Simplify

$$\overrightarrow{PQ}+\overrightarrow{QR}+\overrightarrow{RS}.$$

Join consecutive stages:

$$\overrightarrow{PQ}+\overrightarrow{QR}=\overrightarrow{PR},$$

then

$$\overrightarrow{PR}+\overrightarrow{RS}=\boxed{\overrightarrow{PS}}.$$

The intermediate points cancel conceptually because only the starting and finishing points determine the overall displacement.

Self-check 1

Given

$$\mathbf a=\begin{pmatrix}4\\-6\\1\end{pmatrix}, \qquad \mathbf b=\begin{pmatrix}-7\\2\\5\end{pmatrix},$$

find $\mathbf a+\mathbf b$.

Answer

$$\mathbf a+\mathbf b =\begin{pmatrix}4-7\\-6+2\\1+5\end{pmatrix} =\boxed{\begin{pmatrix}-3\\-4\\6\end{pmatrix}}.$$

The zero vector and additive inverses

The zero vector has every component equal to zero:

$$\mathbf 0=\begin{pmatrix}0\\0\end{pmatrix} \quad\text{or}\quad \mathbf 0=\begin{pmatrix}0\\0\\0\end{pmatrix}.$$

It represents no change and satisfies

$$\mathbf a+\mathbf 0=\mathbf a.$$

The negative of $\mathbf a$ has every direction reversed:

$$-\mathbf a=\begin{pmatrix}-a_1\\-a_2\end{pmatrix}, \qquad \mathbf a+(-\mathbf a)=\mathbf 0.$$

If $\mathbf a$ is represented by $\overrightarrow{AB}$, then

$$-\mathbf a=\overrightarrow{BA}.$$

Negate every component

If $\mathbf a=(3,-5)^{\mathsf T}$, then $-\mathbf a=(-3,5)^{\mathsf T}$. Changing only one sign does not reverse the vector.

Subtracting vectors

Subtraction means adding the opposite vector:

$$\boxed{\mathbf a-\mathbf b=\mathbf a+(-\mathbf b)}.$$

Therefore

$$\begin{pmatrix}a_1\\a_2\end{pmatrix} -\begin{pmatrix}b_1\\b_2\end{pmatrix} =\begin{pmatrix}a_1-b_1\\a_2-b_2\end{pmatrix}.$$

Geometrically, if $\mathbf a$ and $\mathbf b$ start from the same point, $\mathbf a-\mathbf b$ points from the head of $\mathbf b$ to the head of $\mathbf a$. This orientation follows from

$$\mathbf b+(\mathbf a-\mathbf b)=\mathbf a.$$

Worked example 3: subtraction with negative components

Let

$$\mathbf u=\begin{pmatrix}-2\\5\end{pmatrix}, \qquad \mathbf v=\begin{pmatrix}4\\-3\end{pmatrix}.$$

Find $\mathbf u-\mathbf v$ and $\mathbf v-\mathbf u$.

$$\begin{aligned} \mathbf u-\mathbf v &=\begin{pmatrix}-2-4\\5-(-3)\end{pmatrix}\\ &=\boxed{\begin{pmatrix}-6\\8\end{pmatrix}}. \end{aligned}$$

Reversing the subtraction reverses the vector:

$$\begin{aligned} \mathbf v-\mathbf u &=\begin{pmatrix}4-(-2)\\-3-5\end{pmatrix}\\ &=\begin{pmatrix}6\\-8\end{pmatrix} =-(\mathbf u-\mathbf v). \end{aligned}$$

Unlike addition, subtraction is not commutative:

$$\mathbf a-\mathbf b\ne\mathbf b-\mathbf a$$

unless $\mathbf a=\mathbf b$.

Worked example 4: find a missing stage

A three stage journey has total displacement

$$\mathbf r=\begin{pmatrix}7\\1\end{pmatrix}.$$

The first two displacements are

$$\mathbf a=\begin{pmatrix}3\\-4\end{pmatrix}, \qquad \mathbf b=\begin{pmatrix}-2\\6\end{pmatrix}.$$

Find the third displacement $\mathbf c$.

The journey equation is

$$\mathbf a+\mathbf b+\mathbf c=\mathbf r.$$

Hence

$$\begin{aligned} \mathbf c &=\mathbf r-\mathbf a-\mathbf b\\ &=\begin{pmatrix}7\\1\end{pmatrix} -\begin{pmatrix}3\\-4\end{pmatrix} -\begin{pmatrix}-2\\6\end{pmatrix}\\ &=\begin{pmatrix}7-3-(-2)\\1-(-4)-6\end{pmatrix}\\ &=\boxed{\begin{pmatrix}6\\-1\end{pmatrix}}. \end{aligned}$$

Check by addition:

$$\begin{pmatrix}3\\-4\end{pmatrix} +\begin{pmatrix}-2\\6\end{pmatrix} +\begin{pmatrix}6\\-1\end{pmatrix} =\begin{pmatrix}7\\1\end{pmatrix}.$$

Self-check 2

If $\mathbf p=(1,-4,2)^{\mathsf T}$ and $\mathbf q=(-3,5,-1)^{\mathsf T}$, find $\mathbf p-\mathbf q$.

Answer

$$\mathbf p-\mathbf q =\begin{pmatrix}1-(-3)\\-4-5\\2-(-1)\end{pmatrix} =\boxed{\begin{pmatrix}4\\-9\\3\end{pmatrix}}.$$

Multiplying a vector by a scalar

A scalar is an ordinary number. Multiplying a vector by a scalar multiplies every component:

$$\boxed{ \lambda\begin{pmatrix}a_1\\a_2\end{pmatrix} =\begin{pmatrix}\lambda a_1\\\lambda a_2\end{pmatrix}}.$$

The value and sign of $\lambda$ have geometric meaning.

Scalar $\lambda$	Effect on $\mathbf a$
$\lambda>1$	Same direction, longer
$0<\lambda<1$	Same direction, shorter
$\lambda=0$	Becomes $\mathbf 0$
$-1<\lambda<0$	Opposite direction, shorter
$\lambda<-1$	Opposite direction, longer
$\lambda=-1$	Same length, opposite direction

More precisely,

$$|\lambda\mathbf a|=|\lambda|\,|\mathbf a|.$$

The absolute value is essential because a magnitude cannot be negative. Magnitudes are developed fully in vector magnitude and direction.

Worked example 5: combine scalar multiples

Given

$$\mathbf a=\begin{pmatrix}2\\-1\\4\end{pmatrix}, \qquad \mathbf b=\begin{pmatrix}-3\\5\\2\end{pmatrix},$$

find $3\mathbf a-2\mathbf b$.

First calculate each multiple:

$$3\mathbf a=\begin{pmatrix}6\\-3\\12\end{pmatrix}, \qquad 2\mathbf b=\begin{pmatrix}-6\\10\\4\end{pmatrix}.$$

Then subtract:

$$\begin{aligned} 3\mathbf a-2\mathbf b &=\begin{pmatrix}6\\-3\\12\end{pmatrix} -\begin{pmatrix}-6\\10\\4\end{pmatrix}\\ &=\boxed{\begin{pmatrix}12\\-13\\8\end{pmatrix}}. \end{aligned}$$

Notice that subtracting $-6$ gives $12$ in the first component.

Algebraic laws

Vector arithmetic follows familiar distributive and associative laws. For scalars $s,t$,

$$\begin{aligned} s(\mathbf a+\mathbf b)&=s\mathbf a+s\mathbf b,\\ (s+t)\mathbf a&=s\mathbf a+t\mathbf a,\\ s(t\mathbf a)&=(st)\mathbf a,\\ (\mathbf a+\mathbf b)+\mathbf c&=\mathbf a+(\mathbf b+\mathbf c). \end{aligned}$$

These laws allow vector expressions to be expanded, collected and rearranged like algebraic expressions, provided vectors are combined with vectors and scalars with scalars.

Worked example 6: simplify symbolically

Simplify

$$2(\mathbf a-3\mathbf b)-(4\mathbf a-\mathbf b).$$

Expand both brackets:

$$2\mathbf a-6\mathbf b-4\mathbf a+\mathbf b.$$

Collect the $\mathbf a$ and $\mathbf b$ terms separately:

$$\boxed{-2\mathbf a-5\mathbf b}.$$

The vectors $\mathbf a$ and $\mathbf b$ behave like distinct algebraic quantities. They cannot be combined into one coefficient unless they are known to be equal or parallel.

There is no ordinary vector multiplication here

At this stage, $2\mathbf a$ means scalar multiplication. An expression such as $\mathbf a\mathbf b$ is not defined by the rules on this page. Do not multiply corresponding components unless a later topic explicitly defines a product.

Self-check 3

Simplify

$$3(2\mathbf p-\mathbf q)+2(\mathbf q-\mathbf p).$$

Answer

$$\begin{aligned} 3(2\mathbf p-\mathbf q)+2(\mathbf q-\mathbf p) &=6\mathbf p-3\mathbf q+2\mathbf q-2\mathbf p\\ &=\boxed{4\mathbf p-\mathbf q}. \end{aligned}$$

Linear combinations

An expression of the form

$$s\mathbf a+t\mathbf b$$

is a linear combination of $\mathbf a$ and $\mathbf b$. It means take a scaled copy of $\mathbf a$, a scaled copy of $\mathbf b$, then add them.

In two dimensions, two non-parallel vectors can be used to express every vector in the plane. Finding the scalars means comparing components and solving simultaneous equations.

Worked example 7: express one vector using two others

Let

$$\mathbf a=\begin{pmatrix}1\\2\end{pmatrix}, \qquad \mathbf b=\begin{pmatrix}3\\-1\end{pmatrix}, \qquad \mathbf c=\begin{pmatrix}11\\1\end{pmatrix}.$$

Find scalars $s$ and $t$ such that

$$\mathbf c=s\mathbf a+t\mathbf b.$$

Substitute the components:

$$\begin{pmatrix}11\\1\end{pmatrix} =s\begin{pmatrix}1\\2\end{pmatrix} +t\begin{pmatrix}3\\-1\end{pmatrix} =\begin{pmatrix}s+3t\\2s-t\end{pmatrix}.$$

Equal vectors have equal corresponding components, so

$$s+3t=11, \qquad 2s-t=1.$$

From the second equation, $t=2s-1$. Substitute into the first:

$$s+3(2s-1)=11,$$

so

$$7s=14 \quad\Longrightarrow\quad s=2.$$

Then $t=2(2)-1=3$. Therefore

$$\boxed{\mathbf c=2\mathbf a+3\mathbf b}.$$

Check directly:

$$2\begin{pmatrix}1\\2\end{pmatrix} +3\begin{pmatrix}3\\-1\end{pmatrix} =\begin{pmatrix}2\\4\end{pmatrix} +\begin{pmatrix}9\\-3\end{pmatrix} =\begin{pmatrix}11\\1\end{pmatrix}.$$

Why non-parallel matters

If $\mathbf b=2\mathbf a$, then

$$s\mathbf a+t\mathbf b=(s+2t)\mathbf a,$$

which can only produce vectors parallel to $\mathbf a$. The two component equations will either be inconsistent, giving no representation, or dependent, giving infinitely many representations. Non-parallel directions provide genuinely independent choices.

Solving vector equations

A vector equation is true exactly when all corresponding component equations are true. You can often rearrange it as a whole before working componentwise.

Worked example 8: one unknown vector

Given

$$4\mathbf x+\mathbf a=3\mathbf b,$$

where

$$\mathbf a=\begin{pmatrix}2\\-5\end{pmatrix}, \qquad \mathbf b=\begin{pmatrix}-2\\1\end{pmatrix},$$

find $\mathbf x$.

Rearrange the vector equation first:

$$4\mathbf x=3\mathbf b-\mathbf a.$$

Now substitute:

$$\begin{aligned} 4\mathbf x &=3\begin{pmatrix}-2\\1\end{pmatrix} -\begin{pmatrix}2\\-5\end{pmatrix}\\ &=\begin{pmatrix}-6\\3\end{pmatrix} -\begin{pmatrix}2\\-5\end{pmatrix}\\ &=\begin{pmatrix}-8\\8\end{pmatrix}. \end{aligned}$$

Divide every component by $4$:

$$\boxed{\mathbf x=\begin{pmatrix}-2\\2\end{pmatrix}}.$$

Substitution gives $4\mathbf x+\mathbf a=(-6,3)^{\mathsf T}=3\mathbf b$, so the answer is consistent.

Worked example 9: determine scalar parameters

Find $m$ and $n$ if

$$m\begin{pmatrix}2\\3\end{pmatrix} +n\begin{pmatrix}-1\\4\end{pmatrix} =\begin{pmatrix}8\\1\end{pmatrix}.$$

Comparing components gives

$$2m-n=8, \qquad 3m+4n=1.$$

From the first equation, $n=2m-8$. Substitute:

$$3m+4(2m-8)=1.$$

Therefore

$$11m=33 \quad\Longrightarrow\quad m=3,$$

and

$$n=2(3)-8=-2.$$

Thus

$$\boxed{m=3,qquad n=-2}.$$

Worked example 10: three dimensional consistency

Find $k$ if

$$\begin{pmatrix}2k-1\\k+4\\3k\end{pmatrix} =\begin{pmatrix}5\\7\\9\end{pmatrix}.$$

Every component must give the same value of $k$:

$$2k-1=5\Longrightarrow k=3,$$ $$k+4=7\Longrightarrow k=3,$$

and

$$3k=9\Longrightarrow k=3.$$

All three agree, so $\boxed{k=3}$.

If one component produced a different value, there would be no solution. Checking every component matters, especially when a three dimensional equation contains only one unknown.

Self-check 4

Given

$$2\mathbf y-3\mathbf a=\mathbf b, \qquad \mathbf a=\begin{pmatrix}1\\-2\end{pmatrix}, \qquad \mathbf b=\begin{pmatrix}5\\4\end{pmatrix},$$

find $\mathbf y$.

Answer

Rearrange:

$$2\mathbf y=\mathbf b+3\mathbf a.$$

Therefore

$$\begin{aligned} 2\mathbf y &=\begin{pmatrix}5\\4\end{pmatrix} +3\begin{pmatrix}1\\-2\end{pmatrix}\\ &=\begin{pmatrix}8\\-2\end{pmatrix}, \end{aligned}$$

so

$$\boxed{\mathbf y=\begin{pmatrix}4\\-1\end{pmatrix}}.$$

Parallel vectors from scalar multiplication

Two non-zero vectors are parallel precisely when one is a scalar multiple of the other:

$$\boxed{\mathbf a\parallel\mathbf b \quad\Longleftrightarrow\quad \mathbf a=\lambda\mathbf b \text{ for some }\lambda\ne0.}$$

If $\lambda>0$, they point in the same direction. If $\lambda<0$, they point in opposite directions.

Worked example 11: test whether vectors are parallel

Consider

$$\mathbf p=\begin{pmatrix}6\\-9\\12\end{pmatrix}, \qquad \mathbf q=\begin{pmatrix}-2\\3\\-4\end{pmatrix}.$$

Each component of $\mathbf p$ is $-3$ times the corresponding component of $\mathbf q$:

$$\mathbf p=-3\mathbf q.$$

Therefore the vectors are parallel and point in opposite directions.

Do not merely compare one pair of components. The same multiplier must work for every component.

Worked example 12: find a parameter for parallel vectors

Find $r$ if

$$\begin{pmatrix}r+1\\10\end{pmatrix}$$

is parallel to

$$\begin{pmatrix}2\\5\end{pmatrix}.$$

There must be a scalar $\lambda$ such that

$$\begin{pmatrix}r+1\\10\end{pmatrix} =\lambda\begin{pmatrix}2\\5\end{pmatrix}.$$

The second component gives

$$10=5\lambda \quad\Longrightarrow\quad \lambda=2.$$

The first component then gives

$$r+1=2(2)=4,$$

so

$$\boxed{r=3}.$$

This scalar multiple method is safer than informal ratio tests when zero components are present.

Common misconceptions

Adding all the numbers together

Vector addition preserves the component structure:

$$\begin{pmatrix}1\\2\end{pmatrix} +\begin{pmatrix}3\\4\end{pmatrix} =\begin{pmatrix}4\\6\end{pmatrix},$$

not the scalar $10$.

Losing a negative when subtracting

Write the component subtraction explicitly:

$$5-(-2)=7.$$

Putting brackets around negative components prevents many errors.

Scaling only one component

The scalar multiplies the entire vector:

$$-2\begin{pmatrix}3\\-4\end{pmatrix} =\begin{pmatrix}-6\\8\end{pmatrix}.$$

Confusing $-\mathbf a$ with $|\mathbf a|$

$-\mathbf a$ is a vector in the opposite direction. $|\mathbf a|$ is a non-negative scalar length. They are different kinds of object.

Assuming parallel means same direction

$\mathbf a=-4\mathbf b$ makes the vectors parallel but oppositely directed. Same direction requires a positive multiplier.

Dividing by a vector

From $3\mathbf x=\mathbf a$, write $\mathbf x=\frac13\mathbf a$. Ordinary division by a vector is not defined, so do not write $\mathbf x=\mathbf a/3$ as though later you could divide by $\mathbf a$.

Mixed self-check

Try these without looking back at the worked examples.

1. Given $\mathbf a=(-3,4)^{\mathsf T}$ and $\mathbf b=(5,-7)^{\mathsf T}$, find $2\mathbf a+\mathbf b$.
2. Simplify $4(\mathbf p-2\mathbf q)-3(\mathbf p-\mathbf q)$.
3. Find $\mathbf x$ if $3\mathbf x-\mathbf a=2\mathbf b$, where $\mathbf a=(6,-3)^{\mathsf T}$ and $\mathbf b=(-3,9)^{\mathsf T}$.
4. Find $s$ and $t$ if $s(1,3)^{\mathsf T}+t(2,-1)^{\mathsf T}=(8,7)^{\mathsf T}$.
5. Find $k$ so that $(k,12)^{\mathsf T}$ is parallel to $(-2,3)^{\mathsf T}$. State whether the two vectors have the same or opposite directions.
6. Simplify $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA}$.

Answers and guidance

$$2\mathbf a+\mathbf b =\begin{pmatrix}-6\\8\end{pmatrix} +\begin{pmatrix}5\\-7\end{pmatrix} =\boxed{\begin{pmatrix}-1\\1\end{pmatrix}}.$$

2. Expand, then collect like vector terms:

   $$4\mathbf p-8\mathbf q-3\mathbf p+3\mathbf q =\boxed{\mathbf p-5\mathbf q}.$$

3. Rearrange before substituting:

   $$3\mathbf x=\mathbf a+2\mathbf b =\begin{pmatrix}6\\-3\end{pmatrix} +\begin{pmatrix}-6\\18\end{pmatrix} =\begin{pmatrix}0\\15\end{pmatrix}.$$

   Hence

   $$\boxed{\mathbf x=\begin{pmatrix}0\\5\end{pmatrix}}.$$

4. Comparing components gives

   $$s+2t=8, \qquad 3s-t=7.$$

   From $s=8-2t$,

   $$3(8-2t)-t=7 \quad\Longrightarrow\quad t=\frac{17}{7}.$$

   Therefore

   $$s=8-\frac{34}{7}=\frac{22}{7}.$$

   Thus $\boxed{s=22/7, t=17/7}$.

5. Write

   $$\begin{pmatrix}k\\12\end{pmatrix} =\lambda\begin{pmatrix}-2\\3\end{pmatrix}.$$

   Since $12=3\lambda$, $\lambda=4$, so $k=4(-2)=-8$. The positive multiplier means the vectors have the same direction:

   $$\boxed{k=-8}.$$

6. The route returns to its starting point, so

   $$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{DA} =\overrightarrow{AA} =\boxed{\mathbf 0}.$$

What to remember

$$\begin{aligned} \mathbf a+\mathbf b&=\text{add corresponding components},\\ \mathbf a-\mathbf b&=\mathbf a+(-\mathbf b),\\ \lambda\mathbf a&=\text{multiply every component by }\lambda,\\ \overrightarrow{AB}+\overrightarrow{BC}&=\overrightarrow{AC},\\ \mathbf a\parallel\mathbf b&\Longleftrightarrow \mathbf a=\lambda\mathbf b \text{ for some }\lambda\ne0. \end{aligned}$$

Keep the geometric meaning alongside the component calculation. An answer can often be checked by asking whether its direction and approximate size make sense.

Next steps

Continue to vector magnitude and direction to calculate lengths and unit vectors. Then study position vectors and distance to turn geometric routes, line segments and proofs into vector equations. For applied contexts involving displacement, velocity and force, continue to vectors in modelling.