Directed vectors in three dimensions / vector pathways

Overview

Big Picture

A point gives a position; a vector gives a directed change. If $A=(a_1,a_2,a_3)$ and $B=(b_1,b_2,b_3)$ , then

\overrightarrow{AB}=\begin{pmatrix}b_1-a_1\\b_2-a_2\\b_3-a_3\end{pmatrix} =\text{end position}-\text{start position}.

A negative component is meaningful: it means moving in the negative direction of that axis. Do not replace differences by absolute values.

For compact working, this resource also writes a vector as a bold-labelled component triple, for example $\mathbf v=(2,-3,5)$ . Its column form is $\begin{pmatrix}2\\-3\\5\end{pmatrix}$ , and its basis form is $2\mathbf i-3\mathbf j+5\mathbf k$ . The letters $\mathbf i,\mathbf j,\mathbf k$ mean one unit in the positive $x,y,z$ directions respectively. Plain point labels such as $A=(2,-3,5)$ give coordinates, not a displacement instruction.

Method

Step by Step

Mark the requested start and finish.
Choose a route that joins them using known segments.
Write the route as directed vectors whose endpoints join: $A\to P\to B$ means $\overrightarrow{AP}+\overrightarrow{PB}$ .
Reverse any supplied vector that points against your route: $\overrightarrow{BA}=-\overrightarrow{AB}$ .
Substitute the vectors and add matching components. Check that start position plus your result reaches the finish.

Vector addition is commutative: changing the order of the terms does not change their sum. Writing them in travel order makes the route easier to check. The direction of each term, not merely its position in the sum, is crucial.

Key idea

The Why

Let the position vectors of $A,P,B$ , measured from the same origin $O$ , be $\mathbf a,\mathbf p,\mathbf b$ . Then

\overrightarrow{AP}+\overrightarrow{PB} =(\mathbf p-\mathbf a)+(\mathbf b-\mathbf p) =\mathbf b-\mathbf a=\overrightarrow{AB}.

The intermediate position cancels. Going by a longer route gives the same total displacement when the start and finish are unchanged. A closed route back to its starting point has resultant $\mathbf0$ .

Worked example

Worked examples

1. Coordinates, with a negative component

$R=(2,-1,4)$ , $S=(-3,2,6)$ . Find $\overrightarrow{RS}$ .

\overrightarrow{RS}=(-3-2,\ 2-(-1),\ 6-4)=\boxed{(-5,3,2)}.

Check: $(2,-1,4)+(-5,3,2)=(-3,2,6)=S$ . Reversing the journey gives $\overrightarrow{SR}=\boxed{(5,-3,-2)}$ : every component changes sign.

2. A route with a supplied vector pointing the other way

In space, $\overrightarrow{UV}=(1,4,-2)$ and $\overrightarrow{WV}=(3,-1,5)$ . Find $\overrightarrow{UW}$ .

    U ─────→ V ←───── W               supplied arrows
    U ─────→ V ─────→ W               route wanted: U to V to W

Schematic connectivity only; these three-dimensional vectors need not be parallel.

The second leg is $\overrightarrow{VW}=-\overrightarrow{WV}=(-3,1,-5)$ .

\overrightarrow{UW}=\overrightarrow{UV}+\overrightarrow{VW} =(1,4,-2)+(-3,1,-5)=\boxed{(-2,5,-7)}.

An independent route check is $\overrightarrow{UW}+\overrightarrow{WV}=\overrightarrow{UV}$ : $(-2,5,-7)+(3,-1,5)=(1,4,-2)$ .

3. Following a route on a three-dimensional box

The diagram represents a cuboid. $ABCD$ is its lower face; $EFGH$ is directly above it, with $E,F,G,H$ above $A,B,C,D$ respectively.

Redrawn for this page from the resource's text schematic. Not to scale.

Original perspective schematic, not to scale. Lines show edges, not vector directions; each vector's letter order supplies its direction.

Let $\overrightarrow{AB}=\mathbf u$ , $\overrightarrow{AD}=\mathbf v$ , $\overrightarrow{AE}=\mathbf w$ . Opposite edges of a cuboid have the same displacement when travelled in corresponding directions: for example $\overrightarrow{BC}=\mathbf v$ .

Find $\overrightarrow{EC}$ . Choose $E\to A\to B\to C$ :

\overrightarrow{EC}=\overrightarrow{EA}+\overrightarrow{AB}+\overrightarrow{BC} =-\mathbf w+\mathbf u+\mathbf v =\boxed{\mathbf u+\mathbf v-\mathbf w}.

If $\mathbf u=(4,0,0)$ , $\mathbf v=(0,3,0)$ , $\mathbf w=(0,0,5)$ , then $\overrightarrow{EC}=\boxed{(4,3,-5)}=4\mathbf i+3\mathbf j-5\mathbf k$ . The negative third component makes sense: $E$ is on the upper face and $C$ on the lower face.

Common mistake

The Trap

Possible slip	Check to use
Start minus end	Add the proposed vector to the start coordinates. Do you reach the end?
Adding the two position vectors	A displacement from $A$ to $B$ is $\mathbf b-\mathbf a$ , not $\mathbf a+\mathbf b$ .
Making all components positive	Check which coordinates decrease during the journey.
Using a backward arrow unchanged	Write both letter orders, for example $\overrightarrow{VW}=-\overrightarrow{WV}$ .

An incorrect result may fit one of these patterns. Inspect the route and the subtraction before deciding what went wrong.

Practice

Your Turn

Practice only — these are not your fresh-check questions. Show directed routes before doing arithmetic. Q7 is an original exam-style problem, not an official exam item.

$A=(-2,4,1)$ , $B=(3,-1,5)$ . Find $\overrightarrow{AB}$ and $\overrightarrow{BA}$ . Check the forward displacement.
$P=(4,-3,2)$ and $\overrightarrow{PQ}=(-6,5,1)$ . Find the coordinates of $Q$ .
$\overrightarrow{AB}=(2,-3,1)$ , $\overrightarrow{BC}=(-4,1,5)$ . Find $\overrightarrow{AC}$ , then $\overrightarrow{CA}$ .
$\overrightarrow{LM}=(3,1,-2)$ , $\overrightarrow{NM}=(-1,4,3)$ . Find $\overrightarrow{LN}$ , showing why one supplied vector must be reversed.
In a tetrahedron $OPQR$ , $\overrightarrow{OP}=\mathbf p$ , $\overrightarrow{OQ}=\mathbf q$ , $\overrightarrow{OR}=\mathbf r$ . Use routes through $O$ to express $\overrightarrow{QP}$ and $\overrightarrow{PR}$ . Add your answers and explain why their sum is $\overrightarrow{QR}$ .
Use the cuboid and $\mathbf u,\mathbf v,\mathbf w$ from example 3. Write a connected route and find (a) $\overrightarrow{EB}$ , (b) $\overrightarrow{BH}$ , (c) $\overrightarrow{GE}$ . Give vector expressions first, then component vectors for $\mathbf u=(4,0,0)$ , $\mathbf v=(0,3,0)$ , $\mathbf w=(0,0,5)$ .
Exam-style multi-step pathway. The same cuboid has $A=(1,-2,1)$ , with the three edge vectors from Q6. (a) Find the coordinates of $D,F,G$ . (b) Find $\overrightarrow{FD}$ along $F\to B\to A\to D$ , and express it using $\mathbf i,\mathbf j,\mathbf k$ . (c) A route goes $F\to D\to G$ . Find its resultant and verify it directly from $F,G$ . (d) Find the displacement needed to return from $G$ to $F$ , and check that the complete closed route sums to $\mathbf0$ .

Answers

Fully worked solutions

Show the answers

1

\overrightarrow{AB}=(3-(-2),\ -1-4,\ 5-1)=\boxed{(5,-5,4)}.

Reverse all signs: $\overrightarrow{BA}=\boxed{(-5,5,-4)}$ . Check: $(-2,4,1)+(5,-5,4)=(3,-1,5)=B$ .

2

End position equals start position plus displacement:

Q=(4,-3,2)+(-6,5,1)=\boxed{(-2,2,3)}.

Check: $Q-P=(-2-4,2-(-3),3-2)=(-6,5,1)$ .

3

Follow $A\to B\to C$ :

\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC} =(2-4,-3+1,1+5)=\boxed{(-2,-2,6)}.

$\overrightarrow{CA}=-\overrightarrow{AC}=\boxed{(2,2,-6)}$ .

4

The route is $L\to M\to N$ . The given $\overrightarrow{NM}$ points against the second leg, so $\overrightarrow{MN}=-\overrightarrow{NM}$ .

\overrightarrow{LN}=\overrightarrow{LM}-\overrightarrow{NM} =(3,1,-2)-(-1,4,3)=\boxed{(4,-3,-5)}.

Check the onward route: $\overrightarrow{LN}+\overrightarrow{NM}=(3,1,-2)=\overrightarrow{LM}$ .

5

$Q\to O\to P$ gives $\overrightarrow{QP}=-\mathbf q+\mathbf p=\boxed{\mathbf p-\mathbf q}$ .

$P\to O\to R$ gives $\overrightarrow{PR}=-\mathbf p+\mathbf r=\boxed{\mathbf r-\mathbf p}$ . Adding follows $Q\to P\to R$ :

(\mathbf p-\mathbf q)+(\mathbf r-\mathbf p)=\boxed{\mathbf r-\mathbf q}=\overrightarrow{QR}.

The intermediate $\mathbf p$ cancels, exactly as a route through $O$ would give $Q\to O\to R=-\mathbf q+\mathbf r$ .

6

(a) $E\to A\to B$ : $\overrightarrow{EB}=-\mathbf w+\mathbf u=\boxed{\mathbf u-\mathbf w}=\boxed{(4,0,-5)}$ .

(b) $B\to A\to D\to H$ : $\overrightarrow{BH}=-\mathbf u+\mathbf v+\mathbf w=\boxed{(-4,3,5)}$ .

(c) $G\to H\to E$ : $\overrightarrow{GH}=-\mathbf u$ , $\overrightarrow{HE}=-\mathbf v$ , so $\overrightarrow{GE}=\boxed{-\mathbf u-\mathbf v}=\boxed{(-4,-3,0)}$ . Both points are on the upper face, so the third component is zero.

7

(a) Starting at $A$ :

\begin{aligned} D&=A+\mathbf v=(1,-2,1)+(0,3,0)=\boxed{(1,1,1)},\\ F&=A+\mathbf u+\mathbf w=(1,-2,1)+(4,0,0)+(0,0,5)=\boxed{(5,-2,6)},\\ G&=A+\mathbf u+\mathbf v+\mathbf w=\boxed{(5,1,6)}. \end{aligned}

(b) Along the stated route:

\overrightarrow{FD}=\overrightarrow{FB}+\overrightarrow{BA}+\overrightarrow{AD} =-\mathbf w-\mathbf u+\mathbf v=(-4,3,-5) =\boxed{-4\mathbf i+3\mathbf j-5\mathbf k}.

Direct check: $D-F=(1-5,1-(-2),1-6)=(-4,3,-5)$ . (c) $D\to H\to G$ gives $\overrightarrow{DG}=\mathbf w+\mathbf u=(4,0,5)$ . Hence

\overrightarrow{FD}+\overrightarrow{DG}=(-4,3,-5)+(4,0,5)=\boxed{(0,3,0)}.

Directly, $\overrightarrow{FG}=G-F=(5-5,1-(-2),6-6)=(0,3,0)$ , agreeing with the route.

(d) Reverse that displacement: $\overrightarrow{GF}=\boxed{(0,-3,0)}$ . The complete route sums to $(-4,3,-5)+(4,0,5)+(0,-3,0)=(0,0,0)=\mathbf0$ .

Before moving on: can you write and explain the route before calculating? These questions practise directed displacement; success is not evidence of mastery of every vector topic.