Vectors and their operations in space

A classroom is a cuboid $5$ m long, $4$ m wide and $3$ m high. We want to stretch a straight decorative string from one floor corner to the opposite, farthest ceiling corner. In which direction does the string run, and how many metres long must it be?

In the plane we described "in which direction, how far" with vectors. This lesson takes vectors into three dimensions, and by the end we will find the length of the string using vector operations alone.

Vectors in space

Exploring: three journeys instead of one

Call the classroom the cuboid $ABCD.A'B'C'D'$ with $AB=5$ , $AD=4$ , $AA'=3$ (metres). An ant walks from $A$ to $C'$ along the edges: $5$ m along $AB$ , then $4$ m along $BC$ , then $3$ m up $CC'$ . The string goes straight from $A$ to $C'$ .

The two routes have the same start and the same end, so they give the same displacement. If we describe each piece of the route by an arrow, the arrow $AC'$ must be the "sum" of the three arrows $AB$ , $BC$ , $CC'$ . To say this precisely, we need vectors in space.

The cuboid

ABCD.A'B'C'D'

and the diagonal

AC'

Every idea is exactly as in the plane; the only difference is that the vectors may now lie in different planes.

Example Spotting equal vectors in a parallelepiped

Let $ABCD.A'B'C'D'$ be a parallelepiped. Name the vectors other than $\overrightarrow{AA'}$ that are equal to $\overrightarrow{AA'}$ and whose endpoints are vertices of the parallelepiped. Do the same for the vectors opposite to $\overrightarrow{AB}$ .

Solution. The four lateral edges $AA'$ , $BB'$ , $CC'$ , $DD'$ are parallel and equal. We take them from the bottom face to the top face so that they point the same way as $\overrightarrow{AA'}$ :

$\overrightarrow{AA'}=\overrightarrow{BB'}=\overrightarrow{CC'}=\overrightarrow{DD'}.$

The edges parallel to $AB$ are $DC$ , $A'B'$ , $D'C'$ . An opposite vector of $\overrightarrow{AB}$ must point the other way, that is "from right to left":

$\overrightarrow{BA},\quad \overrightarrow{CD},\quad \overrightarrow{B'A'},\quad \overrightarrow{C'D'}.$

So there are $3$ other vectors equal to $\overrightarrow{AA'}$ and $4$ vectors opposite to $\overrightarrow{AB}$ .

Sum and difference of two vectors

The sum of two vectors is defined as in the plane: place them tip to tail and join the first initial point to the last terminal point. Two vectors drawn from a common point always lie in one plane, so all the old rules still hold.

Why is the parallelepiped rule true? The face $ABCD$ is a parallelogram, so $\overrightarrow{AB}+\overrightarrow{AD}=\overrightarrow{AC}$ . The section $ACC'A'$ is also a parallelogram, so $\overrightarrow{AC}+\overrightarrow{AA'}=\overrightarrow{AC'}$ . Putting the two together finishes the proof. This is exactly the ant's route: $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CC'}$ , because $\overrightarrow{BC}=\overrightarrow{AD}$ and $\overrightarrow{CC'}=\overrightarrow{AA'}$ .

The next example shows how to "replace a vector by an equal one" so that vectors join tip to tail.

Example Simplifying a vector sum

Let $ABCD.A'B'C'D'$ be a parallelepiped. Simplify

a) $\overrightarrow{AB}+\overrightarrow{B'C'}+\overrightarrow{DD'}$ ; b) $\overrightarrow{AD'}-\overrightarrow{AB}$ .

Solution. a) We replace the last two vectors by equal vectors that join on to $\overrightarrow{AB}$ : $\overrightarrow{B'C'}=\overrightarrow{BC}$ , $\overrightarrow{DD'}=\overrightarrow{CC'}$ . Then

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CC'}=\overrightarrow{AC'}.$

b) The two vectors share the initial point $A$ , so by the subtraction rule the difference goes from the terminal point of $\overrightarrow{AB}$ to the terminal point of $\overrightarrow{AD'}$ :

$\overrightarrow{AD'}-\overrightarrow{AB}=\overrightarrow{BD'}.$

So a) equals $\overrightarrow{AC'}$ and b) equals $\overrightarrow{BD'}$ .

Multiplying a vector by a number

For a number $k\neq0$ and a vector $\vec{a}\neq\vec{0}$ , the vector $k\vec{a}$ has the same direction as $\vec{a}$ if $k>0$ , the opposite direction if $k<0$ , and length $|k|\cdot|\vec{a}|$ . The familiar properties still hold; the two used most often are:

If $I$ is the midpoint of $AB$ then for every point $M$ : $\overrightarrow{MA}+\overrightarrow{MB}=2\overrightarrow{MI}$ .
If $G$ is the centroid of triangle $ABC$ then $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\vec{0}$ , and for every $M$ : $\overrightarrow{MA}+\overrightarrow{MB}+\overrightarrow{MC}=3\overrightarrow{MG}$ .

In space the points need not lie in one plane, so we often "take a detour" through points we know.

Example The segment joining midpoints of opposite edges

Let $ABCD$ be a tetrahedron and let $M$ , $N$ be the midpoints of $AB$ and $CD$ . Prove that $\overrightarrow{MN}=\frac12\left(\overrightarrow{AD}+\overrightarrow{BC}\right).$

Solution. We go from $M$ to $N$ by two different detours and add them. Through $A$ and $D$ , then through $B$ and $C$ :

$\begin{aligned} \overrightarrow{MN}&=\overrightarrow{MA}+\overrightarrow{AD}+\overrightarrow{DN},\\ \overrightarrow{MN}&=\overrightarrow{MB}+\overrightarrow{BC}+\overrightarrow{CN}. \end{aligned}$

Add the two equalities. Since $M$ is the midpoint of $AB$ , $\overrightarrow{MA}+\overrightarrow{MB}=\vec{0}$ ; likewise $\overrightarrow{DN}+\overrightarrow{CN}=\vec{0}$ :

$2\overrightarrow{MN}=\overrightarrow{AD}+\overrightarrow{BC}.$

Dividing both sides by $2$ gives the result.

The tetrahedron

ABCD

and the segment

MN

The scalar product of two vectors

The angle between two vectors

To measure the angle between $\vec{a}$ and $\vec{b}$ (both non-zero), take any point $O$ and draw $\overrightarrow{OA}=\vec{a}$ , $\overrightarrow{OB}=\vec{b}$ . Then $(\vec{a}, \vec{b})=\angle AOB$ , an angle from $0^\circ$ to $180^\circ$ . The key point is that the two vectors must start at the same point. For instance in a parallelepiped, $\overrightarrow{A'C'}=\overrightarrow{AC}$ , so $(\overrightarrow{AB}, \overrightarrow{A'C'})=(\overrightarrow{AB}, \overrightarrow{AC})=\angle BAC$ .

The scalar product has the same properties as in the plane: it is commutative, distributes over addition, and $\vec{a}^2=|\vec{a}|^2,\qquad \vec{a}\perp\vec{b}\iff\vec{a}\cdot\vec{b}=0,\qquad \cos(\vec{a}, \vec{b})=\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|\cdot|\vec{b}|}.$

Now we answer the opening question. The edges $AB$ , $AD$ , $AA'$ of a cuboid are pairwise perpendicular, so the scalar product of any two different edges is $0$ .

Example The length of the decorative string

The cuboid $ABCD.A'B'C'D'$ has $AB=5$ , $AD=4$ , $AA'=3$ (metres). Find the length $AC'$ .

Solution. By the parallelepiped rule, $\overrightarrow{AC'}=\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA'}$ . Square both sides (scalar product with itself) and expand as for the square of a sum of three numbers:

$AC'^2=AB^2+AD^2+AA'^2+2\left(\overrightarrow{AB}\cdot\overrightarrow{AD}+\overrightarrow{AD}\cdot\overrightarrow{AA'}+\overrightarrow{AA'}\cdot\overrightarrow{AB}\right).$

The three edges are pairwise perpendicular, so the three scalar products in the bracket are all $0$ :

$AC'^2=25+16+9=50,\qquad AC'=\sqrt{50}=5\sqrt2\approx7.07.$

So the string is about $7.07$ m long.

This method works even when the edges are not perpendicular, as long as we know the scalar products. The last example uses it to find an angle in a cube.

Example The angle between two face diagonals

Let $ABCD.A'B'C'D'$ be a cube with edge $a$ . Find the angle $(\overrightarrow{AC}, \overrightarrow{A'D})$ .

Solution. We write both vectors in terms of the three edges $\overrightarrow{AB}$ , $\overrightarrow{AD}$ , $\overrightarrow{AA'}$ (pairwise perpendicular, each of length $a$ ):

$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD},\qquad \overrightarrow{A'D}=\overrightarrow{AD}-\overrightarrow{AA'}.$

Expand; only $\overrightarrow{AD}\cdot\overrightarrow{AD}=a^2$ survives, because the scalar products of different edges are $0$ :

$\overrightarrow{AC}\cdot\overrightarrow{A'D}=\overrightarrow{AB}\cdot\overrightarrow{AD}-\overrightarrow{AB}\cdot\overrightarrow{AA'}+\overrightarrow{AD}^2-\overrightarrow{AD}\cdot\overrightarrow{AA'}=a^2.$

Both vectors are diagonals of a square of side $a$ , so $AC=A'D=a\sqrt2$ :

$\cos(\overrightarrow{AC}, \overrightarrow{A'D})=\frac{a^2}{a\sqrt2\cdot a\sqrt2}=\frac12.$

So $(\overrightarrow{AC}, \overrightarrow{A'D})=60^\circ$ .

Check. Since $\overrightarrow{AC}=\overrightarrow{A'C'}$ , the angle is $\angle C'A'D$ . Triangle $A'C'D$ has three sides that are face diagonals of length $a\sqrt2$ , so it is equilateral and that angle is $60^\circ$ .

Watch out Common mistakes

Adding lengths instead of vectors. Wrong: $|\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA'}|=5+4+3=12$ . Why it is wrong: that is the distance the ant walks, not the displacement. Right: $|\overrightarrow{AC'}|=\sqrt{50}\approx7.07$ .
Measuring an angle before the vectors share a starting point. Wrong: in a square $ABCD$ , writing $(\overrightarrow{AB}, \overrightarrow{BD})=\angle ABD=45^\circ$ . Why it is wrong: $\angle ABD$ is the angle between $\overrightarrow{BA}$ and $\overrightarrow{BD}$ , and $\overrightarrow{BA}=-\overrightarrow{AB}$ . Right: $(\overrightarrow{AB}, \overrightarrow{BD})=180^\circ-45^\circ=135^\circ$ .
One edge in the wrong direction in the parallelepiped rule. Wrong: $\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{A'A}=\overrightarrow{AC'}$ . Why it is wrong: $\overrightarrow{A'A}=-\overrightarrow{AA'}$ points down, not up. Right: $\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{A'A}=\overrightarrow{AC}-\overrightarrow{AA'}=\overrightarrow{A'C}$ .
Treating the scalar product as a vector. $\vec{a}\cdot\vec{b}$ is a number; writing $\overrightarrow{AB}\cdot\overrightarrow{AD}=\overrightarrow{AC}$ is wrong.

Key terms

English	Tiếng Việt
vector in space	vectơ trong không gian
equal vectors	hai vectơ bằng nhau
opposite vectors	hai vectơ đối nhau
zero vector	vectơ-không
parallelepiped rule	quy tắc hình hộp
tetrahedron	tứ diện
coplanar vectors	ba vectơ đồng phẳng
angle between two vectors	góc giữa hai vectơ
scalar product (dot product)	tích vô hướng

Summary

Vectors, length, equal and opposite vectors in space are defined as in the plane.
The triangle and parallelogram rules still hold; add the parallelepiped rule $\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA'}=\overrightarrow{AC'}$ .
Midpoint $I$ of $AB$ : $\overrightarrow{MA}+\overrightarrow{MB}=2\overrightarrow{MI}$ ; centroid $G$ : $\overrightarrow{GA}+\overrightarrow{GB}+\overrightarrow{GC}=\vec{0}$ .
Angle between two vectors: draw them from a common point; the angle is from $0^\circ$ to $180^\circ$ .
$\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos(\vec{a}, \vec{b})$ ; $\vec{a}^2=|\vec{a}|^2$ ; $\vec{a}\perp\vec{b}\iff\vec{a}\cdot\vec{b}=0$ .
To find a length: write the vector in terms of known edges, then square it.

Practice

Each exercise comes with what you need and some steps to start. Try it yourself first, then open the solution.

Exercise 1

Let $ABCD.A'B'C'D'$ be a parallelepiped. Simplify

a) $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD'}$ ; b) $\overrightarrow{AB}+\overrightarrow{A'D'}+\overrightarrow{CC'}$ .

What you need and how to start

What you need: the triangle rule; the parallelepiped rule; parallel edges in the same direction give equal vectors.

Steps:

a) The vectors already join tip to tail.
b) Replace $\overrightarrow{A'D'}=\overrightarrow{AD}$ and $\overrightarrow{CC'}=\overrightarrow{AA'}$ .

Show solution

a) $\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD'}=\overrightarrow{AC}+\overrightarrow{CD'}=\overrightarrow{AD'}$ .

b) $\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA'}=\overrightarrow{AC'}$ by the parallelepiped rule.

Answer. a) $\overrightarrow{AD'}$ ; b) $\overrightarrow{AC'}$ .

Exercise 2

Let $ABCD$ be a tetrahedron. Simplify $\overrightarrow{AB}-\overrightarrow{AC}+\overrightarrow{DC}-\overrightarrow{DB}$ .

What you need and how to start

What you need: the subtraction rule $\overrightarrow{OX}-\overrightarrow{OY}=\overrightarrow{YX}$ .

Steps:

Combine $\overrightarrow{AB}-\overrightarrow{AC}$ .
Combine $\overrightarrow{DC}-\overrightarrow{DB}$ .

Show solution

$\overrightarrow{AB}-\overrightarrow{AC}=\overrightarrow{CB}$ and $\overrightarrow{DC}-\overrightarrow{DB}=\overrightarrow{BC}$ . These two vectors are opposite, so the sum is $\vec{0}$ .

Answer. $\vec{0}$ .

Exercise 3

Let $ABCD.A'B'C'D'$ be a cube with edge $3$ . Find $\overrightarrow{AB}\cdot\overrightarrow{AC}$ , $\overrightarrow{AB}\cdot\overrightarrow{A'C'}$ and $\overrightarrow{AA'}\cdot\overrightarrow{BD}$ .

What you need and how to start

What you need: $\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos(\vec{a}, \vec{b})$ ; draw both vectors from a common point before measuring the angle.

Steps:

$AC=3\sqrt2$ and $\angle BAC=45^\circ$ .
$\overrightarrow{A'C'}=\overrightarrow{AC}$ .
$AA'$ is perpendicular to the base, so it is perpendicular to $BD$ .

Show solution

$\overrightarrow{AB}\cdot\overrightarrow{AC}=3\cdot3\sqrt2\cdot\cos45^\circ=3\cdot3\sqrt2\cdot\frac{\sqrt2}{2}=9.$ Since $\overrightarrow{A'C'}=\overrightarrow{AC}$ , the second product is also $9$ . Since $AA'\perp BD$ , the third product is $0$ .

Answer. $9$ ; $9$ ; $0$ .

Exercise 4

Let $ABCD$ be a regular tetrahedron with edge $2$ (all edges equal $2$ , all faces are equilateral triangles). Find $\overrightarrow{AB}\cdot\overrightarrow{AC}$ , then prove that $AB\perp CD$ .

What you need and how to start

What you need: the angles of an equilateral triangle are $60^\circ$ ; $\overrightarrow{CD}=\overrightarrow{AD}-\overrightarrow{AC}$ .

Steps:

Find $\overrightarrow{AB}\cdot\overrightarrow{AC}$ and $\overrightarrow{AB}\cdot\overrightarrow{AD}$ .
Find $\overrightarrow{AB}\cdot\overrightarrow{CD}$ by replacing $\overrightarrow{CD}$ .

Show solution

$\overrightarrow{AB}\cdot\overrightarrow{AC}=2\cdot2\cdot\cos60^\circ=2$ , and likewise $\overrightarrow{AB}\cdot\overrightarrow{AD}=2$ . Hence $\overrightarrow{AB}\cdot\overrightarrow{CD}=\overrightarrow{AB}\cdot\left(\overrightarrow{AD}-\overrightarrow{AC}\right)=2-2=0,$ so $AB\perp CD$ .

Answer. $\overrightarrow{AB}\cdot\overrightarrow{AC}=2$ ; $\overrightarrow{AB}\cdot\overrightarrow{CD}=0$ , so $AB\perp CD$ .

Exercise 5

Three people pull a crate standing in the corner of a warehouse. The three pulling forces are pairwise perpendicular (along the two walls and straight up), with magnitudes $30$ N, $40$ N and $120$ N. Find the magnitude of the resultant force.

What you need and how to start

What you need: the resultant is the sum of the three force vectors; for three pairwise perpendicular vectors, $|\vec{a}+\vec{b}+\vec{c}|^2=|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2$ .

Steps:

Square the sum of the three forces.
The scalar products of different forces are $0$ .

Show solution

$|\vec{F}|^2=30^2+40^2+120^2=900+1600+14400=16900,\qquad |\vec{F}|=130.$

Answer. $130$ N.

Exercise 6

Let $ABCD.A'B'C'D'$ be a cube with edge $1$ . Is each claim true or false?

a) $\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA'}=\overrightarrow{AC'}$ .

b) $AC'=\sqrt3$ .

c) $\overrightarrow{AB}\cdot\overrightarrow{CD'}=1$ .

d) The angle between the vectors $\overrightarrow{AC}$ and $\overrightarrow{B'D'}$ is $90^\circ$ .

What you need and how to start

What you need: the parallelepiped rule; the three edges at $A$ are pairwise perpendicular with length $1$ ; write every vector in terms of $\overrightarrow{AB}$ , $\overrightarrow{AD}$ , $\overrightarrow{AA'}$ .

Steps:

b) Square $\overrightarrow{AC'}$ .
c) $\overrightarrow{CD'}=\overrightarrow{CD}+\overrightarrow{DD'}=-\overrightarrow{AB}+\overrightarrow{AA'}$ .
d) $\overrightarrow{B'D'}=\overrightarrow{BD}$ ; the diagonals of a square are perpendicular.

Show solution

a) True. This is the parallelepiped rule.

b) True. $AC'^2=1+1+1=3$ .

c) False. $\overrightarrow{AB}\cdot\left(-\overrightarrow{AB}+\overrightarrow{AA'}\right)=-1+0=-1$ .

d) True. $\overrightarrow{B'D'}=\overrightarrow{BD}$ and $AC\perp BD$ (the diagonals of the square $ABCD$ ).

Answer. a) True; b) True; c) False; d) True.

Exercise 7

A storeroom is a cuboid $ABCD.A'B'C'D'$ with $AB=4$ m, $AD=3$ m, $AA'=4$ m. Find the angle between the vectors $\overrightarrow{AC'}$ and $\overrightarrow{AC}$ (in degrees, rounded to the nearest whole number).

What you need and how to start

What you need: $\cos(\vec{a}, \vec{b})=\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$ ; the parallelepiped rule.

Steps:

Write $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{AD}$ and $\overrightarrow{AC'}=\overrightarrow{AB}+\overrightarrow{AD}+\overrightarrow{AA'}$ .
Find the scalar product and the two lengths.

Show solution

The scalar products of different edges are $0$ , so $\overrightarrow{AC'}\cdot\overrightarrow{AC}=AB^2+AD^2=16+9=25,\qquad AC=5,\qquad AC'=\sqrt{16+9+16}=\sqrt{41}.$ $\cos(\overrightarrow{AC'}, \overrightarrow{AC})=\frac{25}{5\sqrt{41}}=\frac{5}{\sqrt{41}}\approx0.781,\quad\text{angle}\approx38.7^\circ.$

Answer. $39$ .

Exercise 8

(Revisiting Multiplying a vector by a number.) Let $ABCD$ be a tetrahedron and $G$ the centroid of triangle $BCD$ . Prove that $\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}=3\overrightarrow{AG}$ .

What you need and how to start

What you need: $G$ is the centroid of triangle $BCD$ $\iff$ $\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}=\vec{0}$ (grade 10).

Steps:

Insert the point $G$ : $\overrightarrow{AB}=\overrightarrow{AG}+\overrightarrow{GB}$ , and so on.
Add the three equalities.

Show solution

$\overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD}=3\overrightarrow{AG}+\left(\overrightarrow{GB}+\overrightarrow{GC}+\overrightarrow{GD}\right)=3\overrightarrow{AG}+\vec{0}=3\overrightarrow{AG}.$

Answer. The identity is proved.

Exercise 9 Challenge

a) Prove that for any four points $A$ , $B$ , $C$ , $D$ : $\overrightarrow{AB}\cdot\overrightarrow{CD}+\overrightarrow{AC}\cdot\overrightarrow{DB}+\overrightarrow{AD}\cdot\overrightarrow{BC}=0.$ b) A tetrahedron $ABCD$ has $AB\perp CD$ and $AC\perp BD$ . Prove that $AD\perp BC$ .

What you need and how to start

What you need: insert the point $A$ : $\overrightarrow{CD}=\overrightarrow{AD}-\overrightarrow{AC}$ ; the scalar product distributes over addition.

Steps:

Put $\vec{b}=\overrightarrow{AB}$ , $\vec{c}=\overrightarrow{AC}$ , $\vec{d}=\overrightarrow{AD}$ and write every vector in terms of them.
Expand; the terms cancel in pairs.
b) The first two products are $0$ .

Show solution

a) With $\overrightarrow{CD}=\vec{d}-\vec{c}$ , $\overrightarrow{DB}=\vec{b}-\vec{d}$ , $\overrightarrow{BC}=\vec{c}-\vec{b}$ : $\vec{b}\cdot(\vec{d}-\vec{c})+\vec{c}\cdot(\vec{b}-\vec{d})+\vec{d}\cdot(\vec{c}-\vec{b})=\vec{b}\cdot\vec{d}-\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{b}-\vec{c}\cdot\vec{d}+\vec{d}\cdot\vec{c}-\vec{d}\cdot\vec{b}=0.$ b) The first two products are $0$ by hypothesis, so $\overrightarrow{AD}\cdot\overrightarrow{BC}=0$ , that is $AD\perp BC$ .

Answer. a) The identity holds for any four points; b) $AD\perp BC$ .

Vectors in space

Exploring: three journeys instead of one

Sum and difference of two vectors

Multiplying a vector by a number

The scalar product of two vectors

The angle between two vectors

Key terms

Summary

Practice

Mini test

Part I. Multiple choice

Part II. True or false

Part III. Short answer

a) $\overrightarrow{AB}\cdot\overrightarrow{AA'}=0$ .	True	False
b) $AC'=7$ .	True	False
c) $\overrightarrow{AC'}\cdot\overrightarrow{AB}=2$ .	True	False
d) The angle between the vectors $\overrightarrow{AC'}$ and $\overrightarrow{AA'}$ is greater than $35^\circ$ .	True	False