Vectors

What you'll learn

How to write 3D vectors using $i\mathbf{i}$ , $j\mathbf{j}$ and $k\mathbf{k}$ .
How to find vectors between points, lengths and midpoints.
How to add vectors and find resultants.
How to prove lines are parallel and identify simple shapes using vectors.

1. Vectors as components

A vector describes a movement: how far, and in which direction. In 3D, we usually split that movement into three perpendicular directions: the $x$ -, $y$ - and $z$ -directions.

For example, $3i−2j+5k3\mathbf{i}-2\mathbf{j}+5\mathbf{k}$ means:

A 3D component diagram showing the vector made from 3 units in the i direction, 2 units in the negative j direction and 5 units in the k direction.

3 units in the $i\mathbf{i}$ direction,
2 units in the negative $j\mathbf{j}$ direction,
5 units in the $k\mathbf{k}$ direction.

The same vector can also be written as a column vector:

(3−25)\begin{pmatrix} 3\\ -2\\ 5 \end{pmatrix}

Definition

Vector, scalar and components

A vector has size and direction.
A scalar has size only.
A unit vector has length one; $i\mathbf{i}$ , $j\mathbf{j}$ and $k\mathbf{k}$ are unit vectors in the positive $x$ , $y$ and $z$ directions.
A component tells you how far the vector goes in one coordinate direction.

Key Idea

Work component by component

To add or subtract vectors, deal with the $i\mathbf{i}$ , $j\mathbf{j}$ and $k\mathbf{k}$ components separately. The resultant is the single vector found by adding several vectors together.

Example

Finding a resultant force

Three forces act on a particle:

F1=−2i+5j+3k,F2=6i−j−4k,F3=i+2j+7k\mathbf{F}_1=-2\mathbf{i}+5\mathbf{j}+3\mathbf{k},\quad \mathbf{F}_2=6\mathbf{i}-\mathbf{j}-4\mathbf{k},\quad \mathbf{F}_3=\mathbf{i}+2\mathbf{j}+7\mathbf{k}

Find the resultant force.

Add the three vectors component by component:
$R=(−2i+5j+3k)+(6i−j−4k)+(i+2j+7k)=(−2+6+1)i+(5−1+2)j+(3−4+7)k=5i+6j+6k\begin{aligned} \mathbf{R} &=(-2\mathbf{i}+5\mathbf{j}+3\mathbf{k})+(6\mathbf{i}-\mathbf{j}-4\mathbf{k})+(\mathbf{i}+2\mathbf{j}+7\mathbf{k})\\ &=(-2+6+1)\mathbf{i}+(5-1+2)\mathbf{j}+(3-4+7)\mathbf{k}\\ &=5\mathbf{i}+6\mathbf{j}+6\mathbf{k} \end{aligned}$
The resultant force is $5i+6j+6k5\mathbf{i}+6\mathbf{j}+6\mathbf{k}$ .

2. Position vectors, displacement and length

A position vector tells you where a point is relative to a fixed origin, usually called $O$ .

If point $A$ has position vector $a\mathbf{a}$ , then $OA⃗=a\vec{OA}=\mathbf{a}$ .

Position vectors are measured from the origin, while displacement vectors go from one point to another.

A displacement vector tells you how to move from one point to another. For example, $AB⃗\vec{AB}$ means “the vector from $A$ to $B$ ”.

The vector from A to B points from the tail at A to the head at B, and reversing the order reverses the direction.

Definition

Position vector and magnitude

If $A$ has position vector $a\mathbf{a}$ and $B$ has position vector $b\mathbf{b}$ , then $AB⃗=b−a\vec{AB}=\mathbf{b}-\mathbf{a}$ .
The magnitude of a vector is its length. For $v=xi+yj+zk\mathbf{v}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ ,
$∣v∣=x2+y2+z2|\mathbf{v}|=\sqrt{x^2+y^2+z^2}$

Key Idea

Head minus tail

To find $AB⃗\vec{AB}$ , subtract the position vector of the starting point $A$ from the position vector of the finishing point $B$ .

Example

Finding a vector and its length

Point $A$ has position vector $2i−3j+4k2\mathbf{i}-3\mathbf{j}+4\mathbf{k}$ and point $B$ has position vector $−5i+j+6k-5\mathbf{i}+\mathbf{j}+6\mathbf{k}$ . Find $AB⃗\vec{AB}$ and $∣AB⃗∣|\vec{AB}|$ .

Use head minus tail:
$AB⃗=b−a\vec{AB}=\mathbf{b}-\mathbf{a}$
Subtract the components:
$AB⃗=(−5i+j+6k)−(2i−3j+4k)=−7i+4j+2k\begin{aligned} \vec{AB} &=(-5\mathbf{i}+\mathbf{j}+6\mathbf{k})-(2\mathbf{i}-3\mathbf{j}+4\mathbf{k})\\ &=-7\mathbf{i}+4\mathbf{j}+2\mathbf{k} \end{aligned}$
Use 3D Pythagoras for the length:
$∣AB⃗∣=(−7)2+42+22=69|\vec{AB}|=\sqrt{(-7)^2+4^2+2^2}=\sqrt{69}$

Common Mistake

Subtracting the wrong way round

If you calculate $a−b\mathbf{a}-\mathbf{b}$ , you have found $BA⃗\vec{BA}$ instead of $AB⃗\vec{AB}$ . The length is the same, but the direction is reversed.

3. Unknown components from a magnitude

Sometimes you are given the length of a vector and asked to find an unknown component. The key move is to use the magnitude formula, then square both sides.

Example

Finding possible values of an unknown component

Given that $∣2i+mj−3k∣=7|2\mathbf{i}+m\mathbf{j}-3\mathbf{k}|=7$ , find the possible values of $m$ .

Apply the magnitude formula:
$22+m2+(−3)2=7\sqrt{2^2+m^2+(-3)^2}=7$
Square both sides:
$2^2+m^2+(-3)^2=7^2$
Simplify:
$m^2+13=49$
Solve for $m$ :
$m2=36⇒m=±6m^2=36\Rightarrow m=\pm 6$
The possible values are $m = 6$ and $m = - 6$ .

Tip

Expect two signs

A magnitude only measures length, not direction. So a positive component and the matching negative component can give the same length.

4. Routes, midpoints and extending a line

Vectors can be chained together like a journey. If you go from $A$ to $B$ , then from $B$ to $C$ , the total journey is from $A$ to $C$ :

Vector addition can be shown as a journey where AB followed by BC gives the resultant AC.

AB⃗+BC⃗=AC⃗\vec{AB}+\vec{BC}=\vec{AC}

A midpoint is the point halfway between two points. If $A$ has position vector $a\mathbf{a}$ and $B$ has position vector $b\mathbf{b}$ , then the midpoint has position vector

a+b2\frac{\mathbf{a}+\mathbf{b}}{2}

Example

Using a route in a triangle

In triangle $A B C$ , suppose $AB⃗=5i+2j−k\vec{AB}=5\mathbf{i}+2\mathbf{j}-\mathbf{k}$ and $AC⃗=7i−4j+3k\vec{AC}=7\mathbf{i}-4\mathbf{j}+3\mathbf{k}$ . Find $BC⃗\vec{BC}$ and the length of $A B$ .

Use the route from $A$ to $C$ through $B$ :
$AB⃗+BC⃗=AC⃗\vec{AB}+\vec{BC}=\vec{AC}$
Rearrange to find $BC⃗\vec{BC}$ :
$BC⃗=AC⃗−AB⃗=(7i−4j+3k)−(5i+2j−k)=2i−6j+4k\begin{aligned} \vec{BC} &=\vec{AC}-\vec{AB}\\ &=(7\mathbf{i}-4\mathbf{j}+3\mathbf{k})-(5\mathbf{i}+2\mathbf{j}-\mathbf{k})\\ &=2\mathbf{i}-6\mathbf{j}+4\mathbf{k} \end{aligned}$
Find the length of $A B$ :
$∣AB⃗∣=52+22+(−1)2=30|\vec{AB}|=\sqrt{5^2+2^2+(-1)^2}=\sqrt{30}$

Example

Extending a line segment

Point $A$ has position vector $i+2j−k\mathbf{i}+2\mathbf{j}-\mathbf{k}$ and point $B$ has position vector $4i−j+5k4\mathbf{i}-\mathbf{j}+5\mathbf{k}$ . Point $D$ is such that $AB⃗=BD⃗\vec{AB}=\vec{BD}$ . Find the position vector of $D$ .

Extending a line segment with equal vectors places D so that B is halfway between A and D.

First find $AB⃗\vec{AB}$ :
$AB⃗=(4i−j+5k)−(i+2j−k)=3i−3j+6k\begin{aligned} \vec{AB} &=(4\mathbf{i}-\mathbf{j}+5\mathbf{k})-(\mathbf{i}+2\mathbf{j}-\mathbf{k})\\ &=3\mathbf{i}-3\mathbf{j}+6\mathbf{k} \end{aligned}$
Since $BD⃗\vec{BD}$ is the same vector, add it to the position vector of $B$ :
$d=(4i−j+5k)+(3i−3j+6k)=7i−4j+11k\begin{aligned} \mathbf{d} &=(4\mathbf{i}-\mathbf{j}+5\mathbf{k})+(3\mathbf{i}-3\mathbf{j}+6\mathbf{k})\\ &=7\mathbf{i}-4\mathbf{j}+11\mathbf{k} \end{aligned}$
So the position vector of $D$ is $7i−4j+11k7\mathbf{i}-4\mathbf{j}+11\mathbf{k}$ .

5. Parallel vectors and shapes

Two vectors are parallel when they point in the same or exactly opposite direction. In component form, this means one vector is a multiple of the other.

Parallel vectors have the same or opposite direction because one is a scalar multiple of the other.

Definition

Parallel vectors

Two non-zero vectors $u\mathbf{u}$ and $v\mathbf{v}$ are parallel if there is a scalar $λ\lambda$ such that $u=λv\mathbf{u}=\lambda\mathbf{v}$ . This means every component is multiplied by the same number.

For coordinate shape questions, calculate side vectors. A quadrilateral is a four-sided shape. A parallelogram has both pairs of opposite sides parallel. A trapezium has one pair of opposite sides parallel.

Side vectors reveal whether opposite sides of a quadrilateral are parallel.

Example

Showing a quadrilateral is a trapezium

Relative to origin $O$ , points $A$ , $B$ and $C$ have position vectors

2i+j−3k,5i−j−2k,6i−4j+2k2\mathbf{i}+\mathbf{j}-3\mathbf{k},\quad 5\mathbf{i}-\mathbf{j}-2\mathbf{k},\quad 6\mathbf{i}-4\mathbf{j}+2\mathbf{k}

respectively. Show that $O A B C$ is a trapezium.

Find $AB⃗\vec{AB}$ :
$AB⃗=(5i−j−2k)−(2i+j−3k)=3i−2j+k\begin{aligned} \vec{AB} &=(5\mathbf{i}-\mathbf{j}-2\mathbf{k})-(2\mathbf{i}+\mathbf{j}-3\mathbf{k})\\ &=3\mathbf{i}-2\mathbf{j}+\mathbf{k} \end{aligned}$
Compare this with $OC⃗\vec{OC}$ :
$OC⃗=6i−4j+2k=2(3i−2j+k)=2AB⃗\vec{OC}=6\mathbf{i}-4\mathbf{j}+2\mathbf{k}=2(3\mathbf{i}-2\mathbf{j}+\mathbf{k})=2\vec{AB}$
So $A B$ is parallel to $O C$ , meaning one pair of opposite sides is parallel.
Check the other pair:
$BC⃗=(6i−4j+2k)−(5i−j−2k)=i−3j+4k\vec{BC}=(6\mathbf{i}-4\mathbf{j}+2\mathbf{k})-(5\mathbf{i}-\mathbf{j}-2\mathbf{k})=\mathbf{i}-3\mathbf{j}+4\mathbf{k}$
$BC⃗\vec{BC}$ is not a scalar multiple of $OA⃗=2i+j−3k\vec{OA}=2\mathbf{i}+\mathbf{j}-3\mathbf{k}$ , so exactly one pair of opposite sides is parallel. Therefore $O A B C$ is a trapezium.

6. A harder mixed geometry problem

Some vector questions combine several ideas: equal lengths, midpoints and area. The trick is to turn each geometric statement into an equation.

If a triangle is isosceles with $A B = A C$ , then the line from $A$ to the midpoint of $B C$ is the perpendicular height.

In an isosceles triangle, the line from the apex to the midpoint of the base is the height used in the area formula.

Example

Using equal lengths and area

Point $A$ has position vector $(ab3)\begin{pmatrix} a\\ b\\ 3 \end{pmatrix}$ , where $a$ and $b$ are positive constants. Points $B$ and $C$ have position vectors

(324)and(102)\begin{pmatrix} 3\\2\\4 \end{pmatrix} \quad\text{and}\quad \begin{pmatrix} 1\\0\\2 \end{pmatrix}

respectively. Given that $A B = A C$ and the area of triangle $A B C$ is $6\sqrt{6}$ , find $a$ and $b$ .

Compare squared lengths to avoid square roots:
$AB2=(3−a)2+(2−b)2+1AC2=(1−a)2+b2+1\begin{aligned} AB^2&=(3-a)^2+(2-b)^2+1\\ AC^2&=(1-a)^2+b^2+1 \end{aligned}$
Since $A B = A C$ , set these equal and simplify:
$(3−a)2+(2−b)2+1=(1−a)2+b2+1⇒a+b=3(3-a)^2+(2-b)^2+1=(1-a)^2+b^2+1\Rightarrow a+b=3$
Find the midpoint $M$ of $B C$ :
$m=12(3+12+04+2)=(213)\mathbf{m}=\frac{1}{2}\begin{pmatrix}3+1\\2+0\\4+2\end{pmatrix}=\begin{pmatrix}2\\1\\3\end{pmatrix}$
Use $b = 3 - a$ to find the base and height:
$∣BC⃗∣=(−2)2+(−2)2+(−2)2=23∣AM⃗∣=(a−2)2+(b−1)2=2∣a−2∣\begin{aligned} |\vec{BC}|&=\sqrt{(-2)^2+(-2)^2+(-2)^2}=2\sqrt{3}\\ |\vec{AM}|&=\sqrt{(a-2)^2+(b-1)^2}=\sqrt{2}|a-2| \end{aligned}$
Apply the area formula:
$6=12(23)(2∣a−2∣)=6∣a−2∣\sqrt{6}=\frac{1}{2}(2\sqrt{3})(\sqrt{2}|a-2|)=\sqrt{6}|a-2|$
Hence $∣ a - 2∣ = 1$ . The candidates are $a = 1, b = 2$ or $a = 3, b = 0$ ; because $a$ and $b$ are positive, $a = 1$ and $b = 2$ .

Exam technique

In the exam

For $AB⃗\vec{AB}$ , always use head minus tail: position vector of $B$ minus position vector of $A$ .
For shape proofs, calculate the relevant side vectors and state the scalar multiple clearly.
For length conditions, square both sides early and remember that unknown components may have two signs.

Self review

Check yourself

If $A$ and $B$ have position vectors $a\mathbf{a}$ and $b\mathbf{b}$ , what is $BA⃗\vec{BA}$ ?
How can you tell from components that two non-zero vectors are parallel?
Why can a magnitude equation give two possible values for an unknown component?

Vectors

What you'll learn

How to write 3D vectors using $i\mathbf{i}$ , $j\mathbf{j}$ and $k\mathbf{k}$ .
How to find vectors between points, lengths and midpoints.
How to add vectors and find resultants.
How to prove lines are parallel and identify simple shapes using vectors.

1. Vectors as components

A vector describes a movement: how far, and in which direction. In 3D, we usually split that movement into three perpendicular directions: the $x$ -, $y$ - and $z$ -directions.

For example, $3i−2j+5k3\mathbf{i}-2\mathbf{j}+5\mathbf{k}$ means:

A 3D component diagram showing the vector made from 3 units in the i direction, 2 units in the negative j direction and 5 units in the k direction.

3 units in the $i\mathbf{i}$ direction,
2 units in the negative $j\mathbf{j}$ direction,
5 units in the $k\mathbf{k}$ direction.

The same vector can also be written as a column vector:

(3−25)\begin{pmatrix} 3\\ -2\\ 5 \end{pmatrix}

Definition

Vector, scalar and components

A vector has size and direction.
A scalar has size only.
A unit vector has length one; $i\mathbf{i}$ , $j\mathbf{j}$ and $k\mathbf{k}$ are unit vectors in the positive $x$ , $y$ and $z$ directions.
A component tells you how far the vector goes in one coordinate direction.

Key Idea

Work component by component

To add or subtract vectors, deal with the $i\mathbf{i}$ , $j\mathbf{j}$ and $k\mathbf{k}$ components separately. The resultant is the single vector found by adding several vectors together.

Example

Finding a resultant force

Three forces act on a particle:

F1=−2i+5j+3k,F2=6i−j−4k,F3=i+2j+7k\mathbf{F}_1=-2\mathbf{i}+5\mathbf{j}+3\mathbf{k},\quad \mathbf{F}_2=6\mathbf{i}-\mathbf{j}-4\mathbf{k},\quad \mathbf{F}_3=\mathbf{i}+2\mathbf{j}+7\mathbf{k}

Find the resultant force.

Add the three vectors component by component:
$R=(−2i+5j+3k)+(6i−j−4k)+(i+2j+7k)=(−2+6+1)i+(5−1+2)j+(3−4+7)k=5i+6j+6k\begin{aligned} \mathbf{R} &=(-2\mathbf{i}+5\mathbf{j}+3\mathbf{k})+(6\mathbf{i}-\mathbf{j}-4\mathbf{k})+(\mathbf{i}+2\mathbf{j}+7\mathbf{k})\\ &=(-2+6+1)\mathbf{i}+(5-1+2)\mathbf{j}+(3-4+7)\mathbf{k}\\ &=5\mathbf{i}+6\mathbf{j}+6\mathbf{k} \end{aligned}$
The resultant force is $5i+6j+6k5\mathbf{i}+6\mathbf{j}+6\mathbf{k}$ .

2. Position vectors, displacement and length

A position vector tells you where a point is relative to a fixed origin, usually called $O$ .

If point $A$ has position vector $a\mathbf{a}$ , then $OA⃗=a\vec{OA}=\mathbf{a}$ .

Position vectors are measured from the origin, while displacement vectors go from one point to another.

A displacement vector tells you how to move from one point to another. For example, $AB⃗\vec{AB}$ means “the vector from $A$ to $B$ ”.

The vector from A to B points from the tail at A to the head at B, and reversing the order reverses the direction.

Definition

Position vector and magnitude

If $A$ has position vector $a\mathbf{a}$ and $B$ has position vector $b\mathbf{b}$ , then $AB⃗=b−a\vec{AB}=\mathbf{b}-\mathbf{a}$ .
The magnitude of a vector is its length. For $v=xi+yj+zk\mathbf{v}=x\mathbf{i}+y\mathbf{j}+z\mathbf{k}$ ,
$∣v∣=x2+y2+z2|\mathbf{v}|=\sqrt{x^2+y^2+z^2}$

Key Idea

Head minus tail

To find $AB⃗\vec{AB}$ , subtract the position vector of the starting point $A$ from the position vector of the finishing point $B$ .

Example

Finding a vector and its length

Use head minus tail:
$AB⃗=b−a\vec{AB}=\mathbf{b}-\mathbf{a}$
Subtract the components:
$AB⃗=(−5i+j+6k)−(2i−3j+4k)=−7i+4j+2k\begin{aligned} \vec{AB} &=(-5\mathbf{i}+\mathbf{j}+6\mathbf{k})-(2\mathbf{i}-3\mathbf{j}+4\mathbf{k})\\ &=-7\mathbf{i}+4\mathbf{j}+2\mathbf{k} \end{aligned}$
Use 3D Pythagoras for the length:
$∣AB⃗∣=(−7)2+42+22=69|\vec{AB}|=\sqrt{(-7)^2+4^2+2^2}=\sqrt{69}$

Common Mistake

Subtracting the wrong way round

If you calculate $a−b\mathbf{a}-\mathbf{b}$ , you have found $BA⃗\vec{BA}$ instead of $AB⃗\vec{AB}$ . The length is the same, but the direction is reversed.

3. Unknown components from a magnitude

Sometimes you are given the length of a vector and asked to find an unknown component. The key move is to use the magnitude formula, then square both sides.

Example

Finding possible values of an unknown component

Given that $∣2i+mj−3k∣=7|2\mathbf{i}+m\mathbf{j}-3\mathbf{k}|=7$ , find the possible values of $m$ .

Apply the magnitude formula:
$22+m2+(−3)2=7\sqrt{2^2+m^2+(-3)^2}=7$
Square both sides:
$2^2+m^2+(-3)^2=7^2$
Simplify:
$m^2+13=49$
Solve for $m$ :
$m2=36⇒m=±6m^2=36\Rightarrow m=\pm 6$
The possible values are $m = 6$ and $m = - 6$ .

Tip

Expect two signs

A magnitude only measures length, not direction. So a positive component and the matching negative component can give the same length.

4. Routes, midpoints and extending a line

Vectors can be chained together like a journey. If you go from $A$ to $B$ , then from $B$ to $C$ , the total journey is from $A$ to $C$ :

Vector addition can be shown as a journey where AB followed by BC gives the resultant AC.

AB⃗+BC⃗=AC⃗\vec{AB}+\vec{BC}=\vec{AC}

A midpoint is the point halfway between two points. If $A$ has position vector $a\mathbf{a}$ and $B$ has position vector $b\mathbf{b}$ , then the midpoint has position vector

a+b2\frac{\mathbf{a}+\mathbf{b}}{2}

Example

Using a route in a triangle

Use the route from $A$ to $C$ through $B$ :
$AB⃗+BC⃗=AC⃗\vec{AB}+\vec{BC}=\vec{AC}$
Rearrange to find $BC⃗\vec{BC}$ :
$BC⃗=AC⃗−AB⃗=(7i−4j+3k)−(5i+2j−k)=2i−6j+4k\begin{aligned} \vec{BC} &=\vec{AC}-\vec{AB}\\ &=(7\mathbf{i}-4\mathbf{j}+3\mathbf{k})-(5\mathbf{i}+2\mathbf{j}-\mathbf{k})\\ &=2\mathbf{i}-6\mathbf{j}+4\mathbf{k} \end{aligned}$
Find the length of $A B$ :
$∣AB⃗∣=52+22+(−1)2=30|\vec{AB}|=\sqrt{5^2+2^2+(-1)^2}=\sqrt{30}$

Example

Extending a line segment

Extending a line segment with equal vectors places D so that B is halfway between A and D.

First find $AB⃗\vec{AB}$ :
$AB⃗=(4i−j+5k)−(i+2j−k)=3i−3j+6k\begin{aligned} \vec{AB} &=(4\mathbf{i}-\mathbf{j}+5\mathbf{k})-(\mathbf{i}+2\mathbf{j}-\mathbf{k})\\ &=3\mathbf{i}-3\mathbf{j}+6\mathbf{k} \end{aligned}$
Since $BD⃗\vec{BD}$ is the same vector, add it to the position vector of $B$ :
$d=(4i−j+5k)+(3i−3j+6k)=7i−4j+11k\begin{aligned} \mathbf{d} &=(4\mathbf{i}-\mathbf{j}+5\mathbf{k})+(3\mathbf{i}-3\mathbf{j}+6\mathbf{k})\\ &=7\mathbf{i}-4\mathbf{j}+11\mathbf{k} \end{aligned}$
So the position vector of $D$ is $7i−4j+11k7\mathbf{i}-4\mathbf{j}+11\mathbf{k}$ .

5. Parallel vectors and shapes

Two vectors are parallel when they point in the same or exactly opposite direction. In component form, this means one vector is a multiple of the other.

Parallel vectors have the same or opposite direction because one is a scalar multiple of the other.

Definition

Parallel vectors

Side vectors reveal whether opposite sides of a quadrilateral are parallel.

Example

Showing a quadrilateral is a trapezium

Relative to origin $O$ , points $A$ , $B$ and $C$ have position vectors

2i+j−3k,5i−j−2k,6i−4j+2k2\mathbf{i}+\mathbf{j}-3\mathbf{k},\quad 5\mathbf{i}-\mathbf{j}-2\mathbf{k},\quad 6\mathbf{i}-4\mathbf{j}+2\mathbf{k}

respectively. Show that $O A B C$ is a trapezium.

Find $AB⃗\vec{AB}$ :
$AB⃗=(5i−j−2k)−(2i+j−3k)=3i−2j+k\begin{aligned} \vec{AB} &=(5\mathbf{i}-\mathbf{j}-2\mathbf{k})-(2\mathbf{i}+\mathbf{j}-3\mathbf{k})\\ &=3\mathbf{i}-2\mathbf{j}+\mathbf{k} \end{aligned}$
Compare this with $OC⃗\vec{OC}$ :
$OC⃗=6i−4j+2k=2(3i−2j+k)=2AB⃗\vec{OC}=6\mathbf{i}-4\mathbf{j}+2\mathbf{k}=2(3\mathbf{i}-2\mathbf{j}+\mathbf{k})=2\vec{AB}$
So $A B$ is parallel to $O C$ , meaning one pair of opposite sides is parallel.
Check the other pair:
$BC⃗=(6i−4j+2k)−(5i−j−2k)=i−3j+4k\vec{BC}=(6\mathbf{i}-4\mathbf{j}+2\mathbf{k})-(5\mathbf{i}-\mathbf{j}-2\mathbf{k})=\mathbf{i}-3\mathbf{j}+4\mathbf{k}$
$BC⃗\vec{BC}$ is not a scalar multiple of $OA⃗=2i+j−3k\vec{OA}=2\mathbf{i}+\mathbf{j}-3\mathbf{k}$ , so exactly one pair of opposite sides is parallel. Therefore $O A B C$ is a trapezium.

6. A harder mixed geometry problem

Some vector questions combine several ideas: equal lengths, midpoints and area. The trick is to turn each geometric statement into an equation.

If a triangle is isosceles with $A B = A C$ , then the line from $A$ to the midpoint of $B C$ is the perpendicular height.

In an isosceles triangle, the line from the apex to the midpoint of the base is the height used in the area formula.

Example

Using equal lengths and area

Point $A$ has position vector $(ab3)\begin{pmatrix} a\\ b\\ 3 \end{pmatrix}$ , where $a$ and $b$ are positive constants. Points $B$ and $C$ have position vectors

(324)and(102)\begin{pmatrix} 3\\2\\4 \end{pmatrix} \quad\text{and}\quad \begin{pmatrix} 1\\0\\2 \end{pmatrix}

respectively. Given that $A B = A C$ and the area of triangle $A B C$ is $6\sqrt{6}$ , find $a$ and $b$ .

Compare squared lengths to avoid square roots:
$AB2=(3−a)2+(2−b)2+1AC2=(1−a)2+b2+1\begin{aligned} AB^2&=(3-a)^2+(2-b)^2+1\\ AC^2&=(1-a)^2+b^2+1 \end{aligned}$
Since $A B = A C$ , set these equal and simplify:
$(3−a)2+(2−b)2+1=(1−a)2+b2+1⇒a+b=3(3-a)^2+(2-b)^2+1=(1-a)^2+b^2+1\Rightarrow a+b=3$
Find the midpoint $M$ of $B C$ :
$m=12(3+12+04+2)=(213)\mathbf{m}=\frac{1}{2}\begin{pmatrix}3+1\\2+0\\4+2\end{pmatrix}=\begin{pmatrix}2\\1\\3\end{pmatrix}$
Use $b = 3 - a$ to find the base and height:
$∣BC⃗∣=(−2)2+(−2)2+(−2)2=23∣AM⃗∣=(a−2)2+(b−1)2=2∣a−2∣\begin{aligned} |\vec{BC}|&=\sqrt{(-2)^2+(-2)^2+(-2)^2}=2\sqrt{3}\\ |\vec{AM}|&=\sqrt{(a-2)^2+(b-1)^2}=\sqrt{2}|a-2| \end{aligned}$
Apply the area formula:
$6=12(23)(2∣a−2∣)=6∣a−2∣\sqrt{6}=\frac{1}{2}(2\sqrt{3})(\sqrt{2}|a-2|)=\sqrt{6}|a-2|$
Hence $∣ a - 2∣ = 1$ . The candidates are $a = 1, b = 2$ or $a = 3, b = 0$ ; because $a$ and $b$ are positive, $a = 1$ and $b = 2$ .

Exam technique

In the exam

For $AB⃗\vec{AB}$ , always use head minus tail: position vector of $B$ minus position vector of $A$ .
For shape proofs, calculate the relevant side vectors and state the scalar multiple clearly.
For length conditions, square both sides early and remember that unknown components may have two signs.

Self review

Check yourself

If $A$ and $B$ have position vectors $a\mathbf{a}$ and $b\mathbf{b}$ , what is $BA⃗\vec{BA}$ ?
How can you tell from components that two non-zero vectors are parallel?
Why can a magnitude equation give two possible values for an unknown component?

Vectors

What you'll learn

1. Vectors as components

2. Position vectors, displacement and length

3. Unknown components from a magnitude

4. Routes, midpoints and extending a line

5. Parallel vectors and shapes

6. A harder mixed geometry problem

Recap questions

You've reached the end

Vectors

What you'll learn

1. Vectors as components

2. Position vectors, displacement and length

3. Unknown components from a magnitude

4. Routes, midpoints and extending a line

5. Parallel vectors and shapes

6. A harder mixed geometry problem

Recap questions

You've reached the end

Vectors

What you'll learn

1. Vectors as components

2. Position vectors, displacement and length

3. Unknown components from a magnitude

4. Routes, midpoints and extending a line

5. Parallel vectors and shapes

6. A harder mixed geometry problem

Recap questions

You've reached the end

Vectors Revision Guide

Vectors

What you'll learn

1. Vectors as components

2. Position vectors, displacement and length

3. Unknown components from a magnitude

4. Routes, midpoints and extending a line

5. Parallel vectors and shapes

6. A harder mixed geometry problem

Recap questions

You've reached the end

Vectors Revision Guide