1.10.7 Problem solving using vectors

What you'll learn

How to translate geometric information into vector equations.
How to prove that points are collinear and divide lines in a given ratio.
How to find resultants and solve equilibrium problems involving forces.
How to interpret your vector answer in the original context.

Vector foundations

A vector is a quantity with both magnitude and direction. Magnitude means size or length. Vectors can represent displacement, velocity, acceleration and force.

A vector is often written as a column vector:

a=(a1a2)\mathbf{a}=\begin{pmatrix}a_1\\a_2\end{pmatrix}

The numbers $a_1$ and $a_2$ are the horizontal and vertical components of the vector.

By contrast, a scalar has magnitude but no direction. Mass, time and temperature are scalars.

Adding and subtracting vectors

Vectors are added and subtracted component by component:

(ab)+(cd)=(a+cb+d).\begin{pmatrix}a\\b\end{pmatrix} + \begin{pmatrix}c\\d\end{pmatrix} = \begin{pmatrix}a+c\\b+d\end{pmatrix}.

Multiplying a vector by a scalar changes its magnitude and, if the scalar is negative, reverses its direction:

k(ab)=(kakb).k\begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}ka\\kb\end{pmatrix}.

Key Idea

Follow a route

If you travel from one point to another through intermediate points, add the vectors for each part of the journey. For example, $AB→+BC→=AC→\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$ .

Example

Finding a missing displacement

A particle moves through displacements $(5−2)\begin{pmatrix}5\\-2\end{pmatrix}$ and $(−17)\begin{pmatrix}-1\\7\end{pmatrix}$ , measured in metres.

Add the corresponding components:
$(5−2)+(−17)=(5−1−2+7).\begin{pmatrix}5\\-2\end{pmatrix} + \begin{pmatrix}-1\\7\end{pmatrix} = \begin{pmatrix}5-1\\-2+7\end{pmatrix}.$
Simplify to obtain the total displacement:
$m.\begin{pmatrix}4\\5\end{pmatrix}\text{ m}.$
Interpret the components: the particle finishes 4 m to the right and 5 m above its starting point.

Position vectors and vectors between points

A position vector describes the position of a point relative to a fixed origin, usually called $O$ . If $OA→=a\overrightarrow{OA}=\mathbf{a}$ , then $a\mathbf{a}$ is the position vector of $A$ .

To move from $A$ to $B$ , subtract the position vector of the starting point from that of the finishing point.

Definition

Vector between two points

If $OA→=a\overrightarrow{OA}=\mathbf{a}$ and $OB→=b\overrightarrow{OB}=\mathbf{b}$ , then

AB→=b−a.\overrightarrow{AB}=\mathbf{b}-\mathbf{a}.

The diagram shows both this subtraction rule and the related idea that balancing forces must be equal and opposite.

Position-vector subtraction from A to B and equilibrium of three concurrent forces

Example

Using position vectors in a triangle

Points $A$ and $B$ have position vectors

a=(2−1),b=(85).\mathbf{a}=\begin{pmatrix}2\\-1\end{pmatrix}, \qquad \mathbf{b}=\begin{pmatrix}8\\5\end{pmatrix}.

Find $AB→\overrightarrow{AB}$ and the position vector of the midpoint $M$ of $A B$ .

Subtract the position vector of $A$ from that of $B$ :
$AB→=b−a=(85)−(2−1)=(66).\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = \begin{pmatrix}8\\5\end{pmatrix} - \begin{pmatrix}2\\-1\end{pmatrix} = \begin{pmatrix}6\\6\end{pmatrix}.$
The midpoint is halfway from $A$ to $B$ , so
$OM→=a+12AB→.\overrightarrow{OM} = \mathbf{a}+\frac12\overrightarrow{AB}.$
Substitute the vectors:
$OM→=(2−1)+12(66)=(52).\overrightarrow{OM} = \begin{pmatrix}2\\-1\end{pmatrix} + \frac12\begin{pmatrix}6\\6\end{pmatrix} = \begin{pmatrix}5\\2\end{pmatrix}.$

Common Mistake

Subtracting in the wrong order

For $AB→\overrightarrow{AB}$ , calculate “position of $B$ minus position of $A$ ”. Reversing the subtraction gives $BA→\overrightarrow{BA}$ , which points in the opposite direction.

Dividing a line in a ratio

Suppose $P$ lies on the line segment from $A$ to $B$ and divides it in the ratio $A P : P B = m : n$ . The fraction of the complete journey from $A$ to $B$ is then $mm+n\frac{m}{m+n}$ .

Therefore,

OP→=a+mm+n(b−a).\overrightarrow{OP} = \mathbf{a} + \frac{m}{m+n}(\mathbf{b}-\mathbf{a}).

This “start plus a fraction of the journey” approach is often easier to remember than a section formula.

Example

Locating a point that divides a line

The position vectors of $A$ and $B$ are $a\mathbf{a}$ and $b\mathbf{b}$ . Point $P$ divides $A B$ in the ratio $A P : P B = 2 : 3$ . Find $OP→\overrightarrow{OP}$ .

The whole segment contains $2 + 3 = 5$ equal parts, and $A P$ contains 2 of them. Hence
$AP→=25AB→.\overrightarrow{AP}=\frac25\overrightarrow{AB}.$
Replace $AB→\overrightarrow{AB}$ with $b−a\mathbf{b}-\mathbf{a}$ :
$OP→=a+25(b−a).\overrightarrow{OP} = \mathbf{a}+\frac25(\mathbf{b}-\mathbf{a}).$
Collect the coefficients of $a\mathbf{a}$ and $b\mathbf{b}$ :
$OP→=35a+25b.\overrightarrow{OP} = \frac35\mathbf{a}+\frac25\mathbf{b}.$

Proving collinearity

Points are collinear if they lie on the same straight line.

Two non-zero vectors are parallel when one is a scalar multiple of the other. Therefore, to prove that $A$ , $B$ and $C$ are collinear, you can show that

AC→=kAB→\overrightarrow{AC}=k\overrightarrow{AB}

for some scalar $k$ .

If $k > 0$ , the vectors point in the same direction. If $k < 0$ , they point in opposite directions.

Example

Proving three points are collinear

The points have position vectors

a=(12),b=(48),c=(714).\mathbf{a}=\begin{pmatrix}1\\2\end{pmatrix}, \quad \mathbf{b}=\begin{pmatrix}4\\8\end{pmatrix}, \quad \mathbf{c}=\begin{pmatrix}7\\14\end{pmatrix}.

Find two vectors beginning at the same point:
$AB→=b−a=(36),AC→=c−a=(612).\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = \begin{pmatrix}3\\6\end{pmatrix}, \qquad \overrightarrow{AC} = \mathbf{c}-\mathbf{a} = \begin{pmatrix}6\\12\end{pmatrix}.$
Compare the vectors:
$AC→=2AB→.\overrightarrow{AC}=2\overrightarrow{AB}.$
Since one vector is a scalar multiple of the other, the vectors are parallel and share point $A$ . Therefore, $A$ , $B$ and $C$ are collinear. Also, $B$ lies between $A$ and $C$ because the scalar lies between 0 and 1 when $AB→\overrightarrow{AB}$ is written as a multiple of $AC→\overrightarrow{AC}$ .

Tip

Make the conclusion explicit

A scalar multiple proves that the relevant vectors are parallel. To complete a collinearity proof, also mention that the vectors pass through a common point.

Magnitude and direction

The magnitude of $v=(xy)\mathbf{v}=\begin{pmatrix}x\\y\end{pmatrix}$ is its length:

∣v∣=x2+y2.|\mathbf{v}|=\sqrt{x^2+y^2}.

Its direction can be found using trigonometry. If $θ\theta$ is measured anticlockwise from the positive horizontal direction, start with

tan⁡θ=yx,\tan\theta=\frac{y}{x},

then use the signs of the components to identify the correct quadrant.

Common Mistake

Trusting inverse tangent without checking

A calculator value of $tan^{-1}(y/x)$ may not give the required quadrant. Sketch the components or inspect their signs before stating the direction.

Vectors and forces

A force is a vector measured in newtons, N. When several forces act on an object, their vector sum is called the resultant force.

If forces are given in component form, add their horizontal components and add their vertical components.

Definition

Equilibrium

An object is in equilibrium when the resultant force acting on it is the zero vector:

∑F=0.\sum\mathbf{F}=\mathbf{0}.

Example

Finding a force that maintains equilibrium

A particle is acted on by forces

N.\mathbf{F}_1=\begin{pmatrix}6\\0\end{pmatrix}\text{ N}, \qquad \mathbf{F}_2=\begin{pmatrix}0\\8\end{pmatrix}\text{ N}.

Find the third force $R\mathbf{R}$ required for equilibrium, including its magnitude.

Add the two known forces:
$N.\mathbf{F}_1+\mathbf{F}_2 = \begin{pmatrix}6\\8\end{pmatrix}\text{ N}.$
For equilibrium, the third force must cancel this resultant:
$N.\mathbf{R} = -\begin{pmatrix}6\\8\end{pmatrix} = \begin{pmatrix}-6\\-8\end{pmatrix}\text{ N}.$
Calculate its magnitude:
$N.|\mathbf{R}| = \sqrt{(-6)^2+(-8)^2} = \sqrt{100} = 10\text{ N}.$
Interpret the answer: the force has magnitude 10 N and acts downwards and to the left, exactly opposite to the resultant of the known forces.

Modelling a problem with vectors

In a context, you must decide what each vector represents and choose positive directions. A reliable process is to:

draw a simple diagram;
label known vectors and unknown components;
translate routes, parallel lines or equilibrium into vector equations;
compare components or coefficients;
interpret the result in context.

A negative component is not automatically wrong. It means that the true direction is opposite to the positive direction you selected.

Key Idea

Translate, solve, interpret

Vector problem solving has three stages: turn the situation into vector equations, solve those equations, and then explain what the resulting numbers or vectors mean.

Exam technique

In the exam

Define any vectors or positive directions that are not already specified, and keep the notation consistent.
For a vector between points, use “finish minus start”; for equilibrium, set the sum of all forces equal to $0\mathbf{0}$ .
Compare components separately, keep exact values unless a decimal is requested, and finish by interpreting magnitude, direction, ratio or position in the context.

Self review

Check yourself

If $OA→=a\overrightarrow{OA}=\mathbf{a}$ and $OB→=b\overrightarrow{OB}=\mathbf{b}$ , how would you express the point three-quarters of the way from $A$ to $B$ ?
What vector relationship could you use to prove that three points are collinear?
Two known forces have resultant $(pq)\begin{pmatrix}p\\q\end{pmatrix}$ N. What force is required for equilibrium, and how would you find its magnitude?

What you'll learn

How to translate geometric information into vector equations.
How to prove that points are collinear and divide lines in a given ratio.
How to find resultants and solve equilibrium problems involving forces.
How to interpret your vector answer in the original context.

Vector foundations

A vector is a quantity with both magnitude and direction. Magnitude means size or length. Vectors can represent displacement, velocity, acceleration and force.

A vector is often written as a column vector:

a=(a1a2)\mathbf{a}=\begin{pmatrix}a_1\\a_2\end{pmatrix}

The numbers $a_1$ and $a_2$ are the horizontal and vertical components of the vector.

By contrast, a scalar has magnitude but no direction. Mass, time and temperature are scalars.

Adding and subtracting vectors

Vectors are added and subtracted component by component:

(ab)+(cd)=(a+cb+d).\begin{pmatrix}a\\b\end{pmatrix} + \begin{pmatrix}c\\d\end{pmatrix} = \begin{pmatrix}a+c\\b+d\end{pmatrix}.

Multiplying a vector by a scalar changes its magnitude and, if the scalar is negative, reverses its direction:

k(ab)=(kakb).k\begin{pmatrix}a\\b\end{pmatrix} = \begin{pmatrix}ka\\kb\end{pmatrix}.

Key Idea

Follow a route

Example

Finding a missing displacement

A particle moves through displacements $(5−2)\begin{pmatrix}5\\-2\end{pmatrix}$ and $(−17)\begin{pmatrix}-1\\7\end{pmatrix}$ , measured in metres.

Add the corresponding components:
$(5−2)+(−17)=(5−1−2+7).\begin{pmatrix}5\\-2\end{pmatrix} + \begin{pmatrix}-1\\7\end{pmatrix} = \begin{pmatrix}5-1\\-2+7\end{pmatrix}.$
Simplify to obtain the total displacement:
$m.\begin{pmatrix}4\\5\end{pmatrix}\text{ m}.$
Interpret the components: the particle finishes 4 m to the right and 5 m above its starting point.

Position vectors and vectors between points

To move from $A$ to $B$ , subtract the position vector of the starting point from that of the finishing point.

Definition

Vector between two points

If $OA→=a\overrightarrow{OA}=\mathbf{a}$ and $OB→=b\overrightarrow{OB}=\mathbf{b}$ , then

AB→=b−a.\overrightarrow{AB}=\mathbf{b}-\mathbf{a}.

The diagram shows both this subtraction rule and the related idea that balancing forces must be equal and opposite.

Position-vector subtraction from A to B and equilibrium of three concurrent forces

Example

Using position vectors in a triangle

Points $A$ and $B$ have position vectors

a=(2−1),b=(85).\mathbf{a}=\begin{pmatrix}2\\-1\end{pmatrix}, \qquad \mathbf{b}=\begin{pmatrix}8\\5\end{pmatrix}.

Find $AB→\overrightarrow{AB}$ and the position vector of the midpoint $M$ of $A B$ .

Subtract the position vector of $A$ from that of $B$ :
$AB→=b−a=(85)−(2−1)=(66).\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = \begin{pmatrix}8\\5\end{pmatrix} - \begin{pmatrix}2\\-1\end{pmatrix} = \begin{pmatrix}6\\6\end{pmatrix}.$
The midpoint is halfway from $A$ to $B$ , so
$OM→=a+12AB→.\overrightarrow{OM} = \mathbf{a}+\frac12\overrightarrow{AB}.$
Substitute the vectors:
$OM→=(2−1)+12(66)=(52).\overrightarrow{OM} = \begin{pmatrix}2\\-1\end{pmatrix} + \frac12\begin{pmatrix}6\\6\end{pmatrix} = \begin{pmatrix}5\\2\end{pmatrix}.$

Common Mistake

Subtracting in the wrong order

For $AB→\overrightarrow{AB}$ , calculate “position of $B$ minus position of $A$ ”. Reversing the subtraction gives $BA→\overrightarrow{BA}$ , which points in the opposite direction.

Dividing a line in a ratio

Suppose $P$ lies on the line segment from $A$ to $B$ and divides it in the ratio $A P : P B = m : n$ . The fraction of the complete journey from $A$ to $B$ is then $mm+n\frac{m}{m+n}$ .

Therefore,

OP→=a+mm+n(b−a).\overrightarrow{OP} = \mathbf{a} + \frac{m}{m+n}(\mathbf{b}-\mathbf{a}).

This “start plus a fraction of the journey” approach is often easier to remember than a section formula.

Example

Locating a point that divides a line

The position vectors of $A$ and $B$ are $a\mathbf{a}$ and $b\mathbf{b}$ . Point $P$ divides $A B$ in the ratio $A P : P B = 2 : 3$ . Find $OP→\overrightarrow{OP}$ .

The whole segment contains $2 + 3 = 5$ equal parts, and $A P$ contains 2 of them. Hence
$AP→=25AB→.\overrightarrow{AP}=\frac25\overrightarrow{AB}.$
Replace $AB→\overrightarrow{AB}$ with $b−a\mathbf{b}-\mathbf{a}$ :
$OP→=a+25(b−a).\overrightarrow{OP} = \mathbf{a}+\frac25(\mathbf{b}-\mathbf{a}).$
Collect the coefficients of $a\mathbf{a}$ and $b\mathbf{b}$ :
$OP→=35a+25b.\overrightarrow{OP} = \frac35\mathbf{a}+\frac25\mathbf{b}.$

Proving collinearity

Points are collinear if they lie on the same straight line.

Two non-zero vectors are parallel when one is a scalar multiple of the other. Therefore, to prove that $A$ , $B$ and $C$ are collinear, you can show that

AC→=kAB→\overrightarrow{AC}=k\overrightarrow{AB}

for some scalar $k$ .

If $k > 0$ , the vectors point in the same direction. If $k < 0$ , they point in opposite directions.

Example

Proving three points are collinear

The points have position vectors

a=(12),b=(48),c=(714).\mathbf{a}=\begin{pmatrix}1\\2\end{pmatrix}, \quad \mathbf{b}=\begin{pmatrix}4\\8\end{pmatrix}, \quad \mathbf{c}=\begin{pmatrix}7\\14\end{pmatrix}.

Find two vectors beginning at the same point:
$AB→=b−a=(36),AC→=c−a=(612).\overrightarrow{AB} = \mathbf{b}-\mathbf{a} = \begin{pmatrix}3\\6\end{pmatrix}, \qquad \overrightarrow{AC} = \mathbf{c}-\mathbf{a} = \begin{pmatrix}6\\12\end{pmatrix}.$
Compare the vectors:
$AC→=2AB→.\overrightarrow{AC}=2\overrightarrow{AB}.$
Since one vector is a scalar multiple of the other, the vectors are parallel and share point $A$ . Therefore, $A$ , $B$ and $C$ are collinear. Also, $B$ lies between $A$ and $C$ because the scalar lies between 0 and 1 when $AB→\overrightarrow{AB}$ is written as a multiple of $AC→\overrightarrow{AC}$ .

Tip

Make the conclusion explicit

A scalar multiple proves that the relevant vectors are parallel. To complete a collinearity proof, also mention that the vectors pass through a common point.

Magnitude and direction

The magnitude of $v=(xy)\mathbf{v}=\begin{pmatrix}x\\y\end{pmatrix}$ is its length:

∣v∣=x2+y2.|\mathbf{v}|=\sqrt{x^2+y^2}.

Its direction can be found using trigonometry. If $θ\theta$ is measured anticlockwise from the positive horizontal direction, start with

tan⁡θ=yx,\tan\theta=\frac{y}{x},

then use the signs of the components to identify the correct quadrant.

Common Mistake

Trusting inverse tangent without checking

A calculator value of $tan^{-1}(y/x)$ may not give the required quadrant. Sketch the components or inspect their signs before stating the direction.

Vectors and forces

A force is a vector measured in newtons, N. When several forces act on an object, their vector sum is called the resultant force.

If forces are given in component form, add their horizontal components and add their vertical components.

Definition

Equilibrium

An object is in equilibrium when the resultant force acting on it is the zero vector:

∑F=0.\sum\mathbf{F}=\mathbf{0}.

Example

Finding a force that maintains equilibrium

A particle is acted on by forces

N.\mathbf{F}_1=\begin{pmatrix}6\\0\end{pmatrix}\text{ N}, \qquad \mathbf{F}_2=\begin{pmatrix}0\\8\end{pmatrix}\text{ N}.

Find the third force $R\mathbf{R}$ required for equilibrium, including its magnitude.

Add the two known forces:
$N.\mathbf{F}_1+\mathbf{F}_2 = \begin{pmatrix}6\\8\end{pmatrix}\text{ N}.$
For equilibrium, the third force must cancel this resultant:
$N.\mathbf{R} = -\begin{pmatrix}6\\8\end{pmatrix} = \begin{pmatrix}-6\\-8\end{pmatrix}\text{ N}.$
Calculate its magnitude:
$N.|\mathbf{R}| = \sqrt{(-6)^2+(-8)^2} = \sqrt{100} = 10\text{ N}.$
Interpret the answer: the force has magnitude 10 N and acts downwards and to the left, exactly opposite to the resultant of the known forces.

Modelling a problem with vectors

In a context, you must decide what each vector represents and choose positive directions. A reliable process is to:

draw a simple diagram;
label known vectors and unknown components;
translate routes, parallel lines or equilibrium into vector equations;
compare components or coefficients;
interpret the result in context.

A negative component is not automatically wrong. It means that the true direction is opposite to the positive direction you selected.

Key Idea

Translate, solve, interpret

Vector problem solving has three stages: turn the situation into vector equations, solve those equations, and then explain what the resulting numbers or vectors mean.

Exam technique

In the exam

Define any vectors or positive directions that are not already specified, and keep the notation consistent.
For a vector between points, use “finish minus start”; for equilibrium, set the sum of all forces equal to $0\mathbf{0}$ .
Compare components separately, keep exact values unless a decimal is requested, and finish by interpreting magnitude, direction, ratio or position in the context.

Self review

Check yourself

If $OA→=a\overrightarrow{OA}=\mathbf{a}$ and $OB→=b\overrightarrow{OB}=\mathbf{b}$ , how would you express the point three-quarters of the way from $A$ to $B$ ?
What vector relationship could you use to prove that three points are collinear?
Two known forces have resultant $(pq)\begin{pmatrix}p\\q\end{pmatrix}$ N. What force is required for equilibrium, and how would you find its magnitude?

What you'll learn

Vector foundations

Adding and subtracting vectors

Position vectors and vectors between points

Dividing a line in a ratio

Proving collinearity

Magnitude and direction

Vectors and forces

Modelling a problem with vectors

Review 1.10.7 Problem solving using vectors by teaching Genie

1.10.7 Problem solving using vectors

What you'll learn

Vector foundations

Adding and subtracting vectors

Position vectors and vectors between points

Dividing a line in a ratio

Proving collinearity

Magnitude and direction

Vectors and forces

Modelling a problem with vectors

Review 1.10.7 Problem solving using vectors by teaching Genie

1.10.7 Problem solving using vectors

What you'll learn

Vector foundations

Adding and subtracting vectors

Position vectors and vectors between points

Dividing a line in a ratio

Proving collinearity

Magnitude and direction

Vectors and forces

Modelling a problem with vectors

Review 1.10.7 Problem solving using vectors by teaching Genie

1.10.7 Problem solving using vectors Revision Guide

1.10.7 Problem solving using vectors

What you'll learn

Vector foundations

Adding and subtracting vectors

Position vectors and vectors between points

Dividing a line in a ratio

Proving collinearity

Magnitude and direction

Vectors and forces

Modelling a problem with vectors

Review 1.10.7 Problem solving using vectors by teaching Genie

1.10.7 Problem solving using vectors Revision Guide