Vector Calculations

Being able to use vectors in calculations is a key skill in A-Level, this section will cover the following areas of vector calculations:

Finding vector magnitudes using Pythagoras’ theorem
Resolving vectors into component form
Finding the distance between points using a vector’s magnitude
Using the cosine rule to find the angle between two vectors

Make sure you are happy with the following topics before continuing.

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Finding Vector Magnitudes Using Pythagoras’ Theorem

Notation: The magnitude of vector $boldsymbol{a}$ is written as $|boldsymbol{a}|$ , similarly the magnitude of $overrightarrow{AB}$ is written as $|overrightarrow{AB}|$ .

As shown on the image on the right, the $boldsymbol{i}$ and $boldsymbol{j}$ components of a vector create a right-angled triangle, this means you can use Pythagoras’ theorem to find the magnitude (length) of the vector.

Using Pythagoras’ theorem we can calculate the magnitude of vector $boldsymbol{a}$ :

$|boldsymbol{a}|=sqrt{4^2+5^2}=sqrt{41}$

Sometimes you may be required to find a unit vector in the direction of a particular vector, which can be found using:

Unit vector in the direction of a vector $boldsymbol{a}=dfrac{boldsymbol{a}}{|boldsymbol{a}|}$

As stated in position vectors, a unit vector has a magnitude of $1$

Example: Find the unit vector in the direction of $boldsymbol{m}=-3boldsymbol{i}+6boldsymbol{j}$

We first need to find the magnitude of $boldsymbol{m}$ :

$|boldsymbol{m}|=sqrt{(-3)^2+6^2}=sqrt{9+36}=3sqrt{5}$

Therefore, the unit vector in the direction of $boldsymbol{m}$ is:

$dfrac{boldsymbol{m}}{|boldsymbol{m}|}=dfrac{boldsymbol{m}}{3sqrt{5}}=dfrac{1}{3sqrt{5}}(-3boldsymbol{i}+6boldsymbol{j})$

Resolving Vectors into Component Form

When calculating with vectors it is easier to work with them when they are separated into $boldsymbol{i}$ and $boldsymbol{j}$ vectors, allowing you to work with one component at a time.

Splitting a vector into components is called resolving the vector, which can be done using both trigonometry and Pythagoras’ theorem.

Example: An object is travelling at $11text{ ms}^{-1}$ at an angle of $38degree$ to the horizontal.

Find the horizontal $(boldsymbol{i})$ and vertical $(boldsymbol{j})$ components of the object’s velocity, $boldsymbol{v}$ .

It is often easier to draw a diagram and make a right angled triangle for the problem:

We can then use trigonometry to find $x$ and $y$ :

$cos{38degree}=dfrac{x}{11} rArr x=11cos{38degree}$

$sin{38degree}=dfrac{y}{11} rArr y=11sin{38degree}$

Thus,

$begin{aligned} boldsymbol{v}&=(11cos{38degree}boldsymbol{i}+11sin{38degree}boldsymbol{j})text{ ms}^{-1} &=(8.67boldsymbol{i}+6.77boldsymbol{j})text{ ms}^{-1} (2text{ dp}) end{aligned}$

Finding the Distance Between Points Using a Vector’s Magnitude

Calculating the distance between two points is fairly straight forward, you need to find the vector between them and then calculate its magnitude using Pythagoras’ theorem.

Example: The position vectors of points $K$ and $L$ are $6boldsymbol{i}-2boldsymbol{j}$ and $-3boldsymbol{i}+5boldsymbol{j}$ respectively.

Find the distance between the points $K$ and $L$ , giving your answer to $2$ decimal places.

First, we need to find the vector $overrightarrow{KL}$

$overrightarrow{KL}=-3boldsymbol{i}+5boldsymbol{j}-(6boldsymbol{i}-2boldsymbol{j})=-9boldsymbol{i}+7boldsymbol{j}$

Then calculate the magnitude of $overrightarrow{KL}$ :

$|overrightarrow{KL}|=sqrt{(-9)^2+7^2}=sqrt{130}=11.40 , (2text{ dp})$

Using the Cosine Rule to Find the Angle Between Two Vectors

To find the angle between two vectors, $boldsymbol{a}$ and $boldsymbol{b}$ , you can construct a triangle with $boldsymbol{a}$ and $boldsymbol{b}$ as two of the sides and then calculate the magnitude of $overrightarrow{AB}$ for the remaining side. You can use the cosine rule to find the angle between the two vectors.

Example: $overrightarrow{OA}=4boldsymbol{i}+boldsymbol{j}$ and $overrightarrow{OB}=-2boldsymbol{i}+5boldsymbol{j}$

Find the angle $theta$ between the vectors $overrightarrow{OA}$ and $overrightarrow{OB}$ , giving your answer to $2$ decimal places.

We first need to calculate the magnitudes of the vectors, $overrightarrow{OA},overrightarrow{OB}$ and $overrightarrow{AB}$ , so we can construct a triangle.

$|overrightarrow{OA}|=sqrt{4^2+1^2}=sqrt{17}$

$|overrightarrow{OB}|=sqrt{(-2)^2+5^2}=sqrt{29}$

$begin{aligned} |overrightarrow{AB}| &=|overrightarrow{OB}-overrightarrow{OA}| &=|-2boldsymbol{i}+5boldsymbol{j}-(4boldsymbol{i}+boldsymbol{j})| &=|-6boldsymbol{i}+4boldsymbol{j}| &=sqrt{(-6)^2+4^2}=2sqrt{13} end{aligned}$

Then using the cosine rule to find angle $theta$ :

$cos{theta}=dfrac{(sqrt{17})^2+(sqrt{29})^2-(2sqrt{13})^2}{2times sqrt{17} times sqrt{29}}=dfrac{-3}{sqrt{493}}$

$rArr theta=cos ^{-1} left(dfrac{-3}{sqrt{493}} right)=97.77 degree , (2text{ dp})$

Vector Calculations Example Questions

Question 1: Find the unit vector in the direction of $boldsymbol{t}=4boldsymbol{i}-5boldsymbol{j}$

[2 marks]

First find the magnitude of $boldsymbol{t}$ :

$|boldsymbol{t}|=sqrt{4^2+(-5)^2}=sqrt{16+25}=sqrt{41}$

Therefore, the unit vector is:

$dfrac{boldsymbol{t}}{|boldsymbol{t}|}=dfrac{boldsymbol{t}}{sqrt{41}}=dfrac{1}{sqrt{41}}(4boldsymbol{i}-5boldsymbol{j})$

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Question 2: Point $S$ has position vector $dbinom{4}{-3}$ and point $T$ has position vector $dbinom{-5}{2}$ . Find the distance between points $S$ and $T$ , giving your answer to $2$ decimal places.

[2 marks]

First, we need to find the vector $overrightarrow{ST}$ between the points.

$overrightarrow{ST}=dbinom{-5}{2}-dbinom{4}{-3}=dbinom{-5-4}{2-(-3)}=dbinom{-9}{5}$

So, $|overrightarrow{ST}|=sqrt{(-9)^2+5^2}=sqrt{106}=10.30 , (2 text{ dp})$

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Question 3: A vector $boldsymbol{w}$ models the movement of an object. The object is travelling at a speed of $8text{ ms}^{-1}$ with direction of $25degree$ above the horizontal.

Write $boldsymbol{w}$ in component form, giving the coefficients of $boldsymbol{i}$ and $boldsymbol{j}$ to $2$ decimal places.

[2 marks]

It is first useful to draw a diagram of the vector $boldsymbol{w}$ , forming a right-angled triangle.

Now we can use trigonometry to find $x$ and $y$ :

$cos{25degree}=dfrac{x}{8} rArr x=8cos{25degree}$

$sin{25degree}=dfrac{y}{8} rArr y=8sin{25degree}$

So,

$begin{aligned} boldsymbol{w}&=8cos{25degree}boldsymbol{i}+8sin{25degree}boldsymbol{j} &=7.25boldsymbol{i}+3.38boldsymbol{j}end{aligned}$

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Question 4: The points $X$ , $Y$ and $Z$ form a triangle, and have position vectors $2boldsymbol{i}+3boldsymbol{j}$ , $-boldsymbol{i}+2boldsymbol{j}$ and $4boldsymbol{i}-2boldsymbol{j}$ respectively. Find the angle, $theta$ between vectors $overrightarrow{YZ}$ and $overrightarrow{ZX}$ , giving your answer to $2$ decimal places.