Vector Calculations

Vector Calculations

A LevelAQAEdexcelOCRAQA 2022Edexcel 2022OCR 2022

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.

A Level AQA Edexcel OCR

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}=-3\boldsymbol{i}+6\boldsymbol{j}

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

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

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

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

 

A LevelAQAEdexcelOCR

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 11\text{ ms}^{-1} at an angle of 38\degree 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{38\degree}=\dfrac{x}{11} \rArr x=11\cos{38\degree}

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

Thus,

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

 

A LevelAQAEdexcelOCR

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 6\boldsymbol{i}-2\boldsymbol{j} and -3\boldsymbol{i}+5\boldsymbol{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}=-3\boldsymbol{i}+5\boldsymbol{j}-(6\boldsymbol{i}-2\boldsymbol{j})=-9\boldsymbol{i}+7\boldsymbol{j}

Then calculate the magnitude of \overrightarrow{KL}:

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

 

A LevelAQAEdexcelOCR

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}=4\boldsymbol{i}+\boldsymbol{j} and \overrightarrow{OB}=-2\boldsymbol{i}+5\boldsymbol{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}| \\ &=|-2\boldsymbol{i}+5\boldsymbol{j}-(4\boldsymbol{i}+\boldsymbol{j})|\\ &=|-6\boldsymbol{i}+4\boldsymbol{j}| \\ &=\sqrt{(-6)^2+4^2}=2\sqrt{13} \end{aligned}

Then using the cosine rule to find angle \theta:

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

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

A LevelAQAEdexcelOCR

Example Questions

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}}(4\boldsymbol{i}-5\boldsymbol{j})

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})

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

 

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

 

So,

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

We first need to find the following vectors: \overrightarrow{XY}, \overrightarrow{YZ} and \overrightarrow{XZ}:

 

\overrightarrow{XY}=-\boldsymbol{i}+2\boldsymbol{j}-(2\boldsymbol{i}+3\boldsymbol{j})=-3\boldsymbol{i}-\boldsymbol{j}

 

\overrightarrow{YZ}=4\boldsymbol{i}-2\boldsymbol{j}-(-\boldsymbol{i}+2\boldsymbol{j})=5\boldsymbol{i}-4\boldsymbol{j}

 

\overrightarrow{XZ}=4\boldsymbol{i}-2\boldsymbol{j}-(2\boldsymbol{i}+3\boldsymbol{j})=2\boldsymbol{i}-5\boldsymbol{j}

 

Then, we need to find the magnitudes of those vectors so we can form a triangle:

 

|\overrightarrow{XY}|=\sqrt{(-3)^2-(-1)^2}=\sqrt{10}

 

|\overrightarrow{YZ}|=\sqrt{5^2+(-4)^2}=\sqrt{41}

 

|\overrightarrow{XZ}|=\sqrt{2^2-(-5)^2}=\sqrt{29}

 

This allows us to draw a triangle:

Finally, using the cosine rule we can calculate the angle \theta:

\cos{\theta}=\dfrac{(\sqrt{29})^2+(\sqrt{41})^2-(\sqrt{10})^2}{2\times\sqrt{29}\times \sqrt{41}}=\dfrac{60}{2\sqrt{1189}}=\dfrac{30}{\sqrt{1189}}

 

\rArr \theta=\cos^{-1}(\dfrac{30}{\sqrt{1189}})=29.54\degree \, (2\text{ dp})

Related Topics

MME

Vector Basics

A Level
MME

Position Vectors

A Level

Additional Resources

MME

Exam Tips Cheat Sheet

A Level
MME

Formula Booklet

A Level

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