Magnitude and DirectionCalculate the magnitude and direction of a vector and convert between forms. 
You can draw on a coordinate grid to help you answer some of the questions below.
This is Vectors  Magnitude and Direction level 2. You can also try:
Vector Connectors
Level 1
Level 3
InstructionsTry your best to answer the questions above. Type your answers into the boxes provided leaving no spaces. As you work through the exercise regularly click the "check" button. If you have any wrong answers, do your best to do corrections but if there is anything you don't understand, please ask your teacher for help. When you have got all of the questions correct you may want to print out this page and paste it into your exercise book. If you keep your work in an ePortfolio you could take a screen shot of your answers and paste that into your Maths file. 




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❎Vector Connectors  A basic set of exercises on vectors which could be done before attempting the following.
Vectors  Addition and subtraction of vectors and multiplication of a vector by a scalar.
Level 1  Magnitude of a vector from a column vector
Level 2  Direction of a vector from a column vector
Level 3  Finding the column vector from magnitudedirection form
Exam Style Questions  A collection of problems in the style of GCSE or IB/Alevel exam paper questions (worked solutions are available for Transum subscribers).
More on this topic including lesson Starters, visual aids, investigations and selfmarking exercises.
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See the National Curriculum page for links to related online activities and resources.
This set of exercises is to provide you with practice calculating the magnitude and direction of a vector and converting between component form and magnitude/direction form.
The magnitude of a vector is the length of the vector. A length cannot be negative. If the horizontal and vertical components are known it can be calculated using Pythagoras' theorem.
For example consider the vector:
$$ \mycolv{9\\12} $$
The magnitude is:
$$ \sqrt{9^2 + 12^2} = 15$$The direction of the vector can be calculated using trigonometry. The angle the vector makes with the horizontal is the inverse tan of the vertical component divide by the horizontal component.
$$ \tan^{1} \frac{12}{9} = 53.1^o$$If the magnitude and direction are given, as in the diagram below, trigonometry can be used to calculate the components for the column vector.
$$ x = 18 \cos 61^o = 8.73$$ $$ y = 18 \sin 61^o = 15.7$$So the column vector, with components expressed to 3 significant figures, is:
$$ \mycolv{8.73\\15.7} $$Don't wait until you have finished the exercise before you click on the 'Check' button. Click it often as you work through the questions to see if you are answering them correctly. You can doubleclick the 'Check' button to make it float at the bottom of your screen.
Answers to this exercise are available lower down this page when you are logged in to your Transum account. If you don’t yet have a Transum subscription one can be very quickly set up if you are a teacher, tutor or parent.
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