What is meant by 'resolving a vector into its components'?

So, let's think of a vector as an 'instruction' - a guide telling us to move in a specific direction for a certain distance. (Draw arbitrary vector A here). This is not the only way to move this distance away in this direction, however! We could also move to the left first, and then move to the point. (Draw two vectors which sum to A). To make things easier for us, we want to look at how far along we must go, and how far we up we must go (draw horizontal and vertical components of A). These are 'components' of A.
So, we know the directions of these components (horizontal and vertical), so now we just need to work out their magnitudes. Let's recall our trigonometry - as we're dealing with a right-angled triangle here - and think about Sin, Cos and Tan. Since we're looking at the opposite and adjacent sides of our triangle, we'll look at Sin and Cos. We know the hypotenuse of our triangle (it's just the magnitude of our original vector), so we can rearrange the trig functions to calculate the opposite and adjacent sides. Our angle is just the direction of the original vector. These horizontal and vertical vectors are the 'components' of our original vector A!

JR
Answered by John R. Physics tutor

2727 Views

See similar Physics GCSE tutors

Related Physics GCSE answers

All answers ▸

I throw a ball straight up with an initial velocity of 2m/s. How high is it after a fifth of a second?


Explain the role of the moderator in a fission reaction.


X-rays and gamma rays have different uses. Describe one use for X-rays and one use for gamma rays.


A rollercoaster carriage of mass 100kg has 45kJ of Kinetic Energy at the lowest point of its ride. Ignoring air resistance and friction between the wheels and the tracks, what is the maximum height above this point it could reach? [Take g as 10m/s/s)


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

© MyTutorWeb Ltd 2013–2025

Terms & Conditions|Privacy Policy
Cookie Preferences