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## Precalculus

### Unit 6: Lesson 7

Vector components from magnitude and direction- Vector components from magnitude & direction
- Vector components from magnitude & direction
- Vector components from magnitude & direction: word problem
- Vector components from magnitude & direction (advanced)
- Converting between vector components and magnitude & direction review

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# Vector components from magnitude & direction (advanced)

Find a vector's components from its magnitude and direction. Direction angle is not given directly.

## Introduction

Each of the two problems below asks you to convert a vector from magnitude and direction form into component form.

But watch out! The

*aren't given for these vectors. You'll need to be careful what you plug into the sine and cosine functions.***direction angles**## Problem 1

## Problem 2

## Want to join the conversation?

- I used my calculator to answer these questions. Is there any way to solve these without a calculator?(42 votes)
- In fact, there is a way to answer (some of) these problems without a calculator - you can use trigonometric angle addition and subtraction properties.

So for example, the first problem is 105 degrees. That is the sum of 45 degrees and 60 degrees. I chose these angles specifically because we know the trigonometric values of these angles based on the unit circle.

cos 45 = sqrt(2)/2

cos 60 = 1/2

sin 45 = sqrt(2)/2

sin 60 = sqrt(3)/2

So let's say we are trying to find the x component, we will use the cosine addition property. If you look that up and plug the above values in you get:

cos 75 = cos (45 + 30) = (sqrt(2) - sqrt(6)) / 4

You can then use this in the SOH CAH TOA definition to get the exact value of the x component which is:

11 * (sqrt(2) - sqrt(6)) / 4

Obviously this will not give pretty answers like this for all angles, just for those which you can manipulate 45, 30, 60 to add up (or subtract) to that very angle. Also, given, this does not give you the numerical value (unless you can calculate all that in your head!).(22 votes)

- I try using the google calculator and I keep getting slightly different results... in the first one, instead of -2.85 and 10.63 I kept getting -2.65 and 10.68. I realy kept trying and retrying, checking and rechecking, and it still remained the same. Anyone else having this problem?(5 votes)
- One thing you should check is to see what mode the calculator is in (radians or degrees) if you are in the wrong mode for the measure of angle you are using you will get the wrong answer. If that isn't the problem the only other issue I can think of would be troubles with rounding and/or significant figures.

Hope this helps!(4 votes)

- Hi, I'm working on some component form of vector questions right now and I understand how to find the components given magnitude and direction as Sal taught in the video. However, what I still don't understand is what the vector components give you... like what does that resultant vector mean? What is it? I would really appreciate it if someone could explain.(3 votes)
- In physics if yoy have two forces pushing from different directions you can use vector addition to solve for the resulting force. For example, if you push a block of wood in one direction and your friend pushes at an angle, the block will go in the resultant direction.(2 votes)

- what is the component form of the vector 'a' with llall =100, 250 dg, approximate each component to the nearest tenth(3 votes)
- =||u⃗ ||cos(θ)should be sin(θ) in the solution of Q2 Finding the y-component of vector u.(3 votes)
- In my physics class, we see the x and y axis inverted and turned in some problems, I have no idea what quadrant I'm in, how do I determine if (x,y) are pos/neg? Thanks(2 votes)
- That's really strange. I've never seen anything like that before.(1 vote)

- if the resultant of 2 equal vectors is twice the magnitude of each of them what will be the angle between the 2 vectors?? THANKS(1 vote)
- If the vectors are equal, they must have the same direction. So the angle between them is 0.(2 votes)

- Law of sines works right?(1 vote)
- Yes, but it takes more time, it is easier to use trig functions so x=cos(angle)*magnitude and y=sin(angle)*magnitude.(2 votes)

- I'm still confused on why for some of the direction angles (in the first step) you would subtract them, but for the others you wouldn't subtract them.(1 vote)
- Hi Hannah,

Measure of the angle is given to us indirectly.

Always remember that the vector (looks like a ray) and the POSITIVE x-axis forms the angle.

Since we are given 75 degrees as our angle, we subtract it from 180 degrees to get 105 degrees. This 75 degrees angle is formed with the vector and**negative**x-axis.

I hope this helps.

Aiena.(1 vote)

- This may sound like a silly question but, is the magnitude and direction worked the same way as if you are looking for the vertical and horizontal components?(1 vote)
- The magnitude and direction are closely tied to the x- and y-components, but they are very different. The x- and y-components are vectors themselves, where one of their coordinates is 0. The magnitude and direction are just real numbers.(1 vote)