# How Do You Calculate Vector Operations for Angled Displacements?

• chocolatelover
In summary: Yes, that's correct. If the x is positive and the y is negative, the angle is in the third quadrant, so you would need to subtract it from 180. If both were negative, then it would be in the fourth quadrant, so you would subtract it from 360.No problem, glad I could help.
chocolatelover

## Homework Statement

Each of the displacement vectors A and B have a magnitude of 3.00 m. Vector B is vertical and is on the y axis, vector A forms a 30 degree angle to the positive x axis. Report all angles counterclockwise from the positve x axis. Find A+B, A-B, B-A, A-2B
This is similar to what the diagram looks like:
http://id.mind.net/~zona/mstm/physics/mechanics/vectors/components/vectorComponents.html

## Homework Equations

magnitude=square root(Ax+Bx)^2+(Ay+By)^2
theta=tan-1(Ay+By/Ax+Bx)

## The Attempt at a Solution

1. A+B
magnitude=square root (3cos30+3cos90)^2+(3sin30+3sin90)^2
=2.7

theta=tan-1(3sin30+3sin90)/(3cos30+3cos90)=69.4

2. A-B
magnitude=square root (3cos30+3cos90)^2+-(3sin30+3sin90)^2=square root (-13.5) This then goes to 3.67, right?
theta=tan-1(3sin30+-3sin90)/(3cos30+-3cos90)=19.05

3. B-A
theta=tan-1(sin90-3sin30)/(3cos30-3cos90)=74.4

4. A-2B
magnitude=square root(3cos30+3cos90)^2+-(2(3sin30+3sin90)^2=5.8
theta=tan-1(3sin30-2(3sin30+3sin90)/(3cos30+-2(3cos30+3cos90)

I would really appreciate if someone could help me with these 4 parts. I've been working on this for two days and I don't understand it. I don't see what I'm doing wrong.

Thank you very much

Last edited by a moderator:
There are a couple of things going on here, one is I'm not quite sure how you're getting the answers you're getting.

First let's simplify things a bit. Vector B is vertical and on the y axis, therefore there is no x component, so you should just use 0 when referring to the x component of B. Also, you can just use 1 as the sin90, so the y component of B is just 3.

And you have A split correctly, the x = 3cos30, the y = 3sin30.

So, for A+B you have:
x component (3cos30 + 0) = 2.6
y component (3sin30 + 3) = 1.5 +3 = 4.5

So the magnitude of A+B =$$\sqrt{2.6^2 + 4.5^2} = \sqrt{27.01} = 5.2$$

Redo the problem and see if you get the same answers, then recalculate the angle.

For A-B, you need to be careful what you subtract from what. It should be:

x component (3cos30 - 0) = 2.6
y component (3sin30 - 3) = 1.5 - 3 = -1.5

From here you can recalculate the magnitude and angle. Once again, be careful where you put the angle as the x is positive, but the y is negative.

Thank you very much

Does this look correct? Aren't the sin angles 90°?

direction of a+b=tan-1(2.6/4.5)=30.02
180-30.02=150 Do I need to subtract it from 180 or 360?

magnitude of a-b=square root of (2.6)^2+(-1.5)^2=3

direction of a-b=tan-1(2.6/-1.5)=-60.02
180-60.02=240.02

b-a
x component=0-3cos30=-2.6
y component=3-3sin30=1.5

magnitude=square root ((-2.6)^2+(1.5)^2)=9.01

direction=tan-1(-2.6/1.5)=-60.1
180-60.1=120.0

a-2b

x component=3cos30-2(0)=2.6
y component=3sin30-2(3)=-4.5

magnitude=square root of (2.6^2+-4.5^2)=5.2
direction=tan-1(2.6/-4.5)=-30.02
180-30.02=149.98

Thank you

chocolatelover said:
Thank you very much
Does this look correct? Aren't the sin angles 90°?
Not sure what you mean by this. The sin 90° = 1, so 3sin90 = 3*1 = 3

direction of a+b=tan-1(2.6/4.5)=30.02
180-30.02=150 Do I need to subtract it from 180 or 360?

Be careful, the x value is 2.6, the y value is 4.5, and tan is y/x (not x/y as you have it).
So, your angle isn't right. As to what to subtract it from, look at where it would be based on the signs of the x and y. Both are positive, so the angle is in the first quadrant, no need to subtract it from anything.

magnitude of a-b=square root of (2.6)^2+(-1.5)^2=3

direction of a-b=tan-1(2.6/-1.5)=-60.02
180-60.02=240.02

The magnitude of a-b is correct, re-do the angle as above. As to where it's located, the x is positive, the y is negative, so where would the angle be?

b-a
x component=0-3cos30=-2.6
y component=3-3sin30=1.5

magnitude=square root ((-2.6)^2+(1.5)^2)=9.01

direction=tan-1(-2.6/1.5)=-60.1
180-60.1=120.0

a-2b

x component=3cos30-2(0)=2.6
y component=3sin30-2(3)=-4.5

magnitude=square root of (2.6^2+-4.5^2)=5.2
direction=tan-1(2.6/-4.5)=-30.02
180-30.02=149.98
The magnitude looks good. Redo the angle.
You seem to be understanding it pretty well, If I were making these mistakes, and believe me I have, it would be because I was going to fast, and not being deliberate with the calculations. Since you're just using two vectors to do all the calculations, you can do the original calculations at the top, and label them such as:
Ax = 3cos30 = 2.6
Ay = 3sin30 = 1.5

etc., then just plug in the values you need to each equation.

Thank you very much

"As to where it's located, the x is positive, the y is negative, so where would the angle be?"

It would be in the third quadrant, right? So, I just need to subtract it from 180, right? If it were in the third or fourth quadrant, I would need to subtract it from 360, right?

thank you

Last edited:

## What is a displacement vector?

A displacement vector is a mathematical representation of the distance and direction between two points in space. It is commonly used in physics and engineering to describe the movement of an object.

## How do displacement vectors help in problem solving?

Displacement vectors help in problem solving by providing a visual representation of the direction and magnitude of movement, making it easier to analyze and calculate various physical quantities such as velocity, acceleration, and force.

## What is the difference between displacement and distance?

Displacement is a vector quantity that refers to the shortest distance between two points and includes direction, while distance is a scalar quantity that refers to the total amount of ground covered by an object.

## Can displacement vectors be negative?

Yes, displacement vectors can be negative. A negative displacement vector indicates movement in the opposite direction of a positive displacement vector. This is commonly seen in situations where an object moves backwards or changes direction.

## How are displacement vectors represented?

Displacement vectors are typically represented by an arrow pointing from the initial position to the final position of an object. The length of the arrow represents the magnitude of displacement, and the direction of the arrow represents the direction of movement.

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