Solve Vector Problems: a + b, a - b, a - 2b

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Homework Help Overview

The discussion revolves around vector addition and subtraction involving two vectors: vector a, which has a speed of 3 m/s directed northwest, and vector b, moving west at 5 m/s. Participants are tasked with determining the resultant vectors a + b, a - b, and a - 2b.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the need to establish a coordinate system to express the vectors in component form. There are attempts to use trigonometry and the Pythagorean theorem to resolve the vectors into their components. Some express frustration over their initial attempts and seek clarification on the process.

Discussion Status

Several participants have provided guidance on setting up a coordinate system and determining the components of the vectors. There is an ongoing exploration of how to express vector a in terms of its components, with some participants asking for examples and further clarification on the setup.

Contextual Notes

Some participants mention constraints such as limited internet access affecting their ability to share their attempts. There is a focus on ensuring that the vectors are correctly represented in a Cartesian coordinate system before proceeding with calculations.

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Homework Statement


We have two vectors: one with a speed of 3 m/s to the northwest, let's call this vector a, and we also have a vector b moving to the west with a speed of 5 m/s.

Determine: a + b, a - b and a - 2b


Homework Equations


I have no idea.


The Attempt at a Solution


I tried to put the vectors in terms of unit vectors, but that didn't work, then I tried to use Pythagoras' Theorem, but that wasn't right either. I'm getting really frustrated with this problem! :cry:
 
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you need a coordinate system so that you can put vectors into component form
a = ( , ) ; b = ( , ) then addition becames simple. so first determine a set of axes and then give your vectors the appropriate coordinates before moving on. shall need some simple trig I think
 
Could you show me how to do this?
 
Basically, what you just said you tried. Since you don't show how you tried, I don't know why it "didn't work".

"one with a speed of 3 m/s to the northwest, let's call this vector a, and we also have a vector b moving to the west with a speed of 5 m/s."
Okay, so [itex]\vec{a}[/itex] has equal [itex]\vec{i}[/itex] and [itex]\vec{j}[/itex] components except that the [itex]\vec{i}[/itex] component is negative. Set it up as a right triangle with legs x and x, hypotenuse of length 3. Use the Pythagorean theorem to determine x. The vector is [itex]-x\vec{i}+ x\vec{j}[/itex].

b is due west with "length" 5 so it should be easy to write it in [itex]x\vec{i}+ y\vec{j}[/itex] form!

Once you have those two, the arithmetic is simple.
 
Last edited by a moderator:
Darkiekurdo said:
Could you show me how to do this?

ok, let me give u an example.

take the N direction as your +ve y-axis direction and E as your +ve x-axis direction. then a velocity vector pointing at S with magnitude 2m/s has a vector form based on this set of coord sys of
v=(0,-2)
and for a velocity vector pointing at say SW with magnitude [tex]\sqrt{2}[/tex] m/s has
u=(-1,-1)
 
I'm sorry I didn't show how I did it, but I don't have internet right now. I will show how I did it when I have internet.
 
As everyone else has said you need to find the components of the vectors and then the calculations are fairly easy. In this situation south-north will be the y-axis and west to east will be the x-axis of a cartesian coordinate system. The unit vectors will then be [itex]\mathbf{\hat{i}}[/itex] and [itex]\mathbf{\hat{j}}[/itex]. You know the magnitude of the vectors so you can work out the components.

[tex]|\mathbf{a}|=\sqrt{x^2+y^2}[/tex]

For vector a you know that the x and y components must be the same and for vector b you know the y component is 0.

Post your attempt when you get a chance.
 

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