Sum of 4 Vectors: Magnitude & Angle

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Homework Help Overview

The discussion revolves around the summation of four vectors in both unit-vector notation and in terms of magnitude and angle. The vectors are given in different forms, including Cartesian coordinates and polar coordinates, which adds complexity to the problem.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore how to set up the problem by converting vectors into component form and summing them. Some express confusion about the inclusion of a third dimension (k component) and question the dimensionality of the problem.

Discussion Status

Some participants have provided hints and guidance on how to approach the summation of the vectors, while others are still grappling with the setup and the dimensional aspects of the problem. There is a mix of understanding and confusion regarding the calculations and the representation of the vectors.

Contextual Notes

There is a noted discrepancy regarding the dimensionality of the vectors, as some participants question the presence of a k component when the original vectors appear to be two-dimensional. Additionally, the angles provided for vectors B and D are discussed in relation to their Cartesian components.

Kp0684
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What is the sum of the following four vectors in (a) unit- vector notation, and as (b) a magnitude and (c) an angle?... A=(2.00m)i + (3.00m)j...B: 4.00m, at +65.0 degrees...C= (-4.00m)i - (6.00m)j...D: 5.00m, at -235 degrees...i understand how to get the magnitude and the angle but how would i set this up... would i start with A and C and find their sum...i believe what's confusing me is B and D...otherwise i know how to set it up...need help...
 
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HINT:Write all 4 vectors in component form.

Then simply add the 4 numbers for the "x" component and the 4 numbers for the "y" component.

Daniel.
 
Kp0684 said:
What is the sum of the following four vectors in (a) unit- vector notation, and as (b) a magnitude and (c) an angle?... A=(2.00m)i + (3.00m)j...B: 4.00m, at +65.0 degrees...C= (-4.00m)i - (6.00m)j...D: 5.00m, at -235 degrees...i understand how to get the magnitude and the angle but how would i set this up... would i start with A and C and find their sum...i believe what's confusing me is B and D...otherwise i know how to set it up...need help...


Here are the formula's you will need to apply.

Given two vectors with components A = (i,j,k) and B = (a,b,c)

magnitude [tex]\sqrt{a^2 + b^2 + c^2}[/tex]
scalar product: A.B = ia + b j + ck
scalar product: A.B = magnitude of A * magnitude of B * cos(t) where t is the angle between the two vectors A and B

sum A+B = (i+a,j+b,k+c) (this is a new vector, the scalar product yields a number)
marlon
 
okay, i get -2.00i - 3.00j - 1.17k when i sum it up...iam still lost on this one...need help again...
 
i, j and k represent the 3 directions of the space, for example high, wide and length. The object will be placed this 3 distances in the space from the point you consider as reference
 
I'm very confused! Where did "k" come from? You original post said
"... A=(2.00m)i + (3.00m)j...B: 4.00m, at +65.0 degrees...C= (-4.00m)i - (6.00m)j...D: 5.00m, at -235 degrees..." with only two dimensions.

I presume that the angles are measured relative to the positive x axis.
B would be 4cos(65)i+ 4sin(65)j and D would be 5cos(-235)i+ 5sin(-235)j

Adding those four vectors does not give you any "k" component.

Of course, once you have found the sum, you find the length by the Pythagorean theorem and the angle is arctan((j component)/(x component)).
 

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