Finding Vector Length & Cosine of Angle for Fixed \theta

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Homework Help Overview

The discussion revolves around three unit vectors a, b, and c, which maintain a fixed angle θ between any two of them. Participants are tasked with finding the length of the vector v = a + b + c, determining the largest possible value of θ, and calculating the cosine of the angle β between vector a and vector v.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the expression for the length of vector v and question the conditions under which this length can exist. There is discussion about the implications of θ being less than 90 degrees and the geometric interpretation of the vectors' arrangement.

Discussion Status

Some participants have provided insights into the calculations and interpretations of the problem. There is recognition of the conditions for the existence of the vector length and the maximum angle θ, with suggestions for visualizing the vectors' arrangement. However, no consensus has been reached on the final outcomes.

Contextual Notes

Participants note that the length of v must satisfy the condition 3 + 6 cos θ ≥ 0, leading to discussions about the ranges of θ that satisfy this inequality.

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Homework Statement


Three units vectors a, b, and c have property that the angle between any two is a fixed angle [tex]\theta[/tex]

(i) find in terms of [tex]\theta[/tex] the length of the vector v = a + b + c
(ii) find the largest possible value of [tex]\theta[/tex]
(iii) find the cosine of the angle [tex]\beta[/tex] between a and v

Homework Equations


unit vector = vector with length 1unit

magnitude of vector = [tex]\sqrt{x^2+y^2+z^2}[/tex]

[tex]\cos \theta = \frac{r_1\cdot r_2}{|r_1||r_2|}[/tex]

The Attempt at a Solution


(i) I think I get it right. The answer is [tex]\sqrt{3+6\cos \theta}[/tex]

(ii) I don't know how to do this. I think [tex]\theta < 90^o[/tex] , but I can't find the exact value

(iii)
[tex]\cos \beta = \frac{a\cdot v}{|a||v|}[/tex]

After some calculation,

[tex]\cos \beta = \frac{2+\cos \theta}{\sqrt{3+6\cos \theta}}[/tex]

Can it be simplified further?

Thanks a lot
 
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i) looks ok

ii) think about the case when they are all in the same plane...

iii) shouldn't this be 1 + 2cos(theta) on the numerator?
 


Look at (i), and ask yourself for what values of theta can that length even exist? You know that v=a+b+c must be an actual vector, which means it must have an actual length
 


though v can be the zero vector, with zero length
 


Hi lanedance and Office_Shredder

Ah yes or (iii) it should be 1 + 2cos(theta). I found it but don't know why I wrote 2 + cos(theta) here...:redface:

For (ii) , The length of v can exist if :

[tex]3+6 \cos \theta \geq 0[/tex]

I found the value for [tex]\theta[/tex] = [0o, 120o] U [240o, 360o] for [tex]0^o\leq \theta \leq 360^o[/tex]

How to continue :confused:

Thanks
 


so you're pretty much there,

first though, the way to visualise this is to consider all the vectors pointing in the same direction, theta = 0. this is where |v| = 3

as the angle is increased, imagine the vectors spreading something like a flower opening, keeping the same angle between each, with |v| decreasing. The maximum angle occurs when they are all in a plane, theta = 120, and |v| = 0. Agreeing with the first range of your solution.

I also think you only need to consider upto 120 (solutions for 120<theta<= 180 do not exist, and above 180 you can just measure the angle the other way)
 
Last edited:


Hi lanedance

Ahh I get it now

Thanks a lot for you both !:smile:
 

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