Finding Vector Length & Cosine of Angle for Fixed \theta

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songoku
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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?
 


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)
 
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Hi lanedance

Ahh I get it now

Thanks a lot for you both !:smile: