Finding Vector Length & Cosine of Angle for Fixed \theta

[songoku]

Homework Statement

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

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

Homework Equations

unit vector = vector with length 1unit

magnitude of vector = \sqrt{x^2+y^2+z^2}

\cos \theta = \frac{r_1\cdot r_2}{|r_1||r_2|}

The Attempt at a Solution

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

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

(iii)
\cos \beta = \frac{a\cdot v}{|a||v|}

After some calculation,

\cos \beta = \frac{2+\cos \theta}{\sqrt{3+6\cos \theta}}

Can it be simplified further?

Thanks a lot

 

[lanedance]

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?

 

[Office_Shredder]

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

 

[lanedance]

though v can be the zero vector, with zero length

 

[songoku]

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...

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

3+6 \cos \theta \geq 0

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

How to continue

Thanks

 

[lanedance]

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)

 

[songoku]

Hi lanedance

Ahh I get it now

Thanks a lot for you both !