Finding Vector Length & Cosine of Angle for Fixed \theta

In summary, the homework statement is that three unit vectors have a property that the angle between any two is a fixed angle \theta. lanedance and Office_Shredder try to find a solution for the equation for \cos \beta, and eventually reach a value of \cos \beta = \frac{2+\cos \theta}{\sqrt{3+6\cos \theta}}.
  • #1
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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  • #2


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?
 
  • #3


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
 
  • #4


though v can be the zero vector, with zero length
 
  • #5


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
 
  • #6


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:
  • #7


Hi lanedance

Ahh I get it now

Thanks a lot for you both !:smile:
 

FAQ: Finding Vector Length & Cosine of Angle for Fixed \theta

1. What is the formula for finding the length of a vector?

The formula for finding the length (or magnitude) of a vector is:
||v|| = √(x² + y² + z²)
where x, y, and z are the components of the vector.

2. How do you find the cosine of the angle for a fixed theta?

The cosine of the angle (θ) for a fixed theta can be calculated using the dot product of the two vectors:
cos(θ) = (v · w) / (||v|| * ||w||)
where v and w are the two vectors.

3. What is the significance of finding the vector length and cosine of angle?

Finding the vector length and cosine of angle is important in many mathematical and scientific applications, such as physics, engineering, and computer graphics. It allows us to measure the size and direction of a vector, and to calculate the angle between two vectors.

4. Can the length of a vector be negative?

No, the length of a vector cannot be negative. It is always a positive value, as it represents the distance from the origin to the tip of the vector.

5. How do you interpret the cosine of angle for two vectors?

The cosine of angle between two vectors can be interpreted as the measure of similarity or alignment between the two vectors. A cosine value of 1 indicates that the two vectors are parallel, a value of 0 indicates they are perpendicular, and a value of -1 indicates they are anti-parallel.

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