Find vectors making a certain angle with given vectors

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Homework Help Overview

The discussion revolves around finding all unit vectors in R4 that make an angle of \(\pi/3\) with three given vectors: A=(1,1,-1,-1), B=(1,-1,1,-1), and C=(1,-1,-1,1). Participants are exploring the implications of the unit vector condition and the geometric relationships involved.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of the dot product and the condition for unit vectors, leading to simultaneous equations. There are questions about the derivation of specific equations and the implications of the unit vector condition on the solutions.

Discussion Status

Multiple interpretations of the equations are being explored, with some participants questioning the algebraic steps taken. There is an ongoing examination of potential solutions, including cases where certain variables are set to zero.

Contextual Notes

Participants note the importance of verifying algebraic manipulations and the implications of the unit vector condition, which adds complexity to the problem. The discussion includes consideration of multiple potential solutions and the need to clarify assumptions about the relationships between the variables.

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Homework Statement



Find all unit vectors in R4 making an angle of \pi/3 with the three vectors
A=(1,1,-1,-1)
B=(1,-1,1,-1)
C=(1,-1,-1,1)

Homework Equations


u.v=|u||v|cos\Theta

The Attempt at a Solution


using V=(w,x,y,z) as the vector we are trying to find, I solved the above equation for all three vectors A,B and C obtaining simultaneous equations to solve, however I obtained
x=y=z and w2-wx-x2=0
I am not sure how to solve this to find all unit vectors.
Also is there a method of doing this that I am unaware of?
 
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How did you get w2-wx-x2=0? Did you use the fact that V is a unit vector?
 
vela said:
How did you get w2-wx-x2=0? Did you use the fact that V is a unit vector?

I haven't yet considered the unit vector part, I was just going to divide the answer by the modulus.

I obtained this results as |u|cos\Theta is 2*1/2=1 for all the vectors A,B&C therefore right hand side of the equation for all the vectors is |v|, so we get v.u=|v| and I took

|v| = \sqrt{}w<sup>2</sup>+x<sup>2</sup>+y<sup>2</sup>+z<sup>2</sup>

I put C.u=B.u to obtain w-x-y+z=w-x+y-z which gave me that y=z

Then A.u=B.u to obtain w+x-y-z=w-x+y-z which gave me x=y

Then A.u=C.u to give w+x-y-z=w-x-y+z which gives x=z

so if you take x, y & z to all equal x (for example)

and use any of the equations we get w-x=\sqrt{}w<sup>2</sup>+3x<sup>2</sup>
from this I got the w2-wx-x2=0
 
LASmith said:
and use any of the equations we get w-x = \sqrt{w^2+3x^2}.
from this I got the w2-wx-x2=0.
Recheck your algebra. The second equation doesn't follow from the first.

Also, if you use the fact that |V|=1, the first equation will give you w=1+x.
 
Last edited:
vela said:
Recheck your algebra. The second equation doesn't follow from the first.

Also, if you use the fact that |V|=1, the first equation will give you w=1+x.


Okay, sorry I got the algebra wrong, I do not obtain w2-wx-x2=0 instead, I get w=-x. Which would be a solution, however, then fact vela pointed out w=1+x contradicts this, so what value am I supposed to use?
 
How'd you get w=-x? I got wx+x2=0. Try again!
 
vela said:
How'd you get w=-x? I got wx+x2=0. Try again!

Yes, I obtained this too, the you get x(w+x)=0
Therefore you get either x=0 or w+x=0 which leads to my answer w=-x
 
Heh. I didn't recognize the root. But there's no problem. You have two equations and two unknowns:
\begin{align*}
w+x & = 0 \\
w-x & = 1
\end{align*}
You can solve that.

Also, don't just ignore the x=0 solution as well.
 
vela said:
w+x & = 0 \\
w-x & = 1
\end{align*}
You can solve that.

Also, don't just ignore the x=0 solution as well.

therefore we get w=1/2 therefore x=-1/2 y=-1/2 and z=-1/2

Also using x=0 we get w=0, y=0 & z=0
so when it asks for all the unit vectors there must only be one, as the second one has a modulus of zero. So are they no more unit vectors apart from
(0.5,-0.5,-0.5,-0.5)?
 
  • #10
You didn't handle the x=0 case correctly. In particular, it doesn't follow that w=0 from x=0.
 
  • #11
vela said:
In particular, it doesn't follow that w=0 from x=0.

Yes, I see it now, if x=0, then the |v|=1 still so we can use the equation w-x=1.
This implies that w=1, and as x=y=z=0
We get the second unit vector (1,0,0,0)

So there are two unit vectors for this answer. Thank you for all your help :)
 

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