# Homework Help: Find vectors making a certain angle with given vectors

1. Aug 8, 2011

### LASmith

1. The problem statement, all variables and given/known data

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)

2. Relevant equations
u.v=|u||v|cos$\Theta$

3. 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?

2. Aug 8, 2011

### vela

Staff Emeritus
How did you get w2-wx-x2=0? Did you use the fact that V is a unit vector?

3. Aug 9, 2011

### LASmith

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| = $\sqrt{}w2+x2+y2+z2$

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=$\sqrt{}w2+3x2$
from this I got the w2-wx-x2=0

4. Aug 9, 2011

### vela

Staff Emeritus
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: Aug 9, 2011
5. Aug 9, 2011

### LASmith

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?

6. Aug 9, 2011

### vela

Staff Emeritus
How'd you get w=-x? I got wx+x2=0. Try again!

7. Aug 9, 2011

### LASmith

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

8. Aug 9, 2011

### vela

Staff Emeritus
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.

9. Aug 9, 2011

### LASmith

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. Aug 9, 2011

### vela

Staff Emeritus
You didn't handle the x=0 case correctly. In particular, it doesn't follow that w=0 from x=0.

11. Aug 10, 2011

### LASmith

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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