I have a question that I think is meant to be bookwork..but I can't seem to find the proof anywhere and so have tried to work it out for myself.
I will write out the question and then the attempt at a proof of the first part.
Any help would be gratefully recieved!
I will denote *as powers and use ^ for the vector products to avoid confusion.
a) Suppose that a.b = 0
Find the solution r of the pair of equations:
r^a=b and r.c=n where a.c doesnt=0

Under what conditions does a solution exist if a.c=0? Find the most general solution in this case. Interpret your results geometrically.

b) Show that, if x doesnt=0 the equation r^d=xr+e has a unique solution and find it.

My attempt at a)

take vector product with a.
a^(r^a)=(a.a)r - (a.r)a=a^b

Hence r={1/|a|^2}a^b +ta where t = a.r / |a|*2
subsitute into r.c = n and we have
t =( n/a.c ) - [a,b,c]/(a.c)|a|*2

Therefore the solution is a single point.

I think that part is correct but I'm not sure, and have no idea how to continue with the problem! Please help!
Thank you!

2. Originally Posted by reeha
I have a question that I think is meant to be bookwork-..but I can't seem to find the proof anywhere and so have tried to work it out for myself.
I will write out the question and then the attempt at a proof of the first part.
Any help would be gratefully recieved!
I will denote *as powers and use ^ for the products to avoid confusion.
Since it is standard to use * as product and ^ as power, this increases the confusion!

a) Suppose that a.b = 0 [/quote]
And, now, having said you would use ^ for products, I have no idea what you could mean by a.b!
Assuming you mean 'a times b= 0" then either a= 0 or b= 0.
Find the solution r of the pair of equations:

r^a=b and r.c=n where a.c doesnt=0
if "a times c" is NOT 0, then a cannot be 0 so b must be 0. But r^a apparently means "r times a" and if that is equal to 0, r must be equal to 0.[/quote]

Under what conditions does a solution exist if a.c=0? Find the most general solution in this case. Interpret your results geometrically. [/quote]
if a.c= 0 then either a= 0 or c= 0. If c= 0 and a is not, then the problem is just as before. If a= 0, we have r^0= 0 which is false for all r except 0 so we still must have r= 0. Of course, since you have made your notation as confusing as possible, I have no idea if this is what you mean!

b) Show that, if x doesnt=0 the equation r^d=xr+e has q unique and find it.
This is a joke, right? There is NO q in your equation so we cannot say anything about it!

My attempt at a)

take vector product with a.
Vector product? This is the fist time you have said anything about vectors!

[tex]a^(r^a)=(a.a)r - (a.r)a=a^b

Hence r={1/|a|^2}a^b +ta where t = a.r / |a|*2
subsitute into r.c = n and we have
t =( n/a.c ) - [a,b,c]/(a.c)|a|*2

Therefore the solution is a single point.

I think that part is correct but I'm not sure, and have no idea how to continue with the problem! Please help!
Thank you![/QUOTE]

3. Im very sorry that the notation seems to have annoyed you, my computer doesnt seem to be able to cope with latex so I have to write it all out in normal text and there are sometimes some errors. Again, very sorry! Also I used ^ for vector product because thats the closest symbol on my keyboard to what we use in class, so therefore decided to change *to powers. Also a.c means scalar product as again I have used the closest symbol on my keyboard thinking that it would be obvious, but I appreciate now that it wasn't. Again, sorry.

If you think you might be able to help by me re-writing it out in another way I would be more than happy to do so. However if you read again I think it makes sense now.
Thank you.