[RESOLVED] 3 variables + 3 equations... should be solvable?

[NickThissen]

Hey...

Usually I don't have any problems with this stuff but this one's got me stuck...

The question was to solve
u x x = v
for x (u and v are both known vectors)

I'm pretty sure my answer is correct but it might not be... Let's assume it is...

I made the vector x = (x, y, z).
After doing the calculation I came up with:

2z + y = 0
z + x = -1
y - 2x = 2

I know that for 3 variables, you need 3 seperate equations... So I should be able to solve this?

The problem is... I can't do it :S

I tried many things but it always ends up with something I already know!

For example:

Multiply z+x=-1 with 2 : 2z + 2x = -2

Substract it from 2z + y = 0:

2z + y = 0
2z + 2x = -2 -
-------------
y - 2x = 2 <-- And this I already knew... (it's one of the equations).


Once I ended up with 0=0 also...:
2z + y = 0 and y = 2 + 2x
2z + (2 + 2x) = 0 and z = -1 - x
-2 - 2x + 2 + 2x = 0
0 = 0 ... *Sigh*


What am I doing wrong? :S

[zaza]

The reason is, you don't really have 3 separate equations. As you have found out, your equations reduce to each other.

Combining 1) and 3) gives you -2z - 2x = 2

If 2) was anything else, then you'd now be able to solve it:

eg 2) z + 2x = 4

Now you have

z + x = -1 from 1) and 3)
z + 2x = 4 from 2)

-> 4 - x = -1
-> x = 5
-> z = -6
-> y = 12



But you don't, and hence you can't solve it.


zaza

[NickThissen]

Hm... I suppose your right... Although, how should I solve the original question then?:


Solve for x:
u x x = v

where u = i + 2j - k
and v = j + 2k

?

My answer is this:

x = xi + yj + zk


u x x =

┌ ┐ ┌ ┐ ┌ ┐ ┌ ┐
│ i j k │ │2 -1│ │1 -1│ │1 2│
│ 1 2 -1 │ = i │y z│ - j │x z│ + k │x y│
│ x y z │ └ ┘ └ ┘ └ ┘
└ ┘
= (2z + y)i - (z+x)j + (y-2x)k

This is thus equal to v = 0i + 1j + 2k

So:
2z+y = 0
z+x = -1
y-2x = 2... and where back at the first post :p

If I can't solve this... what's the answer to the question then?

[zaza]

The answer is; have another look at what you've got and what you are doing. What does the cross product do? It takes two vectors and returns a third that is perpendicular to them both in a right-handed sense.

Now, your resultant v is in the jk plane, i.e. it has no i component. Therefore it should come as no surprise that a valid vector x would be one pointing along the i axis, because any vector perpendicular to this will have no i component.

So, we try filling in z = 0 and y = 0, which gives us x = -1. All 3 equations are satisfied. And if you do the cross product you will find that it comes out exactly as you require.

Of course, there are other vectors which will solve this problem. The general rule is, when you get a situation as the above where it doesn't reduce, you can pick a value, any value, for one of the components. You have no particular solution, you have a multitude of solutions. That's why you can't reduce it any further and get a particular set of 3 values - you have reduced it to the equation that will generate any values that you need.
Try it yourself above - pick some value for x, y or z and from the equations work out what the other two are. Then fill it in and check.


zaza

[NickThissen]

I think I understand what you're saying... After looking at the original question on my paper again (I did this off the top of my head earlier lol) the question was actually "Find all solutions for x...".

But I still don't know what the actual answer is now... I thought "(2z + y)i - (z+x)j + (y-2x)k" at first, but that's v...?

So the answer is just xi + yj + zk? :S

[zaza]

No. As I was explaining before, you can pick one value and then calculate the other two. This means that the general solution can be expressed in terms of that one value. Let's say you choose x. Then your vector which is a solution is:

(x, [y in terms of x], [z in terms of x])


In this way, you are saying "Choose a value of x, and you will get the other two components, which will give you a vector that fulfils the requirements".

You could do the same in terms of y or z, but this is what the general solution means; you have found a way to determine all solutions.


zaza

[NickThissen]

Ah ofcourse, I understand now. Thanks :D