# Vector equations, possible to solve for x?

1. ### Jonsson

36
Hello there,

In scalar algebra, I find solving for variables a useful tool. Say ohms law, I want to find ##R## so:

$$U=RI \iff R = \frac{U}{I}$$

Can I do something analogous in vector equations? I.e. May I solve for ##\vec{\omega}## in equations using cross or dot products?

$$\vec{v} = \vec{\omega} \times \vec{r} \iff \vec{\omega} = ?$$

or:

$$\vec{\alpha} \cdot \vec{\beta} = \gamma \iff \vec{\beta} = ?$$

It would be fantastic if I could solve for vectors in some way.

Hope you are able to help.

Kind regards,
Marius

2. ### maajdl

379
Solving v=wxr makes sense, since this can be seen as solving 3 equations with 3 unknowns (each components).
You can find the solution easily by "multiplying" both sides by r: rxv = rx(wxr) = w (r.r) - r (w.r) .

Solving a.b=c makes less sense, since this can be seen as solving 1 equation with three unknowns.
Actually there is an infinite number of solutions to this equation.
If you have one solution b=bo, then you can add any vector b1 such that a.b1=0 to this solution, and you will get another solution b = bo+b1. Any b1 just need to be perpendicular to a.

3. ### Jonsson

36
What exactly do you mean by three equations?

4. ### HomogenousCow

468
With V=W X R you end up with three equations corresponding to the three components of V.
With v=W o R you end up with one equation because v is a scalar, and three unknowns for W.

5. ### jbriggs444

1,936
That's one solution. There are infinitely many others.

Given any solution w for v = w cross r and any scalar k then w + kr is another solution.

6. ### Jonsson

36

1. What is the topic called?

2. Is this correct? And how then do I solve for ##\vec{\omega}##?

$$\vec{v} = \vec{\omega} \times \vec{r} \iff \vec{r} \times \vec{v} = \vec{r} \times (\vec{\omega} \times \vec{r}) = \vec{\omega} (\vec{r}^2) - \vec{r}(\vec{r} \cdot \vec{\omega})$$
Now what?

3. Does it hold for compex vectors?

Thanks.M

7. ### HomogenousCow

468
You have a linear equation for the components of w there.

8. ### jbriggs444

1,936
The above is false. The implication is unidirectional, not bidirectional. The operation of multiplying both sides by r is not reversible.

The reason it's not reversible is that you can't uniquely divide by r. Since the purpose of the exercise was to demonstrate that you can uniquely divide by r, asserting that you can divide by r would amount to a circular argument.

9. ### HallsofIvy

40,960
Staff Emeritus
And, you have the difficulty that the solutions will seldom be unique. As obvious example if $\vec{a}\times\vec{x}= \vec{0}$ or $\vec{a}\cdot\vec{x}= 0$, $\vec{x}$ can be any vector perpendicular to $\vec{a}$