# Vector equations, possible to solve for x?

#### Jonsson

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

#### maajdl

Gold Member
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.

#### Jonsson

Solving v=wxr makes sense, since this can be seen as solving 3 equations with 3 unknowns (each components).
What exactly do you mean by three equations?

#### HomogenousCow

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.

#### jbriggs444

Homework Helper
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) .
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.

#### Jonsson

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

#### HomogenousCow

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

#### jbriggs444

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

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

#### HallsofIvy

Homework Helper
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 $\vec{a}$

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