Electromagnetism and the principle of relativity

Question

I'm reading the book "Fundamental Physics 2: Electromagnetism" by Alonso and Finn. I understand everything up to the point where everything is "unified".

The following example is given in the book:

Take 2 observers $O$ and $O'$ that are moving with a constant velocity $v$ relative to each other and 2 charges $q$ and $Q$ that are at rest relative to $O'$. For observer $O'$ there's only an electrical interaction between $Q$ and $q$; he measures the force $\vec{F'}=q\vec{E'}$ where $\vec{E'}$ is the electrical field intensity caused by $Q$ in $q$, as measured by $O'$. Because $O$ sees the charge $Q$ moving, he observes an electrical field $\vec{E}$ and a magnetic field $\vec{B}$ caused by $Q$ and because he sees $q$ move with a velocity $\vec{v}$, the force exerted on $q$ by $Q$ according to $O$ equal to $\vec{F} = q(\vec{E}+\vec{v}\times\vec{B})$.

Now a bit further it says that we choose the x and x' axis so that $\vec{v}=\vec{e_x}v$.
Because of this: $\vec{v}\times\vec{B} = -\vec{e_y}v\vec{B_z}+\vec{e_z}v\vec{B_y}$.

I don't understand how this last conclusion is made.

Answer 1

It is immediate from the definition of the cross product. Write

$$\mathbf{B} = B_x \mathbf{e_x} + B_y \mathbf{e_y} + B_z \mathbf{e_z}$$

and use that

$$\mathbf{e_x} \times \mathbf{e_y} = \mathbf{e_z}$$

and

$$\mathbf{e_x} \times \mathbf{e_z} = -\mathbf{e_y}$$

If you can't get an identity by applying the physics it's sometimes useful to make sure you haven't missed some simple maths!