Prove electric and magnetic waves are perpendicular

Question

So I have $2$ electric waves and $2$ magnetic waves, and I need to prove mathematically that the total electric and total magnetic waves are perpendicular. $E_1$ is in the $x$-direction and $E_2$ is in the $y$-direction. $B_1$ is in the $y$-direction and $B_2$ is in the $-x$-direction. The two waves are are cosines but have different phases - represented by $\phi_1$ and $\phi_2$:

\begin{align*} E_1 &= x (E)\cos(kz-wt-\phi_1)\\ E_2 &= y (E)\cos(kz-wt-\phi_2)\\ B_1 &= y (B)\cos(kz-wt-\phi_1)\\ B_2 &= -x(B)\cos(kz-wt-\phi_2) \end{align*}

The letters in front indicate the unit vector direction.

Total $E$ field is $E_t = E_1+E_2$, total magnetic field is $B_t = B_1+B_2$.

I need to show that $E_t$ and $B_t$ are perpendicular. I think to do this I need to show that their dot product is zero, but how do I take the dot product of $E_t$ and $B$?

Answer 1

You have your definition wrong. The total electric field is the vector whose components are given by $E_1$ and $E_2$. Similarly, the total magnetic field is the vector whose components are $B_1$ and $B_2$. So, the dot product will simply be $$ E \cdot B = \langle E_1,E_2 \rangle \cdot \langle B_1, B_2 \rangle = E_1 B_1 + E_2 B_2 $$ Edit: sorry, I misunderstood. You said that your $B_1$ is in the $y$ direction and your $B_2$ is in the "$-x$" direction. That would give you $$ E \cdot B = \langle E_1,E_2 \rangle \cdot \langle -B_2,B_1 \rangle = -E_1 B_2 + E_2 B_1=E_2B_1 - E_1B_2 $$ $E_1,E_2,B_1,$ and $B_2$ are scalar quantities, which is why we can multiply them. You cannot multiply vectors; you can only take the dot-product (or cross-product) of the which is not the same thing. From this point, you'll have to go through the math to show that $E_2B_1 - E_1B_2 = 0$ for any value of $t$.

Answer 2

Note: This answer is based on the actual problem as provided in a comment by the OP above.

To simplify notation, let $c_1(z,t) = \cos(kz-wt-\phi_1), c_2(z,t) = \cos(kz-wt-\phi_2)$.

The $E$ field is given by $E_0(c_1(z,t), c_2(z,t))$, or, $E_0c_1(z,t)i + E_0 c_2(z,t)) j$.

The $B$ field is given by $B_0(-c_2(z,t), c_1(z,t))$, or, $-B_0 c_2(z,t)i + B_0 c_1(z,t)) j$.

The dot or inner product is given by $E \cdot B = E_0 B_0 c_1(z,t) c_2(z,t) + E_0 B_0 (-c_2(z,t)) c_1(z,t) = 0$.