I was trying to solve this problem below:
And this is Example 5 the problem refereed to:
The solution from the solution manual is
Am I missing something or the solution manual is wrong? Shouldn't $F$ be the same as the one mentioned in Example $5$? They are dividing some of the coefficients by $c$... Now this is my attempt, this is what I got:
\begin{align*} F&= B_{x}e_{1} + B_{y} e_{2} + B_{z} e_{3} +E_{x}e_{4}+E_{y} e_{5}+E_{z}e_{6} \\ G&= \dfrac{E_{x}}{c^{2}}e_{1} + \dfrac{E_{y}}{c^{2}}e_{2}+ \dfrac{E_{z}}{c^{2}}e_{3} -B_{x}e_{4}- B_{y} e_{5} - B_{z} e_{6} \end{align*} What I did was I replaced the wedge products with $e_{n}$,so now both have the same coefficients as mentioned in the problem, then for the dot product it should be zero. Am I missing something? Is my solution correct or the one in the solution manual? Also, is it correct to assume that $c=1$ so I can definitely obtain zero when applying the dot product? Because without letting $c=1$ it cannot be zero.