I was reading Zee's QFT book. At the very beginning of the chapter 2 on Dirac equation, he mentions, that Dirac desired for a relativistic wave equation that is linear in spacetime gradient. He goes on to say that for that to happen, a term of the form $c^\mu \partial_\mu$ was required. I understand this. But then  he says, had the coefficients $c^\mu$ been numbers, this would form a vector and the term, $c^\mu \partial_\mu$ will no longer be Lorentz invariant. I understand why $c^\mu$ cannot be just numbers, but, I don't understand why the term would violate Lorentz invariance, had $c^\mu$ been a vector with the components being simple numbers. What I am looking for is to understand why $c^\mu\partial_\mu$ would violate Lorentz invariance, when $c^\mu$ is vector s formed from simple numbers.
I have a question about the following passage on pg. 89 of Zee's QFT in a nutshell:     At first sight, what Dirac wanted does not make sense. The equation is supposed to have the form "some linear combination of $\partial_\mu$ acting on some field $\psi$ is equal to some constant times the field." Denote the linear combination by $c^\mu\partial_\mu$. If the $c^\mu$'s are four ordinary numbers, then the four-vector $c^\mu$ defines some direction and the equation cannot be Lorentz invariant.   The simplest equation I can construct that he is referring to is  $$(c^{\mu}\partial_{\mu}-m)\phi(x)=0, $$  for $m&gt;0$. Taking $\phi(x)\rightarrow\phi(\Lambda^{-1}x)$, and $c^{\mu}\rightarrow\Lambda^{\mu}_{\ \ \nu}c^{\nu}$. Then, the left term transforms like   \begin{align*} \Lambda^{\mu}_{\ \ \nu}c^{\nu}(\Lambda^{-1})^{\sigma}_{\ \ \mu}\partial_{\sigma}\phi(\Lambda^{-1}x) &amp;= \delta^{\sigma}_{\ \ \nu}c^{\nu}\partial_{\sigma}\phi(\Lambda^{-1}x)\\ &amp;=c^{\sigma}\partial_{\sigma}\phi(\Lambda^{-1}x). \end{align*}  Then, we have that   $$ (c^{\sigma}\partial_{\sigma}-m)\phi(\Lambda^{-1}x)=0.$$  But doesn't this show that the equation is Lorentz-invariant?
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If $\vec{a},\vec{b},\vec{c},\vec{d}$are unit vectors, $(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})=1$ and $\vec{a}.\vec{c}=\frac{1}{2}$,then If $\vec{a},\vec{b},\vec{c},\vec{d}$are unit vectors such that $(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})=1$ and $\vec{a}.\vec{c}=\frac{1}{2}$,then $(A)\vec{a},\vec{b},\vec{c}$ are non-coplanar $(B)\vec{a},\vec{b},\vec{d}$ are non-coplanar $(C)\vec{b},\vec{d}$ are non-parallel. $(D)\vec{a},\vec{d}$ are parallel and $\vec{b}$ and $\vec{c}$ are parallel.    My Attempt: $(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})=1$ $(\vec{a}.\vec{c})(\vec{b}.\vec{d})-(\vec{a}.\vec{d})(\vec{b}.\vec{c})=1$ $\frac{1}{2}(\vec{b}.\vec{d})-(\vec{a}.\vec{d})(\vec{b}.\vec{c})=1$    Book's Solution which i did not understand: $(\vec{a}\times\vec{b}).(\vec{c}\times\vec{d})=1$ is possible only when $|\vec{a}\times\vec{b}|=|\vec{c}\times\vec{d}|=0$ and $(\vec{a}\times\vec{b})$ is parallel to $(\vec{c}\times\vec{d})$ And the correct option given is $(C)$.  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