Exercise 18b in Schutz's First course in GR The question is as follows:

Show that a timelike vector and a non-zero null vector cannot be orthogonal.

So we have a timelike vector $\vec{A}$, s.t $\vec{A}^2<0$; and a non-zero null vector, $\vec{B}$ s.t $\vec{B}^2=0$.
So we get that $\vec{A}\cdot \vec{B} = \frac{(\vec{A}+\vec{B})^2-\vec{A}^2}{2}$.
I need to show that $\vec{A}\cdot \vec{B} \ne 0$, if it were otherwise then:
$$(\vec{A}+\vec{B})^2 = \vec{A}^2$$
How to continues from here?
 A: If $\vec{A}$ is timelike then we can find an inertial frame in which:
$$ \vec{A} = (a, 0, 0, 0) $$
Can you take it from here?
A: It is tempting to generalize OP's question to the following proposition.

Proposition. Given an $n$-dimensional vector space $V$ over a field $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$ with (possibly degenerate) indefinite sesquilinear form $\beta:V\times V \to \mathbb{F}$ of signature $(n_+,n_0,n_-)$, with  $n_+ + n_0 + n_-=n$, where it is assumed that $$n_-~=~1.$$ Then 
  $$ \forall x,y~\in~ V :\quad \beta(x,x)~\leq~0\quad \wedge\quad \beta(y,y)~\leq~0\quad \wedge\quad \beta(x,y)~=~0$$ 
  $$\quad \Rightarrow\quad  \beta(x,x)~=~0~=~\beta(y,y)\quad\vee\quad x~\in~V^{\perp}\quad\vee\quad y~\in~V^{\perp}. $$

In the non-degenerate case, the proposition states that two non-zero, mutually orthogonal, non-spacelike vectors are both lightlike. 
Sketched proof: 


*

*Pick an arbitrary (but fixed) positive definite, sesquilinear form $\langle \cdot, \cdot \rangle: V\times V \to \mathbb{F}$. 

*Find the unique Hermitian linear operator $B:V\to V$ such that 
$$\forall x,y~\in~ V:\quad\beta( x, y )~=~\langle x, By \rangle.$$

*Decompose the vector space $$V~=~V_+\oplus V_0 \oplus V_-$$ in eigenspaces for $B$, which are mutually orthogonal wrt. $\langle \cdot, \cdot \rangle$. Here $V_0=V^{\perp}$ and
$$n_+ ~=~\dim V_+,\quad n_0 ~=~\dim V_0,\quad n_- ~=~\dim V_-~=~1.$$

*Note that, in a hopefully obvious notation,
$$\beta(x,y)~=~\beta(x_+,y_+)+\beta(x_-,y_-),$$
etc.

*Assume that $x,y\notin V^{\perp}$. From assumptions, we get that
$$ 0~\leq~\beta( x_+, x_+)~\leq~ -\beta( x_-, x_-)~>~0,$$ 
$$ 0~\leq~\beta( y_+, y_+)~\leq~ -\beta( y_-, y_-)~>~0, $$ 
$$ 0~\leq~\beta( x_+, y_+)~=~ -\beta( x_-, y_-). $$

*From Cauchy-Schwarz inequality for a positive definite sesquilinear form, we get that 
$$|\beta( x_+, y_+ )|^2~\leq~\beta( x_+, x_+)\beta( y_+, y_+)$$
and
$$ \beta( x_-, x_-)\beta( y_-, y_-) ~=~|\beta( x_-, y_- )|^2.$$
The latter since $ x_- \parallel y_- $. 

*Combine the above to conclude that 
$$ \beta( x_+, x_+)~=~ -\beta( x_-, x_-)$$ 
and
$$ \beta( y_+, y_+)~=~ -\beta( y_-, y_-). $$ 
