Let plane $\alpha$ be $4x-z+1=0$. Let $r$ be a line perpendicular to plane. Which of the following could be the equation of this line? 
Consider, in a referential $0xyz$, the plane $\alpha$, defined by
$4x-z+1=0$. Let $r$ be a line that's perpendicular to the plane. Which of the
following could define $r$?
(A) $\frac x4=y\wedge z=-1$
(B) $x=4\wedge z=-1$
(C) $x-3=\frac z4 \wedge y=0$
(D) $\frac{x-3}4=-z\wedge y=1$

I thought it would be (B) because the direction vector is $(4;0;-1)$, but my book says it's (D).
Why is it not (B) and why is it (A)?
 A: Write the lines in parametric form:
(A) $x=0+4t$, $y=0+1t$, $z=-1+0t$; direction vector $(4,1,0)$
(B) $x=4+0t$, $y=0+1t$, $z=-1+0t$; direction vector $(0,1,0)$
(C) $x=0+1t$, $y=0+0t$, $z=-12+4t$; direction vector $(1,0,4)$
(D) $x=3-4t$, $y=1+0t$, $z=0+1t$; direction vector $(-4,0,1)$
The plane is parallel to $4x-z=0$, so the direction vector of the line should be orthogonal to every vector of the form $(\alpha,\beta,4\alpha)=\alpha(1,0,4)+\beta(0,1,0)$, that is, it must be orthogonal to both $(1,0,4)$ and $(0,1,0)$.
You can see that the only answer is (D).
The lines in (B) and (C) are parallel to the plane. You're computing wrongly the direction vectors.

How do we write a line in parametric form? Basically you have to determine the solutions of the linear system
\begin{cases}
x-4y=0 \\
z=-1
\end{cases}
for the first line; the complete matrix is
$$
\left[\begin{array}{ccc|c}
1 & -4 & 0 & 0 \\
0 & 0 & 1 & -1
\end{array}\right]
$$
which corresponds to a particular solution $x=0$, $y=0$, $z=-1$; the solutions of the homogeneous system are of the form
$$
t(4,1,0)
$$
so the points on the line are of the form
$$
(x,y,z)=(0,0,-1)+t(4,1,0)
$$
Do similarly for the other three lines. There are shortcuts that avoid considering the linear system, but this method surely works.
A: a line perpendicular to the plane is given by
$$[x,y,z]=[3;1;0]+t[-4;0;1]$$ and $t$ is a real number
A: Normal of the plane:  $\vec{n}$ = $(4,0,-1)$;
Let's  rewrite the lines in parameter form:
$\vec{r} = t(a,b,c) + (d,e,f) $ where
$(a,b,c) $ is the direction vector and $ (a,b,c) $ is the point it passes through for$ t=0$.
A) $x = t, y =  4t$, and $z = -1$.
$\vec{r}$= $t(1,4,0) + (0,0,-1)$.
Direction vector: $(1,4,0)$ ,  
B) $x = 4; y = t, z = -1$;
$\vec{r}$ =  $t(0,1,0) +  (4,0,-1)$
Direction vector: $ (0,1,0)$.
C) $x = t, y =0, z = 4t - 12$;
$\vec{r}$ = $t(1,0,4) +(0,0,-12)$.
Direction vector: $(1,0,4)$.
D) $x = t, y =1, z = 3/4 - t/4$.
$ \vec{r}$ = $ t(1,0,-1/4) + (0,1,3/4)$.
Direction vector: $(1,0,-1/4)$.
For a line to be perpendicular to the plane:
The direction vector is parallel to the normal of the plane 
$\vec{n} = (4,0,-1)$.
A) $(1,4,0)$ is not parallel to $\vec{n} = (4,0,-1)$.
B)and C) have direction vectors perpendicular to $ \vec{n}$.
D) Direction vector $ (1,0,-1/4)$ is parallel to $\vec{n} = (4,0,-1)$.
So it is D).
