Cross product (high school level) I know that the cross product has this property: $v \times u = u \times -v$
Yet I still am struggling getting the order right. I am also not sure if I fully understand the concept of normal vectors. For instance, this question in my maths book:

Line $l$ passes through point $A (-1, 1, 4)$ has has direction vector 
$${\bf d} = 
\begin{pmatrix}
        6 \\
        1 \\
        5 \\
        \end{pmatrix} .
$$
   Point $B$ has coordinates $(3,3,1)$. Plane $\Pi$ has normal vector $\bf n$, and contains the line $l$ and the point $B$.
a) Write down a vector equation for $l$
b) Explain why $\overrightarrow {AB}$ and $\bf d$ are both perpendicular to $\bf n$
c) Hence find one possible vector $\bf n$

The answer to a) is $l$: $${\bf r} = 
\begin{pmatrix}
        -1 \\
        1 \\
        4 \\
        \end{pmatrix} + \lambda
\begin{pmatrix}
        6 \\
        1 \\
        5 \\
        \end{pmatrix} 
$$
The answer to b) is less straightforward for me. I know that $$\overrightarrow {AB} =
\begin{pmatrix}
        4\\
        2\\
        -3\
        \end{pmatrix} 
$$ but I'm not sure how to answer the question. 
Normally I would have calculated the scalar product to show that it equals zero if two vectors are perpendicular but I don't have the normal vector and there is no Cartesian equation of the plane $\Pi$
For c) though, I understand that the normal vector is the cross product of $\overrightarrow {AB}$ and $\bf d$, however I am unsure of the order. I did $\overrightarrow {AB}\times {\bf d}$ but the answer indicated that it should have been ${\bf d} \times \overrightarrow {AB}$.
Could someone please help me answer b) and c)?

For (b), the normal of a plane is perpendicular to all vectors that lie in that plane, by definition. Since $\Pi$ contains $l$, it contains $\mathbf d$; since it contains $A$ and $B$ it contains the vector $AB$. Then $AB$ and $\mathbf d$ are both perpendicular to $\mathbf n$.
Both $AB\times\mathbf d$ and $\mathbf d\times AB$ are fine for the cross product in (c) – they just give opposite-facing vectors. In other words, $\mathbf u\times\mathbf v=-(\mathbf v\times\mathbf u)$. These two complementary vectors correspond to the two directions a given plane's normal vector can face. You're good for this – just remember the signs!
