Getting a 2D triangular lattice as BCC lattice projection In a book that I am reading about atomistic simulations, it is claimed that if we take a body-centered cubic lattice in 3D, which is defined as
$$
\Lambda_1 :=A\mathbb{Z}^3,
$$
where 
$$
A = \begin{bmatrix}1/2&1/2&1/2\\-1/2&1/2&1/2\\-1/2&-1/2&1/2\end{bmatrix}
$$
and project it along $(1,1,1)$-direction, then we end with a 2D homogeneous triangular lattice, defined as
$$
\Lambda_2 := B\mathbb{Z}^2,
$$
where 
$$
B = \begin{bmatrix}1&1/2\\0&\sqrt{3}/2\end{bmatrix}.
$$
I am really bad at visualising 3D things and it's been ages since I last used any first-year geometry, so I really struggle here and would appreciate any potential help. The book says that we should take $\Lambda_1$ and rotate it so that the $(1/2,1/2,1/2)$ direction becomes $(0,0,\sqrt{3}/2)$ and that lattice point $(1,1,0)$ becomes $(\sqrt{2},0,0)$. Then apparently we get
$$
\sqrt{2}\Lambda_2 = \{(l_1,l_2)\,|\,l\in\Lambda_1\}.
$$
I am include to say that there is a typo here, as there seems to be no rotation which accommodates both conditions (I guess?). I understand the idea, as $(1/2,1/2,1/2)$ is the closest point to the origin lying on a plane defined via normal $(1,1,1)$ (that's why we call it projection along this direction?). So that after this rotation our projection along $(1,1,1)$ will just be dropping third coordinate. Using formalue available and consulting some questions here (in particular Transform a plane to the xy plane. ) I get a rotation matrix which takes $(1/2,1/2,1/2)$ to $(0,0,\sqrt{3}/2)$. What next though? My naive idea was for any other lattice site is to get a shifted plane still with normal $(1,1,1)$, i.e. $x+y+z = D$ for different constants $D$ so that a given lattice site belongs to this plane, then translate it onto the plane $x+y+z=0$, then use the rotation matrix and then translate it back to its actual place. 
I am rather lost in the procedure though and have already spent way too much time on it, so I decided to ask. Many thanks for any hints.
 A: I enjoy Mathematica visualizations, so here are some that may be helpful. First, we generate the lattice as the image of integer vectors $\{j e_1+k e_2+l e_3\}$ under the action of $A$: 
lattice=Flatten[Table[{{j, k, l}}.Transpose[{{1, 1, 1}, {-1, 1, 1}, {-1, -1, 1}}],
{j, -10, 10}, {k, -10, 10}, {l, -10, 10}], 3];

(I use row vectors here and consider the action of $A$ on the right for Mathematica convenience. The Flatten command is used to render the result as an array of row vectors.)
With the lattice at hand, we can create a few different visualizations. For instance, we can obtain the unit cell for the BCC lattice:
Show[Graphics3D[{Sphere[lattice, .15], Opacity[0.2, Blue],Cuboid[{0, 0, 0}, 2{1, 1, 1}]}],
PlotRange -> {{-.1, 1.1}, {-.1, 1.1}, {-.1, 1.1}} 2, Boxed -> False]


Or we can just take a bunch of lattice points:
Show[Graphics3D[{Sphere[lattice,.15]}], PlotRange->{{-.6, 1.1}, {-.6, 1.1}, {-.6, 1.1}} 3,
Boxed -> True]


From the first image we can already see somewhat the six-fold symmetry that we're looking for. To make this more evident, we include the option ViewPoint->{1,1,1}10 in the Show command for the second picture. (The {1,1,1}10 makes it so the viewpoint is set along the $(1,1,1)$ direction and from far away from the origin.)

From this we see that the projection onto the $(1,1,1)$ direction is indeed a hexagonal lattice.
