The base of a triangular prism is $ABC$. $A'B'C'$ is an equilateral triangle with lengths $a$... The base of a triangular prism is $ABC$. $A'B'C'$ is an equilateral triangle with lengths $a$, and the lengths of its adjacent sides also equal $a$. Let $I$ be the midpoint of $AB$ and $B'I \perp (ABC)$. Find the distance from the $B'$ to the plane ($ACC'A'$) in term of $a$.
 A: Let $BB'$ be along $x$ axis and $BC$ be along $y$ axis ($B$ being the origin). Given that $B'I$ is perpendicular to $BA$, $\angle{ABC}$ will be $\pi/3$ (as $\Delta BIB'$ is a $(1,\sqrt{3},2)$ right-triangle). The co-ordinates of $A$ will then be of the form $\left(a\cos{\pi/3},a\cos{\pi/3},h\right)$. As the length of $AB$ is $a$, it leads to 
$$
\left(a\cos{\pi/3}\right)^{2}+\left(a\cos{\pi/3}\right)^{2}+h^{2}=a^{2} \\
h=\frac{a}{\sqrt{2}}
$$
Due to symmetry the height of $A$ from plane $BCC'B'$ is the same as $B'$ to $ACC'A'$, which is $a/\sqrt{2}$.
A: Assume
$$I=(0,0,0),\quad A=(-{a\over2},0,0), \quad B=({a\over2},0,0), \quad C=(0,{\sqrt{3}\over2}a,0)\ .$$
Then 
$$ B'=(0,0,{\sqrt{3}\over2}a),\quad AA'=BB'=(-{a\over2},0,{\sqrt{3}\over2}a)\ .$$
The normal $n$ to the plane $\pi$ containing $A$,$C$, $A'$, $C'$ is then parallel to
$$AA'\times AC=(-{a\over2},0,{\sqrt{3}\over2}a)\times({a\over2},{\sqrt{3}\over2}a,0)=(-{3\over4}a^2,{\sqrt{3}\over4}a^2,-{\sqrt{3}\over4}a^2)\ ,$$
so that we may take $n=(\sqrt{3},-1,1)$. The equation of the plane $\pi$ then reads $n\cdot r=n\cdot A$, where $r=(x,y,z)$ denotes the generic point on $\pi$.
We now have to solve the equation
$$n\cdot(B'+t n)=n\cdot A$$
for $t$ and obtain
$$t={n\cdot B'A\over n\cdot n}=-{\sqrt{3}\over 5}a\ .$$
The quantity we are looking for is
$$|t n|=\sqrt{3\over5}\>a\ .$$
