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The base of a triangular prism 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

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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 $a/\sqrt{2}$.

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but I think you are wrong read… – LevanDokite Aug 21 '12 at 22:00
can anyone help me :D – LevanDokite Aug 23 '12 at 14:57

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 $$|t n|=\sqrt{3\over5}\>a\ .$$

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