Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
add comment

2 Answers

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 $a/\sqrt{2}$.

share|improve this answer
    
but I think you are wrong read more:artofproblemsolving.com/Forum/… –  LevanDokite Aug 21 '12 at 22:00
    
can anyone help me :D –  LevanDokite Aug 23 '12 at 14:57
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.