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The base of a triangular prism ABC.A'B'C' is an equilateral triangle with lengths a,

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$begingroup$

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

geometry

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asked Aug 19 '12 at 13:09

LevanDokiteLevanDokite

789

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add a comment | 

1

1

$begingroup$

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

geometry

share|cite|improve this question

asked Aug 19 '12 at 13:09

LevanDokiteLevanDokite

789

$endgroup$

add a comment | 

$begingroup$

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

geometry

share|cite|improve this question

asked Aug 19 '12 at 13:09

LevanDokiteLevanDokite

789

$endgroup$

share|cite|improve this question

asked Aug 19 '12 at 13:09

LevanDokiteLevanDokite

789

share|cite|improve this question

asked Aug 19 '12 at 13:09

LevanDokiteLevanDokite

789

asked Aug 19 '12 at 13:09

LevanDokiteLevanDokite

789

asked Aug 19 '12 at 13:09

LevanDokiteLevanDokite

789

2 Answers
2

active

oldest

votes

0

$begingroup$

Let $BB'$ be along $x$ axis and $BC$ be along $y$ axis ($B$ being the origin). Given that $B'I$ is perpendicular to $BA$, $angleABC$ will be $pi/3$ (as $Delta BIB'$ is a $(1,sqrt3,2)$ right-triangle). The co-ordinates of $A$ will then be of the form $left(acospi/3,acospi/3,hright)$. As the length of $AB$ is $a$, it leads to
$$
left(acospi/3right)^2+left(acospi/3right)^2+h^2=a^2 \
h=fracasqrt2
$$
Due to symmetry the height of $A$ from plane $BCC'B'$ is the same as $B'$ to $ACC'A'$, which is $a/sqrt2$.

share|cite|improve this answer

answered Aug 19 '12 at 17:25

kashyap1123kashyap1123

36625

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  but I think you are wrong read more:artofproblemsolving.com/Forum/…
  $endgroup$
  – LevanDokite
  Aug 21 '12 at 22:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  can anyone help me :D
  $endgroup$
  – LevanDokite
  Aug 23 '12 at 14:57
  

  
  
  
  

add a comment | 

0

$begingroup$

Assume
$$I=(0,0,0),quad A=(-aover2,0,0), quad B=(aover2,0,0), quad C=(0,sqrt3over2a,0) .$$
Then
$$ B'=(0,0,sqrt3over2a),quad AA'=BB'=(-aover2,0,sqrt3over2a) .$$
The normal $n$ to the plane $pi$ containing $A$,$C$, $A'$, $C'$ is then parallel to
$$AA'times AC=(-aover2,0,sqrt3over2a)times(aover2,sqrt3over2a,0)=(-3over4a^2,sqrt3over4a^2,-sqrt3over4a^2) ,$$
so that we may take $n=(sqrt3,-1,1)$. The equation of the plane $pi$ then reads $ncdot r=ncdot A$, where $r=(x,y,z)$ denotes the generic point on $pi$.
We now have to solve the equation
$$ncdot(B'+t n)=ncdot A$$
for $t$ and obtain
$$t=ncdot B'Aover ncdot n=-sqrt3over 5a .$$
The quantity we are looking for is
$$|t n|=sqrt3over5>a .$$

share|cite|improve this answer

answered Oct 17 '13 at 12:34

Christian BlatterChristian Blatter

176k8115328

$endgroup$

add a comment | 

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0

$begingroup$

Let $BB'$ be along $x$ axis and $BC$ be along $y$ axis ($B$ being the origin). Given that $B'I$ is perpendicular to $BA$, $angleABC$ will be $pi/3$ (as $Delta BIB'$ is a $(1,sqrt3,2)$ right-triangle). The co-ordinates of $A$ will then be of the form $left(acospi/3,acospi/3,hright)$. As the length of $AB$ is $a$, it leads to
$$
left(acospi/3right)^2+left(acospi/3right)^2+h^2=a^2 \
h=fracasqrt2
$$
Due to symmetry the height of $A$ from plane $BCC'B'$ is the same as $B'$ to $ACC'A'$, which is $a/sqrt2$.

share|cite|improve this answer

answered Aug 19 '12 at 17:25

kashyap1123kashyap1123

36625

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  but I think you are wrong read more:artofproblemsolving.com/Forum/…
  $endgroup$
  – LevanDokite
  Aug 21 '12 at 22:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  can anyone help me :D
  $endgroup$
  – LevanDokite
  Aug 23 '12 at 14:57
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $BB'$ be along $x$ axis and $BC$ be along $y$ axis ($B$ being the origin). Given that $B'I$ is perpendicular to $BA$, $angleABC$ will be $pi/3$ (as $Delta BIB'$ is a $(1,sqrt3,2)$ right-triangle). The co-ordinates of $A$ will then be of the form $left(acospi/3,acospi/3,hright)$. As the length of $AB$ is $a$, it leads to
$$
left(acospi/3right)^2+left(acospi/3right)^2+h^2=a^2 \
h=fracasqrt2
$$
Due to symmetry the height of $A$ from plane $BCC'B'$ is the same as $B'$ to $ACC'A'$, which is $a/sqrt2$.

share|cite|improve this answer

answered Aug 19 '12 at 17:25

kashyap1123kashyap1123

36625

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  but I think you are wrong read more:artofproblemsolving.com/Forum/…
  $endgroup$
  – LevanDokite
  Aug 21 '12 at 22:00
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  can anyone help me :D
  $endgroup$
  – LevanDokite
  Aug 23 '12 at 14:57
  

  
  
  
  

add a comment | 

$begingroup$

Let $BB'$ be along $x$ axis and $BC$ be along $y$ axis ($B$ being the origin). Given that $B'I$ is perpendicular to $BA$, $angleABC$ will be $pi/3$ (as $Delta BIB'$ is a $(1,sqrt3,2)$ right-triangle). The co-ordinates of $A$ will then be of the form $left(acospi/3,acospi/3,hright)$. As the length of $AB$ is $a$, it leads to
$$
left(acospi/3right)^2+left(acospi/3right)^2+h^2=a^2 \
h=fracasqrt2
$$
Due to symmetry the height of $A$ from plane $BCC'B'$ is the same as $B'$ to $ACC'A'$, which is $a/sqrt2$.

share|cite|improve this answer

answered Aug 19 '12 at 17:25

kashyap1123kashyap1123

36625

$endgroup$

share|cite|improve this answer

answered Aug 19 '12 at 17:25

kashyap1123kashyap1123

36625

answered Aug 19 '12 at 17:25

kashyap1123kashyap1123

36625

answered Aug 19 '12 at 17:25

kashyap1123kashyap1123

36625

0

$begingroup$

Assume
$$I=(0,0,0),quad A=(-aover2,0,0), quad B=(aover2,0,0), quad C=(0,sqrt3over2a,0) .$$
Then
$$ B'=(0,0,sqrt3over2a),quad AA'=BB'=(-aover2,0,sqrt3over2a) .$$
The normal $n$ to the plane $pi$ containing $A$,$C$, $A'$, $C'$ is then parallel to
$$AA'times AC=(-aover2,0,sqrt3over2a)times(aover2,sqrt3over2a,0)=(-3over4a^2,sqrt3over4a^2,-sqrt3over4a^2) ,$$
so that we may take $n=(sqrt3,-1,1)$. The equation of the plane $pi$ then reads $ncdot r=ncdot A$, where $r=(x,y,z)$ denotes the generic point on $pi$.
We now have to solve the equation
$$ncdot(B'+t n)=ncdot A$$
for $t$ and obtain
$$t=ncdot B'Aover ncdot n=-sqrt3over 5a .$$
The quantity we are looking for is
$$|t n|=sqrt3over5>a .$$

share|cite|improve this answer

answered Oct 17 '13 at 12:34

Christian BlatterChristian Blatter

176k8115328

$endgroup$

add a comment | 

0

$begingroup$

Assume
$$I=(0,0,0),quad A=(-aover2,0,0), quad B=(aover2,0,0), quad C=(0,sqrt3over2a,0) .$$
Then
$$ B'=(0,0,sqrt3over2a),quad AA'=BB'=(-aover2,0,sqrt3over2a) .$$
The normal $n$ to the plane $pi$ containing $A$,$C$, $A'$, $C'$ is then parallel to
$$AA'times AC=(-aover2,0,sqrt3over2a)times(aover2,sqrt3over2a,0)=(-3over4a^2,sqrt3over4a^2,-sqrt3over4a^2) ,$$
so that we may take $n=(sqrt3,-1,1)$. The equation of the plane $pi$ then reads $ncdot r=ncdot A$, where $r=(x,y,z)$ denotes the generic point on $pi$.
We now have to solve the equation
$$ncdot(B'+t n)=ncdot A$$
for $t$ and obtain
$$t=ncdot B'Aover ncdot n=-sqrt3over 5a .$$
The quantity we are looking for is
$$|t n|=sqrt3over5>a .$$

share|cite|improve this answer

answered Oct 17 '13 at 12:34

Christian BlatterChristian Blatter

176k8115328

$endgroup$

add a comment | 

$begingroup$

Assume
$$I=(0,0,0),quad A=(-aover2,0,0), quad B=(aover2,0,0), quad C=(0,sqrt3over2a,0) .$$
Then
$$ B'=(0,0,sqrt3over2a),quad AA'=BB'=(-aover2,0,sqrt3over2a) .$$
The normal $n$ to the plane $pi$ containing $A$,$C$, $A'$, $C'$ is then parallel to
$$AA'times AC=(-aover2,0,sqrt3over2a)times(aover2,sqrt3over2a,0)=(-3over4a^2,sqrt3over4a^2,-sqrt3over4a^2) ,$$
so that we may take $n=(sqrt3,-1,1)$. The equation of the plane $pi$ then reads $ncdot r=ncdot A$, where $r=(x,y,z)$ denotes the generic point on $pi$.
We now have to solve the equation
$$ncdot(B'+t n)=ncdot A$$
for $t$ and obtain
$$t=ncdot B'Aover ncdot n=-sqrt3over 5a .$$
The quantity we are looking for is
$$|t n|=sqrt3over5>a .$$

share|cite|improve this answer

answered Oct 17 '13 at 12:34

Christian BlatterChristian Blatter

176k8115328

$endgroup$

share|cite|improve this answer

answered Oct 17 '13 at 12:34

Christian BlatterChristian Blatter

176k8115328

answered Oct 17 '13 at 12:34

Christian BlatterChristian Blatter

176k8115328

answered Oct 17 '13 at 12:34

Christian BlatterChristian Blatter

176k8115328