2006 Paper 1 Q8

Year: 2006
Paper: 1
Question Number: 8

Course: LFM Pure
Section: Introduction to trig

Difficulty: 1500.0 Banger: 1484.0

Problem

{\it Note that the volume of a tetrahedron is equal to \(\frac1 3\) \(\times\) the area of the base \(\times\) the height.} The points \(O\), \(A\), \(B\) and \(C\) have coordinates \((0,0,0)\), \((a,0,0)\), \((0,b,0)\) and \((0,0,c)\), respectively, where \(a\), \(b\) and \(c\) are positive.
  1. Find, in terms of \(a\), \(b\) and \(c\), the volume of the tetrahedron \(OABC\).
  2. Let angle \(ACB = \theta\). Show that \[ \cos\theta = \frac {c^2} { { \sqrt{\vphantom{ \dot b} (a^2+c^2)(b^2+c^2)} } ^{\vphantom A} \ } \] and find, in terms of \(a\), \(b\) and \(c\), the area of triangle \(ABC\). % is %\(\displaystyle \tfrac12 \sqrt{ \vphantom{\dot A } a^2b^2 +b^2c^2 + c^2 a^2 \;} \;\).
Hence show that \(d\), the perpendicular distance of the origin from the triangle \(ABC\), satisfies \[ \frac 1{d^2} = \frac 1 {a^2} + \frac 1 {b^2} + \frac 1 {c^2} \,. \]

No solution available for this problem.

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Problem source
{\it Note that the volume of a 
tetrahedron is equal to  $\frac1  3$ $\times$ 
the area of the base $\times$ the height.}
The points $O$, $A$, $B$ and $C$ have coordinates $(0,0,0)$, $(a,0,0)$, $(0,b,0)$
and $(0,0,c)$, respectively, where $a$, $b$ and $c$ are positive.
\begin{questionparts}
\item Find, in terms of $a$, $b$ and $c$,  the volume of the tetrahedron
  $OABC$.
\item
 Let angle $ACB = \theta$. Show that
\[
\cos\theta = 
\frac {c^2}
{ 
{ \sqrt{\vphantom{ \dot b}
(a^2+c^2)(b^2+c^2)} }
^{\vphantom A}    
\  }
\]
and find, in terms of $a$, $b$ and $c$, the  area of triangle $ABC$.
% is 
%$\displaystyle \tfrac12 \sqrt{ \vphantom{\dot A } a^2b^2 +b^2c^2 + c^2 a^2 \;} \;$.
\end{questionparts}
Hence show that $d$,  the perpendicular distance of the origin from
the  triangle $ABC$, satisfies
\[
\frac 1{d^2} = \frac 1 {a^2} + \frac 1 {b^2} + \frac 1 {c^2} \,.
\]