\documentclass[a4paper,12pt]{article}
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{\bf Question}

Show that the triangle with vertices $(4,0,5)$, $(3,6,4)$ and
$(1,2,3)$ has a right angle. (HINT: consider the edges of the
triangle as vectors and calculate the angle between them.)


{\bf Answer}

Let the triangle have vertices $A$, $B$ and $C$:


\setlength{\unitlength}{.5in}
\begin{picture}(10,4)
\put(3,0){\line(1,0){5}} \put(3,0){\line(1,1){3}}
\put(8,0){\line(-2,3){2}} \put(1,0){$C=(1,2,3)$}
\put(8.2,0){$A=(4,0,5)$} \put(6.1,3.1){$B=(3,6,4)$}
\put(3.5,0.1){$\theta_3$} \put(7.4,0.1){$\theta_1$}
\put(5.9,2.5){$\theta_2$}
\end{picture}

\vspace{.2in} The vertices $A$, $B$ and $C$ have position vectors

$\left( \begin{array}{c}
     4 \\
     0 \\
     5
\end{array}  \right),\ \left( \begin{array}{c}
     3 \\
     6 \\
     4
\end{array}  \right),\ \left( \begin{array}{c}
     1 \\
     2 \\
     3
\end{array}  \right)$ respectively.

Find the vectors corresponding to the edges of the triangle:

$\stackrel{\longrightarrow}{AB}= \left( \begin{array}{c}
     3 \\
     6 \\
     4
\end{array}  \right)-
\left( \begin{array}{c}
     4 \\
     0 \\
     5
\end{array}  \right)=
\left( \begin{array}{r}
     -1 \\
     6 \\
     -1
\end{array}  \right)\ \ \ \ \ \ \stackrel{\longrightarrow}{BA}=
\left( \begin{array}{r}
     1 \\
     -6 \\
     1
\end{array}  \right)$

$\stackrel{\longrightarrow}{AC}= \left( \begin{array}{c}
     1 \\
     2 \\
     3
\end{array}  \right)-
\left( \begin{array}{c}
     4 \\
     0 \\
     5
\end{array}  \right)=
\left( \begin{array}{r}
     -3 \\
      2 \\
     -2
\end{array}  \right)\ \ \ \ \ \ \stackrel{\longrightarrow}{CA}=
\left( \begin{array}{r}
     3 \\
     -2 \\
     2
\end{array}  \right)$

$\stackrel{\longrightarrow}{BC}= \left( \begin{array}{c}
     1 \\
     2 \\
     3
\end{array}  \right)-
\left( \begin{array}{c}
     3 \\
     6 \\
     4
\end{array}  \right)=
\left( \begin{array}{r}
     -2 \\
     -4 \\
     -1
\end{array}  \right)\ \ \ \ \ \ \stackrel{\longrightarrow}{CB}=
\left( \begin{array}{r}
     2 \\
     4 \\
     1
\end{array}  \right)$

\underline {Remember}: Two vectors ${\bf u}$ and ${\bf v}$ are
orthogonal if and only if the dot product ${\bf u}\cdot{\bf v}=0$.

Test $\displaystyle \theta_1$:
$\stackrel{\longrightarrow}{AB}\cdot\stackrel{\longrightarrow}{AC}=\left(
\begin{array}{r}
     -1 \\
     6 \\
     -1
\end{array}  \right)\cdot\left( \begin{array}{r}
     -3 \\
     2 \\
     -2
\end{array}  \right)=3+12+2=17\ne 0$ so angle $\displaystyle \theta_1$ is not a
right angle.

Test $\displaystyle \theta_2$:
$\stackrel{\longrightarrow}{BA}\cdot\stackrel{\longrightarrow}{BC}=\left(
\begin{array}{r}
     1 \\
     -6 \\
     1
\end{array}  \right)\cdot\left( \begin{array}{r}
     -2 \\
     -4 \\
     -1
\end{array}  \right)=-2+24-1=21\ne 0$ so angle $\displaystyle \theta_2$ is not a
right angle.

Test $\displaystyle \theta_3$:
$\stackrel{\longrightarrow}{CA}\cdot\stackrel{\longrightarrow}{CB}=\left(
\begin{array}{r}
     3 \\
     -2 \\
     2
\end{array}  \right)\cdot\left( \begin{array}{r}
     2 \\
     4 \\
     1
\end{array}  \right)=6-8+2=0$ so angle $\displaystyle \theta_3$ is a
right angle as required, and hence the triangle does have a right
angle.


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