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\begin{document}
% TTh class
\centerline{\Large \bf Homework \vspace{.1in} \#2 Solutions}
%% MWF class
%\centerline{\Large \bf Homework \vspace{.1in} \#3 Solutions}
\centerline{\em Math 311}
%\centerline{\bf Autumn \vspace{0.8in} 2000}
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\vspace{5mm}
\noindent{\bf p. 51, \#3:} Find the area of the triangle with vertices
$(1,1,2),\,(2,3,5),\,(1,5,5)$.

\vspace{3mm}
\noindent{\bf Solution:} Set
\[
{\bf A}=(2-1)\I+(3-1)\J+(5-2)\K,\quad {\bf B}=(1-1)\I+(5-1)\J+(5-2)\K.
\]
Check that ${\bf A}\times{\bf B}=-6\I-3\J+4\K.$
Since the area, $A$, of the triangle is $A=(1/2)|{\bf A}\times{\bf B}|$, one gets
that $A=\sqrt{61}/2$.
\qed

\vspace{7mm}
\noindent{\bf p. 51, \#11:} By taking the vector cross product of
${\bf A}=\cos\theta\I+\sin\theta\J$ with ${\bf B}=\cos\psi\I+\sin\psi\J$ and
intepreting geometrically, derive a well-known geometric identity.

\vspace{3mm}
\noindent{\bf Solution:} It is straightforward to compute that
\[
{\bf A}\times{\bf B}=(\cos\theta\sin\psi-\cos\psi\sin\theta)\K.
\]
Furthermore, one can check that $|{\bf A}|=|{\bf B}|=1$.
Assuming that $\psi>\theta$, the positive angle
between ${\bf A}$ and ${\bf B}$ is $\phi=\psi-\theta$.
By definition,
\[
\begin{array}{lll}
{\bf A}\times{\bf B}&=&|{\bf A}|\,|{\bf B}|\sin\phi\,{\bf n} \\
&=&\sin\phi\,{\bf n},
\end{array}
\]
where ${\bf n}$ is the unit vector perpendicular to both ${\bf A}$ and ${\bf
B}$ which follows the right-hand rule.
Since $\phi>0$, one has that ${\bf n}=\K$; hence, by equating the coefficients
of the two expressions for ${\bf A}\times{\bf B}$ one gets that
\[
\sin(\psi-\theta)=\sin\phi=\cos\theta\sin\psi-\cos\psi\sin\theta.
\qed
\]

\vspace{7mm}
\noindent{\bf p. 51, \#21:} Express ${\bf B}=2\I-\J+3\K$ as the sum of a vector
parallel, plus a vector perpendicular, to ${\bf A}=2\I+4\J-2\K$.

\vspace{3mm}
\noindent{\bf Solution:} The vector parallel to ${\bf A},\,{\bf B}_{\|}$, is
given by
\[
{\bf B}_{\|}=\frac{{\bf A}\cdot{\bf B}}{{\bf A}\cdot{\bf A}}\,{\bf A}=
-\frac14\,{\bf A}=-\frac12\I-\J+\frac12\K.
\]
The vector perpendicular to ${\bf A},\,{\bf B}_{\perp}$, is given by
\[
{\bf B}_{\perp}={\bf B}-{\bf B}_{\|}=\frac52\I+\frac52\K.
\qed
\]

\vspace{7mm}
\noindent{\bf p. 57, \#2:} Find the volume of the parallelepiped whose
coterminal edges are arrows representing the vectors ${\bf A}=3\I+4\J,\,
{\bf B}=2\I+3\J+4\K,\,{\bf C}=5\K$.

\vspace{3mm}
\noindent{\bf Solution:} The volume, $V$, is given by the triple scalar
product, i.e.,
\[
V=|[{\bf A},{\bf B},{\bf C}]|=|{\bf A}\times{\bf B}\cdot{\bf C}|=5.
\qed
\]

\vspace{7mm}
\noindent{\bf p. 57, \#6:} Find the equation of the plane passing through the
origin parallel to the vectors ${\bf A}=3\I+\J-2\K$ and ${\bf B}=\I-\J+5\K$.

\vspace{3mm}
\noindent{\bf Solution:} A vector perpendicular to the plane, ${\bf n}$, is
given by
\[
{\bf n}={\bf A}\times{\bf B}=3\I-17\J-4\K.
\]
The equation of the plane is then
\[
3(x-0)-17(y-0)-4(z-0)=0,
\]
i.e., $3x-17y-4z=0$.
\qed

\vspace{7mm}
\noindent{\bf p. 60, \#11:} Simplify $[{\bf A}\times({\bf A}\times{\bf
B})]\times{\bf A}\cdot{\bf C}$.

\vspace{3mm}
\noindent{\bf Solution:} Upon using vector identity (1.30), it is seen that
\[
{\bf A}\times({\bf A}\times{\bf B})=({\bf A}\cdot{\bf B})\,{\bf A}-
({\bf A}\cdot{\bf A})\,{\bf B}.
\]
Since ${\bf A}\times{\bf A}={\bf 0}$, one then gets that
\[
\begin{array}{llr}
[{\bf A}\times({\bf A}\times{\bf B})]\times{\bf A}&=&
-({\bf A}\cdot{\bf A})\,{\bf B}\times{\bf A}\,\, \\
&=&({\bf A}\cdot{\bf A})\,{\bf A}\times{\bf B}.
\end{array}
\]
Upon using the definition of the scalar triple product, one finally sees that
\[
[{\bf A}\times({\bf A}\times{\bf B})]\times{\bf A}\cdot{\bf C}=
({\bf A}\cdot{\bf A})\,[{\bf A},{\bf B},{\bf C}].
\qed
\]

\end{document}
