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\begin{document}

{\bf Question}

Find the vector product ${\bf{a}} \times {\bf{b}}$ if:

\begin{description}
\item[(i)]
${\bf{a}}={\bf{i}}-2{\bf{j}}-4{\bf{k}}\
{\bf{b}}=2{\bf{i}}-2{\bf{j}}+3{\bf{k}}$

\item[(ii)]
${\bf{a}}=-{\bf{i}}+4{\bf{j}}-{\bf{k}}\
{\bf{b}}=3{\bf{i}}+2{\bf{j}}+4{\bf{k}}$

\item[(iii)]
${\bf{a}}=-2{\bf{i}}+4{\bf{k}}\ {\bf{b}}=3{\bf{j}}-2{\bf{k}}$

\end{description}

What are the corresponding vector products ${\bf{b}} \times
{\bf{a}}$.

\medskip

{\bf Answer}
\begin{description}
\item[(i)]
\begin{eqnarray*} & & {\bf{a}}\times{\bf{b}}\\ & = & \left|\begin{array}{ccc} {\bf{i}} &
{\bf{j}} & {\bf{k}}\\ 1 & -2 & 4\\ 2 & -2 & 3\end{array}\right|\\
& = & {\bf{i}}(-2 \times 3)-{\bf{i}}(-2 \times 4)-{\bf{j}}(1
\times 3)\\ & & +{\bf{k}}(1 \times -2)+{\bf{j}}(4 \times
2)-{\bf{k}}(2 \times -2)\\ & = &
-6{\bf{i}}+8{\bf{i}}-3{\bf{j}}-2{\bf{k}}+8{\bf{j}}+4{\bf{k}}\\ & =
& 2{\bf{i}}+5{\bf{j}}+2{\bf{k}} \end{eqnarray*}

[This should be perpendicular to both ${\bf{a}}$ and ${\bf{b}}$.
Hence the dot product should be 0.

Check:

${\bf{a}} \cdot (2{\bf{i}}+5{\bf{j}}+2{\bf{k}})=(1,-2,4) \cdot
(2,5,2)=2-10+8=0 \surd$

${\bf{b}} \cdot (2{\bf{i}}+5{\bf{j}}+2{\bf{k}})=(2,-2,3) \cdot
(2,5,2)=4-10+6=0 \surd$ ]

\item[(ii)]
\begin{eqnarray*} & & {\bf{a}}\times{\bf{b}}\\ & = & \left|\begin{array}{ccc} {\bf{i}} &
{\bf{j}} & {\bf{k}}\\ -1 & 4 & -1\\ 3 & 2 & 4\end{array}\right|\\
& = & {\bf{i}}(4 \times 4)-{\bf{i}}(2 \times -1)-{\bf{j}}(-1
\times 4)\\ & & +{\bf{k}}(-1 \times -2)+{\bf{j}}(3 \times
-1)-{\bf{k}}(3 \times 4)\\ & = &
16{\bf{i}}+2{\bf{i}}+4{\bf{j}}-2{\bf{k}}-3{\bf{j}}-12{\bf{k}}\\ &
= & 18{\bf{i}}+{\bf{j}}-14{\bf{k}} \end{eqnarray*}

[again check that ${\bf{a}} \cdot
(18{\bf{i}}+{\bf{j}}-14{\bf{k}})=0$ so they're perpendicular

$\begin{array}{l} =(-1,4,-1) \cdot (18,1,-14)\\ =-18+4+14=0 \surd
\end{array}$

also: ${\bf{b}} \cdot (18{\bf{i}}+{\bf{j}}-14{\bf{k}})=0$

$\begin{array}{l} =(3,2,4) \cdot (18,1,-14)\\ =54+2-56=0 \surd
\end{array}$]

\item[(iii)]
\begin{eqnarray*} & & {\bf{a}}\times{\bf{b}}\\ & = & \left|\begin{array}{ccc} {\bf{i}} &
{\bf{j}} & {\bf{k}}\\ -2 & 0 & 4\\ 0 & 3 & -2\end{array}\right|\\
& = & {\bf{i}}(0 \times -2)-{\bf{i}}(3 \times 4)-{\bf{j}}(-2
\times -2)\\ & & +{\bf{k}}(-2 \times 3)+{\bf{j}}(0 \times
-2)-{\bf{k}}(0 \times 0)\\ & = & -12{\bf{i}}-4{\bf{j}}-6{\bf{k}}
\end{eqnarray*}

[again check that ${\bf{a}} \cdot
(-12{\bf{i}}-4{\bf{j}}-6{\bf{k}})=0$ so they're perpendicular

$\begin{array}{l} =(-2,0,4) \cdot (-12,-4,-6)\\ =24+0+-24=0 \surd
\end{array}$

also: ${\bf{b}} \cdot (-12{\bf{i}}-4{\bf{j}}-6{\bf{k}})=0$

$\begin{array}{l} =(0,3,-2) \cdot (-12,-4,-6)\\ =0-12+12=0 \surd
\end{array}$]

\end{description}
The corresponding vector products ${\bf{b}}\times{\bf{a}}$ are
just the same as above $\times -1$. Hence

\begin{description}
\item[(i)]
$({\bf{b}}\times{\bf{a}})=-12{\bf{i}}-5{\bf{j}}-2{\bf{k}}$

\item[(ii)]
$({\bf{b}}\times{\bf{a}})=-18{\bf{i}}-{\bf{j}}+14{\bf{k}}$

\item[(iii)]
$({\bf{b}}\times{\bf{a}})=12{\bf{i}}+4{\bf{j}}+6{\bf{k}}$

\end{description}

\end{document}