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{\bf Question}

\begin{description}
\item[(a)]
Let \begin{eqnarray*} {\bf a } &amp; = &amp; 2{\bf i} - 3{\bf j} + {\bf k}
\\ {\bf b} &amp; = &amp; {\bf i} - 2{\bf j} + 4{\bf k} \\ {\bf c} &amp; = &amp;
-{\bf i} + {\bf j} - 3{\bf k} \end{eqnarray*} Evaluate the
following: $${\bf a \cdot b, \hspace{.1in} a \cdot c,
\hspace{.1in} b \times c, \hspace{.1in} a \cdot b \times c }$$
$${\bf c \cdot b \times a,  \hspace{.1in}c \times (b \times c),
\hspace{.1in} a \times (b \times c)}$$

Are ${\bf a, \, b, \, c, }$ linearly independent?

\item[(b)] Find the equations of the two planes which contain the
line $$x - 5 = \frac{y-1}{-1} = \frac{z+3}{3}$$ and which make an
angle of $60^\circ$ witht the plane $y-z = 0$.

\end{description}

\vspace{.25in}

{\bf Answer}


\begin{description}
\item[(a)]
\begin{eqnarray*} {\bf a } &amp; = &amp; 2{\bf i} - 3{\bf j} + {\bf k}
\\ {\bf b} &amp; = &amp; {\bf i} - 2{\bf j} + 4{\bf k} \\ {\bf c} &amp; = &amp;
-{\bf i} + {\bf j} - 3{\bf k} \\ {} \\ {\bf a \cdot b} &amp; = &amp; 12 \\
{\bf  a \cdot c} &amp; = &amp; -8 \\ {\bf  b \times c} &amp; = &amp; (2,-1,-1) \\
{\bf  a \cdot b \times c} &amp; = &amp; 6 \\ {\bf c \cdot b \times a  =
 -a \cdot b \times c} &amp; = &amp; -6 \\ {\bf c \times (b \times c)} &amp; =
 &amp;(-4,-7,-1)  \\ {\bf  a \times (b \times c)} &amp; = &amp; (4,4,4) \end{eqnarray*}


Since ${\bf  a \times (b \times c)} \not=0 \hspace{.2in} {\bf
a,\, b, \, c,}$ are independent.

\item[(b)] Suppose the plane has equation $ax+by+cz =k$

Then $(a,b,c)\cdot (1,-1,3) =0$

So $a-b+3c =0$

Also  $5a +b -3c = k \hspace{.2in} {\rm So\ } k = 6a$

Then $(a,b,c) \cdot (0,1,-1) = b-c$

So $b - c = \sqrt{a^2 +b^2 +c^2} \cdot \sqrt 2 \frac{1}{2}$

So $2b^2 - 4bc +2c^2 = a^2 +b^2 +c^2 $

i.e. $b^2 +c^2 -a^2 -4bc = 0$

But $a = b-3c$

giving $2c(b-4c)=0$ So $c=0$ or $b=4c$

If $b=4c$ then $a = c$ and $b = 4a$

Giving $$x+4y +z =6$$

If $c = 0$ then $ a=b$

Giving $$x+y=6$$

\end{description}
\end{document}
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