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\noindent {\bf 21-259 \hfill Calculus in 3-D \hfill Spring 2003\\

\vspace{.15in}

\centerline{\bf Answers to Review \#1} 

\vspace{.15in} 

\begin{enumerate}
\item \begin{enumerate}
\item  $\sqrt{11}$  
\item  $-1$   
\item $\theta = \cos^{-1}\left(1 / \sqrt{11}\ \sqrt{35}\right)$
\item ${\displaystyle \frac{10}{11}} \left<1,3,-1\right>$
\item $\cos \alpha = 5/\sqrt{35} \cos \beta = -1/\sqrt{35},\ \cos \gamma = 3/\sqrt{35}$
\item $64$ 
\end{enumerate} 

\item[2.] \begin{enumerate} 
\item $7(x-3) - 2(y+6) -3 (z-8) = 0$

\item ${\displaystyle \frac{x-1}{7}} =  {\displaystyle  \frac{y-3}{-2}} =  {\displaystyle \frac{z+5}{-3}};$\ 

$\left\{\begin{array}{ll}
x = 1+7t\\
y= 3-2t\\
z = -5-3t\end{array}\right.    $
\end{enumerate} 

\item[3.] ${\displaystyle \frac{x-1}{6}} = {\displaystyle  \frac{y-3}{-3}} = {\displaystyle \frac{z}{-3}}$

\item[4.] $d = -2 -3\sqrt{6}$ or $d = 3\sqrt{6} -2 $

\item[5.] $\sqrt{10}/\sqrt{3}$

\item[6.] $\vec{b}_1 = \left< {\displaystyle \frac{4}{5}},\ 0,\ {\displaystyle\frac{-2}{5}} \right> , \vec{b}_2 = \left< {\displaystyle \frac{1}{5}},\ 2,\ {\displaystyle \frac{2}{5}}\right>$

%%%%%%%%%%%%%%%%%5
\item[7.] $ \left( {\displaystyle \frac{-1}{2}},\ {\displaystyle \frac{7}{4}},\ {\displaystyle \frac{\sqrt{3}}{4}}\right),\ \left({\displaystyle\frac{1}{2}},\ {\displaystyle \frac{1}{4}},\ {\displaystyle \frac{-\sqrt{3}}{4}}\right)$  

\item[8.] $x -2y + z = 0 $ 


\item[9.] ${\displaystyle \frac{1}{2}}$

\item[10.] ${\displaystyle \frac{-1}{3}} \left<4,1,-5\right>$.

\item[11.] $\sqrt{\displaystyle \frac{1}{2}(A^2+B^2+C^2)}$. 

\item[12.] \begin{enumerate} 
\item $\vec{T} (t) = {\displaystyle \frac{1}{\sqrt{2}}} \left< \cos 2t,\ \sin 2t, 1\right>.$

$\vec{N} (t) = \left<- \sin 2t,\ \cos 2t, 0\right>$.

$\vec{B}(t) = {\displaystyle \frac{1}{\sqrt{2}}} \left<-\cos 2t,\ - \sin 2t,1\right>,\ \kappa (t) = e^{-t}$ 

\item 1
\end{enumerate} 
\item[13.] $2(8-2^{3/2})$.

\item[14.]\begin{enumerate} 
\item 1 

\item DNE
\end{enumerate} 


\item[15.]

\item[16.] $ \begin{tabular}{lll} 
$f_x =$ & $\cos (x-+3y) + 2xy^3$\\
$f_y = $& $3 \cos (x+3y) + 3x^2y^2 + 4y$\\
$f_{xx} =$ &$ - \sin (x+3y) + 2y^3$\\
$f_{xy} = $&$ - 3 \sin (x+3y) + 6xy^2$\\
$f_{yy} =$& $-9 \sin (x+3y) + 6x^2y + 4$
\end{tabular}$ 




\end{enumerate}






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