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QUESTION
\begin{description}

\item[(a)]
Find a unit normal to the surface $xyz+3x^2=4$ at the point
(1,1,1).

\item[(b)]
Find the equation of the tangent plane to the surface
$3yz^2+2x^2-4xy^2=3$ at the point $(0,1,1).$

\item[(c)]
Find the angle between the surfaces $2x^2+y^2+z^2=4$ and
$x^2+y^2-z=1$ at the point $(1,1,1).$

\end{description}

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ANSWER
\begin{description}

\item[(a)]
$f=xyz+3x^2-4\ \ \textbf{x}_0=(1,1,1)\\ \nabla f=(yz+6x,xz,xy)\ \
\nabla f|_{\mathbf{x}_0}=(7,1,1)\ \ \textbf{n}=\pm
\frac{1}{\sqrt{51}}(7,1,1)$

\item[(b)]
 $f=3yz^2+2x^2-4xy^2-3\ \ \textbf{x}_0=(0,1,1)\\
 \nabla f=(4x-4y^2,3z^2-4x,6yz)\ \
 \nabla f|_{\mathbf{x}_0}=(-4,3,6)$\\
 Equation of tangent plane is
 $(\textbf{x}-\textbf{x}_0)\cdot\textbf{n}=0$

\setlength{\unitlength}{.5in}

\begin{center}
\begin{picture}(5,3)
\put(0,0){\line(1,0){3}} \put(0,0){\line(2,3){1}}
\put(3,0){\line(2,3){1}} \put(1,1.5){\line(1,0){3}}
\put(2.4,.9){$\bullet$} \put(2.5,1){\vector(-2,-1){1}}
\put(2.5,1){\vector(0,1){1.5}} \put(2.6,2.4){\bf n}
\put(2,0.5){$\mathbf{x}-\mathbf{x}_0$}
\end{picture}
\end{center}
\medskip

 $-4(x-0)+3(y-1)+6(z-1)=0\\
 -4x+3y+6z=9$

\item[(c)]
 $f_1=2x^2+y^2+z^2-4\ \ f_2=x^2+y^2-z-1\ \ \textbf{x}_0=(1,1,1)\\
 \nabla f_1=(4x,2y,2z)\ \
 \textbf{n}_1=\frac{1}{\sqrt{24}}(4,2,2)=\frac{1}{\sqrt{6}}(2,1,1)\\
 \nabla f_2=(2x,2y,-z)\
 \textbf{n}_2=\frac{1}{3}(4,2,2)$\\
 Angle between surfaces = angle between normals = $\theta\\
 \cos \theta
 =\textbf{n}_1\cdot\textbf{n}_2=\frac{1}{\sqrt{6}}\frac{1}{3}5\ \\
 \theta=\cos^{-1}\left(\frac{5}{3\sqrt{6}}\right)$

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