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\newcommand{\ds}{\displaystyle}
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\begin{document}

{\bf Question}

Find the angle between the planes $2x+6y+z=4$ and $x+3y+2z=3$.

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{\bf Answer}

$2x+6y+z=4$ can be written:

$(x{\bf{i}}+y{\bf{j}}+z{\bf{k}}) \cdot
(2{\bf{i}}+6{\bf{j}}+{\bf{k}})=4$

or ${\bf{r}} \cdot {\bf{n}}_1=D\ \ (1)$

${\bf{r}}=(x{\bf{i}}+y{\bf{j}}+z{\bf{k}}),\
{\bf{n}}_1=(2{\bf{i}}+6{\bf{j}}+{\bf{k}}),\ D=4$

Now we know ${\bf{r}} \cdot \hat {\bf{n}}_1=d\ \ (2)$ is the
vector equation of a plane with unit normal $\hat{\bf{n}}_1$. Thus
comparing $(1)$ and $(2)$ we have with
$\hat{\bf{n}}_1|{\bf{n}}_1|={\bf{n}}_1,\
\ds\frac{d}{|{\bf{n}}|}=d$.

In any case ${\bf{n}}_1$ is normal to the plane $2x+6y+z=4$.
Similarly ${\bf{n}}_2={\bf{i}}+3{\bf{j}}+2{\bf{k}}$ is normal to
the plane $x+3y+2z=3$.

The angle between two planes is given by $\theta$, PICTURE
\vspace{1in}

where ${\bf{n}}_1 \cdot
{\bf{n}}_2=|{\bf{n}}_1||{\bf{n}}_2|\cos\theta$

Thus \begin{eqnarray*} \cos\theta & = & \ds\frac{(2,6,1) \cdot
(1,3,2)}{\sqrt{2^2+6^2+1^2}\sqrt{1^2+3^2+2^2}}\\ & = &
\ds\frac{2+18+2}{\sqrt{41}\sqrt{14}}\\ & = &
\ds\frac{22}{\sqrt{41}\sqrt{14}}\\ \Rightarrow \theta & = &
\arccos \left(\ds\frac{22}{{\sqrt{41}\sqrt{14}}}\right)
\end{eqnarray*}

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