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{\bf Question}

With respect to a set of orthogonal axes, origin $O$ the
coordinates of $A,B,C,$

$D$ are $(3,2,1),\,(1,-2,1),\,(-1,1,2)$ and $(-2,2,0)$
respectively.

${}$

Find the angle between $OA$ and $OD$, the direction cosines of a
vector

perpendicular to $OA$ and $OD$ and the perpendicular distance from
$C$ to $AB$.

${}$

Find the equation of the plane through $D$ parallel to the plane
containing $A,\,B$ and $C$.


\vspace{.25in}

{\bf Answer}

${\bf a} = (3,2,1) \hspace{.2in} {\bf b} = (1,-2,1) \hspace{.2in}
{\bf c} - (-1,1,2) \hspace{.2in} {\bf d} = (-2,2,0)$

${}$

$\ds |a| = \sqrt{14} \hspace{.2in} |d| = \sqrt{8} \hspace{.2in}
{\bf a \cdot d} = -2 \hspace{.2in} \cos \theta =
-\frac{1}{\sqrt{28}}\Rightarrow \theta = 100^\circ, \, 1.761 rad$

${\bf a \times d} = (-2,-2,10) \hspace{.2in} |{\bf a \times d}| =
\sqrt{108} = 6\sqrt3$

direction cosines are $\ds \left( -\frac{1}{3\sqrt3} , \,
-\frac{1}{3\sqrt3} , \, \frac{5}{3\sqrt3}\right)$

${}$

 ${}$

Distances of $c$ from the line $AB$ is $\ds \frac{|(c - a)\times
|(b-a)|}{|b-a|} = \frac{|(4,-2,14)|}{2\sqrt5} =
\frac{3\sqrt6}{\sqrt5}$

${}$

${}$

A normal to the plane is $(a-c)\times(a-b) = (4,-2,14)$ and the
equation of the plane is $2x-y+7z=-6$


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