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


{\bf Question}

\begin{description}
\item[(a)] Obtain the value(s) of $\lambda$ which make the vectors
${\bf i - j} , \, 2{\bf i + j - \lambda k}, \, $

$\lambda{\bf i - j + \lambda k}$ coplanar.
\item[(b)] Show that the four points $(5,2 -1), \, (6,1,4), \,
(-1,-3,6), \, (-3, -2, -1)$ lie in a plane.
\end{description}

\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)] The three vectors ${\bf a \, b \, c}$ are coplanar if
${\bf a \cdot b \times c} = 0$

\begin{eqnarray*} \left| \begin{array}{ccc} 1 & -1 & 0 \\ 2 & 1 & -\lambda \\
\lambda & -1 & \lambda \end{array} \right| = 0 & \Rightarrow &
\left| \begin{array}{ccc} 1 & 0 & 0 \\ 2 & 3 & -\lambda \\ \lambda
& \lambda-1 & \lambda \end{array} \right| = 0 \\ & \Rightarrow &
3\lambda +\lambda^2 - \lambda = 0 \\ & \Rightarrow & \lambda^2 +
2\lambda =0 \\ &\Rightarrow & \lambda = 0, {\rm \ or\ } \lambda =
-2\end{eqnarray*}


\item[(b)] The 4 points all lie in a plane if there exist $\alpha, \, \beta, \,
\gamma, \, \delta$ are not all zero such that $\alpha{\bf a}+
\beta{\bf b}+ \gamma{\bf c}+ \delta{\bf d} = 0$ and
$\alpha+\beta+\gamma+\delta=0$

\begin{eqnarray} 5\alpha + 6\beta - 2\gamma - 3\delta & = & 0 \\ 2\alpha + \beta -
3\gamma - 2\delta & = & 0 \\ -3\alpha + 4\beta + 6\gamma - \delta
& = & 0 \\ \alpha + \beta + \gamma + \delta & = & 0 \end{eqnarray}

Eliminating $\delta$ from equations (1) to (3) using equation (4)
gives:

\begin{eqnarray} 8\alpha + 9\beta + \gamma & = & 0 \\ 4\alpha + 3\beta -
\gamma & = & 0 \\ -2\alpha + 5\beta - 3\gamma - \delta & = & 0
\end{eqnarray}

\newpage
Eliminating $\gamma$ from equations (5) and (7) using equation (6)
gives:

\begin{eqnarray*} 12\alpha +129\beta & = & 0 \\ 26\alpha + 16\beta & = & 0
\end{eqnarray*}

[or use the triple product]

So $$\beta=-\alpha\hspace{.3in} \gamma=\alpha \hspace{.3in}
\delta=-\gamma$$ so they lie in a plane.


\end{description}

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