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{\bf Question}

Let ${\bf a}={\bf i}+{\bf j}+{\bf k}$ and ${\bf b}=2{\bf j}-{\bf
k}$. Find the relation that must hold between $x_1$, $x_2$ and
$x_3$ if the vector ${\bf x}=x_1{\bf i}+x_2{\bf j}+x_3{\bf k}$ is
to be written as $${\bf x}=s{\bf a}+t{\bf b}$$ where $s$ and $t$
are scalars. Show that the vector ${\bf c}=3{\bf i}+{\bf j}+4{\bf
k}$ can be written as ${\bf c}=s{\bf a}+t{\bf b}$ and find $s$ and
$t$ in this case.



{\bf Answer}

${\bf a} \times {\bf b} =\left| \begin{array}{ccc} {\bf{i}} &
{\bf{j}} & {\bf{k}}\\ 1 & 1 & 1 \\ 0 & 2 & -1 \end{array} \right|
= \left( \begin{array}{c} -3 \\ 1 \\ 2 \end{array} \right) $,
${\bf b} \times {\bf a} = -({\bf a} \times {\bf b})= \left(
\begin{array}{c} 3 \\ -1 \\ -2 \end{array} \right) $

Note that ${\bf a}\cdot({\bf a} \times {\bf b}) = 0 = {\bf b}
\cdot ({\bf a} \times {\bf b})$

Take the dot product of the equation ${\bf x} = s{\bf a} + t{\bf
b}$ with the vector ${\bf a} \times {\bf b}$

${\bf x} \cdot ({\bf a} \times {\bf b}) = s{\bf a} \cdot ({\bf a}
\times {\bf b}) + t{\bf b} \cdot ({\bf a} \times {\bf b}) = 0$

Hence the relation is: $\left( \begin{array}{c} x_1 \\ x_2 \\ x_3
\end{array} \right) \cdot \left( \begin{array}{c} -3 \\ 1 \\ 2
\end{array} \right) = 0$

or $ -3x_1 + x_2 + 2x_3 = 0$

Note that the points which can be written as ${\bf x} = s{\bf a} +
t{\bf b}$ all lie on a plane through the origin with equation $
-3x + y + 2z = 0$.  The vector ${\bf c} = \left(
\begin{array}{c} 3 \\ 1 \\ 4 \end{array} \right)$ can be written
in the form ${\bf c} = s{\bf a} + t{\bf b}$ because the components
of $\bf c$ satisfy the relation: $-3(3) + (1) + 2(4) = -9 + 1 + 8
= 0$

${\bf c} \times {\bf a} =\left| \begin{array}{ccc} {\bf{i}} &
{\bf{j}} & {\bf{k}}\\ 3 & 1 & 4 \\ 1 & 1 & 1 \end{array} \right| =
\left( \begin{array}{c} -3 \\ 1 \\ 2 \end{array} \right) $, ${\bf
c} \times {\bf b} =\left| \begin{array}{ccc} {\bf{i}} & {\bf{j}} &
{\bf{k}}\\ 3 & 1 & 4 \\ 0 & 2 & -1 \end{array} \right| = \left(
\begin{array}{c} -9 \\ 3 \\ 6 \end{array} \right) $

To find $s$ take the cross product of ${\bf c} = s{\bf a} + t{\bf
b}$ with ${\bf b}$: ${\bf c}\times {\bf b}  = s ({\bf a} \times
{\bf b}) + t ({\bf b} \times {\bf b}) = s ({\bf a} \times {\bf
b})$ since
 $ {\bf b} \times {\bf b} = 0$

$\left( \begin{array}{c} -9 \\ 3 \\ 6 \end{array} \right) = s
\left( \begin{array}{c} -3 \\ 1 \\ 2 \end{array} \right)$ and so $
s = 3$


To find $t$ take the cross product of ${\bf c} = s{\bf a} + t{\bf
b}$ with ${\bf a}$: ${\bf c}\times {\bf a}  = s ({\bf a} \times
{\bf a}) + t ({\bf b} \times {\bf a}) = t ({\bf b} \times {\bf
a})$ since
 $ {\bf a} \times {\bf a} = 0$

$\left( \begin{array}{c} -3 \\ 1 \\ 2 \end{array} \right) =
t\left(\begin{array}{c} 3 \\ -1 \\ -2 \end{array} \right)$ and so
$ t = -1$

hence ${\bf c} = 3{\bf a} - {\bf b}$.



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