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QUESTION


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\item[(a)]
Find the vector equation of the plane $\Pi_1$ which passes through
the three points $L=(1,3,4),\ M=(2,-3,0)$ and $N=(1,0,1)$. What is
the equation of the plane in terms of $x,y,z$ coordinates?

A second plane $\Pi_2$ is parallel to $\Pi_1$ and passes through
the point $Q=(3,1,1)$. Find the equation of $\Pi_2$ in terms of
$x,y,z$ coordinates.

\item[(b)]
(You should state clearly any properties of the cross product that
you use.) Let $\mathbf{a}=\mathbf{i}-3\mathbf{j}+2\mathbf{k}$ and
$\mathbf{b}=2\mathbf{i}+2\mathbf{j}-5\mathbf{k}$ be vectors.
Calculate $\mathbf{a}\times\mathbf{b}$ and hence write down
$\mathbf{b}\times\mathbf{a}$. Find the relation that must hold
between $x_1,x_2$ and $x_3$ if the vector
$\mathbf{x}=x_1\mathbf{i}+x_2\mathbf{j}+x_3\mathbf{k}$ is to be
written in the form

$$\mathbf{x}=s\mathbf{a}+t\mathbf{b}$$

where $s$ and $t$ are scalars.

If $\mathbf{c}=7\mathbf{i}-5\mathbf{j}-4\mathbf{k}$,  calculate
$\mathbf{c}\times\mathbf{a}$ and $\mathbf{c}\times\mathbf{b}$.
Show that $\mathbf{c}$ can be expressed in the form
$\mathbf{c}=s\mathbf{a}+t\mathbf{b}$ and find $s$ and $t$ in this
case.

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ANSWER


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\item[(a)]
Vector equation of plane: $\mathbf{n}(\mathbf{r}-\mathbf{r}_0)=0$
where

$\mathbf{r}=\left(\begin{array}{c}x\\y\\z\end{array}\right)$ is
the position vector of a general point on the plane;

$\mathbf{r}_0$ is the position vector of a known point on the
plane;

$\mathbf{n}$ is a normal vector to the plane.

The point $L=(1,3,4)$ lies on the plane, so let
$\mathbf{r}_0=\left(\begin{array}{c}1\\3\\4\end{array}\right).$
two vectors in the plane are $\vec{LM}$ and $\vec{LN}$.

$$\vec{LM}=\left(\begin{array}{c}2\\-3\\0\end{array}\right)-
\left(\begin{array}{c}1\\3\\4\end{array}\right)=
\left(\begin{array}{c}1\\-6\\-4\end{array}\right),\
\vec{LN}=\left(\begin{array}{c}1\\0\\1\end{array}\right)-
\left(\begin{array}{c}1\\3\\4\end{array}\right)=
\left(\begin{array}{c}0\\-3\\-3\end{array}\right)$$

A vector normal to the plane is $\vec{LM}\times\vec{LN}$ when

$$\vec{LM}\times\vec{LN}=
\left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&-6&-4\\0&-3&-3\end{array}\right|=
\left(\begin{array}{c}6\\3\\-3\end{array}\right)$$

so let
$\mathbf{n}=\left(\begin{array}{c}2\\1\\-1\end{array}\right).$

Vector equation of plane $\Pi_1$ is therefore
$\left(\begin{array}{c}2\\1\\-1\end{array}\right)
\cdot\left[\left(\begin{array}{c}x\\y\\z\end{array}\right)
-\left(\begin{array}{c}1\\3\\4\end{array}\right)\right]=0$

Standard equation of $\Pi_1:\ 2(x-1)+(y-3)-(z-4)=0$ or
$2x+y-z=2+3-4=1$

Plane $\Pi_2$ is parallel to $\Pi_1$ and so has the equation
$2x+y-z=d$ for some constant $d$.

Since $\Pi_2$ contains the point $Q=(3,1,1)$ we have
$2(3)+(1)-(1)=d\Rightarrow d=6$ so the standard equation of
$\Pi_2$ is $2x+y-z=6$.

\item[(b)]
$$\mathbf{a}\times\mathbf{b}=\left|\begin{array}{ccc}\mathbf{i}
&\mathbf{j}&\mathbf{k}\\1&-3&2\\2&27&-5\end{array}\right|=
\left(\begin{array}{c}11\\9\\8\end{array}\right),\
\mathbf{b}\times\mathbf{a}=-(\mathbf{a}\times\mathbf{b})
=\left(\begin{array}{c}-11\\-9\\-8\end{array}\right)$$

Note that $\mathbf{a}.(\mathbf{a}\times\mathbf{b})=0=\mathbf{b}.
(\mathbf{a}\times\mathbf{b})$

Take dot product of $\mathbf{x}=s\mathbf{a}+t\mathbf{b}$ with
vector $(\mathbf{a}\times\mathbf{b})=0$ so
$\left(\begin{array}{c}x_1\\x_2\\x_3\end{array}\right)
\left(\begin{array}{c}11\\9\\8\end{array}\right)=0$ and relation
is $11x_1+9x_2+8x_3=0$

$$\mathbf{c}\times\mathbf{a}=
\left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k}
\\7&-5&-4\\1&-3&2\end{array}\right|
=\left(\begin{array}{c}-22\\-18\\-16\end{array}\right),\
\mathbf{c}\times\mathbf{b}=
\left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k
}\\7&-5&-4\\2&2&-5\end{array}\right|
=\left(\begin{array}{c}33\\27\\24\end{array}\right)$$

$\mathbf{c}=\left(\begin{array}{c}7\\-5\\-4\end{array}\right)$ so
if $\mathbf{c}=s\mathbf{a}+t\mathbf{b}$ then we must have the
relation

$$11(7)+9(-5)+8)-4)=77-45-32=0.$$

so $\mathbf{c}$ can be expanded in the required form.

Note $\mathbf{a}\times\mathbf{a}=0=\mathbf{b}\times\mathbf{b}$.

To fins $s$: take cross product of
$\mathbf{c}=s\mathbf{a}+t\mathbf{b}$ with $\mathbf{b}$

$$\mathbf{c}\times\mathbf{b}=
s\mathbf{a}\times\mathbf{b}+t\mathbf{b}\times\mathbf{b}=
s\mathbf{a}\times\mathbf{b}$$

then $\left(\begin{array}{c}33\\27\\24\end{array}\right)=
s\left(\begin{array}{c}11\\9\\8\end{array}\right)$ and so $s=3$.

To find $t$: Take the cross product of
$\mathbf{c}=s\mathbf{a}+t\mathbf{b}$ with $\mathbf{a}$

$$\mathbf{c}\times\mathbf{a}=
s\mathbf{a}\times\mathbf{a}+t\mathbf{b}\times\mathbf{a}=
t\mathbf{b}\times\mathbf{a}$$

then $\left(\begin{array}{c}-22\\-18\\-16\end{array}\right)=
t\left(\begin{array}{c}-11\\-9\\-8\end{array}\right)$ and $t=2$.

Therefore, $\mathbf{c}=3\mathbf{a}+2\mathbf{b}$.

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