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QUESTION

Find the vector equation of the plane $\Pi_1$ which passes through
the points $L=(1,1,0),\ M=(1,-2,2)$ and $N=(3,0,3)$. 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 $Q=(1,1,1)$. Find the equation of $\Pi_2$ in terms of
$x,y,z$ coordinates.

Give the parametric equation for the line $\ell$ through the point
$Q$ orthogonal to $\Pi_1$, and find the point $A$ where it
intersects the plane $\Pi_1$. Write down the vector joining the
point $L$ to $A$, and verify that this is orthogonal to the line
$\ell$.




ANSWER


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

$$\mathbf{u}\times\mathbf{v}=\left|\begin{array}{ccc}\mathbf{i}
&amp;\mathbf{j}&amp;\mathbf{k}\\0&amp;-3&amp;2\\2&amp;-1&amp;3\end{array}\right|=
-7\mathbf{i}+4\mathbf{j}+6\mathbf{k}=\left(\begin{array}{c}
-7\\4\\6\end{array}\right)$$

 so the equation of $\Pi_1$ is
$\left(\begin{array}{c}-7\\4\\6\end{array}\right).\mathbf{w}=
\left(\begin{array}{c}-7\\4\\6\end{array}\right)
\left(\begin{array}{c}1\\1\\0\end{array}\right)=-3$

In co-ordinates this is $-7x+4y+6z=-3$.

$\Pi_2$ has equation $-7x+4y+6z=3$. $\ell$ has equation
$(1,1,1)+t(-7,4,6)$ or $(x,y,z)=(1-7t,1+4t,1+6t)$ This point lies
in $\Pi_1\Leftrightarrow
-7(1-7t)+4(1+4t)+6(1+6t)=-3\Leftrightarrow-7+4+6+3=(-49-16-36)t$
i.e. $t=\frac{-6}{101}$ hence
$A=\left(\frac{143}{101},\frac{77}{101},\frac{65}{101}\right)$.

$\vec{LA}$ is orthogonal to
$\ell\Leftrightarrow\vec{LA}.(\mathbf{u}\times\mathbf{v})=0$

$\vec{LA}=\left(\frac{42}{101},-\frac{24}{101},\frac{65}{101}\right)^T,\
\vec{LA}.\left(\begin{array}{c}-7\\4\\6\end{array}\right)=-284-96+390=0$.




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