## Vector Equation of a Plane

The vector equation of a plane can be written as
$r.n=a.n$
where $$a$$ is a position vector of a point on the plane and $$n$$ is a normal to the plane.
LESSON 1
Determine the equation of the plane, in vector form and Cartesian form, which contains the point $$(2,-3,1)$$ with normal $$i-2j+3k$$.
SOLUTION
The equation of a plane can be written in the form
$r.n=a.n$
where $$n$$ is a vector perpendicular to the plane and $$a$$ is a position vector of a point on the plane.
State the position vector
$\begin{pmatrix} 2\\-3 \\1 \end{pmatrix}$
State the normal to the plane

$n=\begin{pmatrix} 1\\-2 \\3 \end{pmatrix}$

Substitute the vectors $$a$$ and $$n$$ into the general form of the vector equation of the plane, $$r.n=a.n$$
$r.\begin{pmatrix} 1\\-2 \\3 \end{pmatrix}=\begin{pmatrix} 2\\-3 \\1 \end{pmatrix}.\begin{pmatrix} 1\\-2 \\3 \end{pmatrix}$
Determine the dot product $$a.n$$
$r.\begin{pmatrix} 1\\-2\\3 \end{pmatrix}=2(1)+(-3)(-2)+1(3)$
$r.\begin{pmatrix} 1\\-2\\3 \end{pmatrix}=11$
CARTESIAN FORM
Replace $$r$$ with $$\begin{pmatrix} x\\y\\z \end{pmatrix}$$ in the vector equation of the plane
$\begin{pmatrix} x\\y\\z \end{pmatrix}.\begin{pmatrix} 1\\-2\\3 \end{pmatrix}=11$
Determine the dot product, $$r.n$$
$x-2y+3z=11$

TRY THIS
Complete Questions 13 and 14 from worksheet.