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, \(r.n\)
\[x-2y+3z=11\]

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Complete Questions 13 and 14 from worksheet.