M408M Learning Module Pages
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Chapter 10: Parametric Equations
and Polar Coordinates
Chapter 12: Vectors and the Geometry of Space
Learning module LM 12.1:
3dimensional rectangular coordinates:
Learning module LM 12.2: Vectors:
Learning module LM 12.3: Dot products:
Learning module LM 12.4: Cross products:
Learning module LM 12.5: Equations of Lines and Planes:
Equations of a line
Equations of planes
Finding the normal to a plane
Distances to lines and planes
Learning module LM 12.6: Surfaces:
Chapter 13: Vector Functions
Chapter 14: Partial Derivatives
Chapter 15: Multiple Integrals


Finding the normal to a plane
Finding the normal to a plane
As promised, we return the the question of finding the equation for a
plane from the location of three points, say $$Q(x_1,\,y_1,\,z_1)\,,
\qquad R(x_2,\,y_2,\,z_2)\,, \qquad S(x_3,\,y_3,\,z_3)$$ The fact that
the crossproduct ${\bf a}\times {\bf b}$ is perpendicular to both
${\bf a}$ and ${\bf b}$ makes it very useful when dealing with normals
to planes.
Let
$${\bf b}=\langle\,x_1,\,y_1,\,z_1\, \rangle\,, \ \ {\bf r} = \langle\,x_2,\,y_2,\,z_2\, \rangle\,, \ \ {\bf s} = \langle\,x_3,\,y_3,\,z_3\, \rangle.$$
The vectors
$$\overrightarrow{QR} \ = \ {\bf r}  {\bf b} \,, \qquad \overrightarrow{QS} \ = \ {\bf s}  {\bf b} \,, $$
then lie in the plane.
The normal to the plane is given by the cross product ${\bf n} = (
{\bf r}  {\bf b})\times ( {\bf s}  {\bf b})$. Once this normal has
been calculated, we can then use the pointnormal form to get the
equation of the plane passing through $Q,\,R,\, $ and $S$.


In practice, it's usually easier to work out ${\bf n}$ in a given example rather than try to set up some general equation for the plane.
Example: Find an equation for the plane passing through the points
$$Q(1,\,1,\,2)\,, \ \ R(4,\,2,\,2)\,, \ \ S(2,\,1,\,5)\,.$$
Solution: when the plane passes through $Q,\, R$, and $S$, then the vectors
$$\overrightarrow{QR} = \langle\, 3,\, 1,\, 0\, \rangle\,,\qquad \overrightarrow{QS}= \langle\, 1,\, 0,\, 3\, \rangle\,,$$
lie in the plane. Thus the crossproduct
$${\bf n} \ = \ \left\begin{array}{ccc}{\bf i} & {\bf j} & {\bf k} \\ 3 & 1 & 0 \\ 1 & 0 & 3 \end{array}\right \ = \ \ = \ 3{\bf i} +9{\bf j} +{\bf k}$$
is normal to the plane.

If ${\bf r} = \langle \,x,\,y,\, z\,\rangle$ determines an arbitrary
point $P$ in the plane, the vector
$${\bf v} \ = \ \overrightarrow{QP}\ = \ \langle \,x+1,\,y1,\, z2\,\rangle$$
lies in the plane and so is perpendicular to ${\bf n}$.
In this case,
$${\bf n}\cdot{\bf v}\ = \ 3(x+1) +9(y1) +(z2) \ = \ 0\,,$$
which after simplification becomes
$$3x +9y +z  8 \ = \ 0\,.$$
Consequently, the plane $3x +9y +z = 8 $
passes through $Q,\, R$, and $S$.

