Planes in space

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We discuss how to find implicit and explicit formulas for planes.

Planes are the three-dimensional analogue of lines in two-dimensions.

Implicit planes

Remember an implicit function in $\R ^3$ is one of the form: $F(x,y,z) = 0$ We would like to know the implicit formula for a plane. Here the dot product saves the day. Recall that if $\vec {n} = \vector {a,b,c}$ is any vector, and $\vec {x}= \vector {x,y,z}$ , then the equation $\vec {n}\dotp \vec {x} = 0$ is solved by all vectors $\vec {x}$ that are orthogonal to $\vec {n}$ . We plotted several such vectors below:

From this we see that $\begin{align*} \vec{n}\dotp\vec{x} &=0\\ ax+by+cz &= 0 \end{align*}$ gives the formula for a plane. Since $\vec {0} = \vec {x}$ is a solution, this plane must pass through the origin. If we want our plane to be located anywhere in space, we must know a point on the plane, call it $\vec {p}=\vector {x_0,y_0,z_0}$ . Putting this together, we can now see:

If you know

a vector $\vec {n} = \vector {a,b,c}$ and
a point (given by a vector) $\vec {p} = \vector {x_0,y_0,z_0}$

then,

$\begin{align*} \vec{n}\dotp(\vec{x}-\vec{p}) &= 0\\ a(x-x_0) + b(y-y_0) + c(z-z_0) &= 0 \end{align*}$ is an implicit equation for a plane passing through the point $(x_0,y_0,z_0)$ with normal vector $\vec {n}$ .

Find the implicit equation of a plane that passes through the point $(5,-5,-1)$ and with normal vector $\vector {-5,5,5}$ . Check your answer by modifying the definition of $F(x,y,z)$ in the SAGE code below:

xxxxxxxxxx
 
vector=arrow3d((3,-5,-1),(-2,0,4),5,color='blue');
​
F(x,y,z) = 1*(x+2)+1*(y+2)+1*(z+2)
​
plane=implicit_plot3d(F(x,y,z)==1,(x,-10,10),(y,-10,10),(z,-10,10),color='red');
plane+vector

Normal vectors not only allow us to define equations for planes but also they help us describe properties of planes.

Two planes are said to be parallel if their normal vectors are parallel. Two planes are said to be orthogonal if their normal vectors are orthogonal.

What is the (most obvious) normal vector for the plane $-6x+10y-2z = 4?$ $\vec {n} = \vector {\answer {-6},\answer {10},\answer {-2}}$

Which of the following planes are parallel to the plane $-6x+10y-2z = 4$ ?

Which of the following planes are orthogonal to the plane $-6x+10y-2z = 4$ ?

Parametric planes

Given any two nonzero vectors, $\vec {v}$ and $\vec {w}$ such that $\vec {v}\cross \vec {w} \ne \vec {0}$ we can produce a parametric formula for a plane by writing $\vec {L}(s,t) = \vec {p} + s\cdot \vec {v} + t\cdot \vec {w},$ where $\vec {p}$ is a vector whose “tip” is on the plane, and $\vec {v}$ and $\vec {w}$ are in the plane.

Given two nonzero vectors, $\vec {v}$ and $\vec {w}$ , what does it mean for $\vec {v}\cross \vec {w} \ne \vec {0}$ ?

The vector-valued formula for a plane $\vec {L}(s,t) = \vec {p} + s\cdot \vec {v} + t\cdot \vec {w},$ is very similar to our formula for a line, $\vecl (t) = \vec {p} + t\vec {v}$ where $\vec {v}$ is a vector that points in the direction of the line, both represent linear relationships, and hence we use similar notation for both.

Now that we have two methods of graphing planes, let’s use both of the representations at once!

Let $\vec {v} = \vector {1,-2,1}$ and $\vec {w} = \vector {-1,-2,1}$ . Compute $\vec {v}\cross \vec {w}$ . $\vec {v}\cross \vec {w} = \vector {\answer {0},\answer {-2},\answer {-4}}$

Use your answer above to give an implicit equation for the plane that passes through the point $(1,2,3)$ that is normal to $\vec {v}\cross \vec {w}$ . Check your answer by modifying the SAGE code below to show appropriate normal vector and modifying the definition of $F(x,y,z)$ :

xxxxxxxxxx
 
vector=arrow3d((1,2,3),(1,-2,-4),5,color='blue');
​
F(x,y,z) = 1*(x+2)+1*(y+2)+1*(z+2)
                                                                  
​
                                                                  
​
plane=implicit_plot3d(F(x,y,z)==1,(x,-10,10),(y,-10,10),(z,-10,10),color='red');
plane+vector

Now give a parametric formula for the same plane using the vectors given above.

xxxxxxxxxx
 
var('t s')
normal=arrow3d((1,2,3),(1,0,-1),5,color='blue')
​
point = vector([1,2,3])
​
dir1 = vector([t,t,t])
​
dir2 = vector([0,s,s])
​
paraPlane = parametric_plot(point+dir1+dir2,(s,-2,3),(t,-2,3))
paraPlane+normal

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