2.6: Affine lines and planes in ℝ³

import numpy as np
import plotly.graph_objects as go
from pathlib import Path
import sys

_notes_root = next(p for p in [Path.cwd(), *Path.cwd().parents] if (p / 'myst.yml').exists())
if str(_notes_root) not in sys.path:
    sys.path.insert(0, str(_notes_root))
from plot_style import style_plotly

BLUE, ORANGE, PINK = '#3d81f6', 'orange', '#d81a60'


def base3():
    fig=style_plotly(go.Figure(),renderer='plotly_mimetype')
    axis=dict(range=[-7,10],dtick=2,tickfont=dict(size=11),showbackground=True,showspikes=False,
              backgroundcolor='white',gridcolor='#e5e7eb',zerolinecolor='#9ca3af')
    fig.update_layout(autosize=True,height=520,showlegend=False,font=dict(size=16),
                      margin=dict(l=0,r=0,t=15,b=0),
                      scene=dict(bgcolor='white',xaxis=dict(title=dict(text='x',font=dict(size=13)),**axis),yaxis=dict(title=dict(text='y',font=dict(size=13)),**axis),
                                 zaxis=dict(title=dict(text='z',font=dict(size=13)),**axis),aspectmode='cube',
                                 camera=dict(eye=dict(x=1.6,y=-2.1,z=1.3))))
    for direction in np.eye(3):
        line3(fig,-6*direction,8*direction,'#9ca3af',width=2)
    return fig


def line3(fig,start,end,color=BLUE,width=5,dash='solid'):
    fig.add_trace(go.Scatter3d(x=[start[0],end[0]],y=[start[1],end[1]],z=[start[2],end[2]],
                              mode='lines',line=dict(color=color,width=width,dash=dash),
                              hoverinfo='skip',showlegend=False))


def vec3(fig,end,label,color=BLUE,start=(0,0,0),offset=(0.25,0.25,0.35)):
    start,end=np.asarray(start,float),np.asarray(end,float)
    direction=(end-start)/np.linalg.norm(end-start)
    line3(fig,start,end-0.15*direction,color,width=7)
    fig.add_trace(go.Cone(x=[end[0]],y=[end[1]],z=[end[2]],u=[direction[0]],v=[direction[1]],w=[direction[2]],
                         anchor='tip',sizemode='absolute',sizeref=0.45,colorscale=[[0,color],[1,color]],
                         showscale=False,hoverinfo='skip'))
    pos=end+offset
    fig.add_trace(go.Scatter3d(x=[pos[0]],y=[pos[1]],z=[pos[2]],mode='text',text=[label],
                              textfont=dict(family='Palatino',color=color,size=20),hoverinfo='skip'))


def plane3(fig,normal,color=BLUE,opacity=0.24,d=0,bounds=(-5,7)):
    # Draw a complete rectangle in the plane, with room around every edge.
    n=np.asarray(normal,float)
    unit=n/np.linalg.norm(n)
    reference=np.eye(3)[np.argmin(np.abs(unit))]
    u=np.cross(unit,reference)
    u=u/np.linalg.norm(u)
    v=np.cross(unit,u)
    center=d*n/np.dot(n,n)
    lo,hi=bounds
    corners=np.array([center+a*u+b*v for a,b in
                      [(lo,lo),(hi,lo),(hi,hi),(lo,hi)]])
    fig.add_trace(go.Mesh3d(x=corners[:,0].tolist(),y=corners[:,1].tolist(),z=corners[:,2].tolist(),
                            i=[0,0],j=[1,2],k=[2,3],
                            color=color,opacity=opacity,flatshading=True,
                            lighting=dict(ambient=1,diffuse=0,specular=0,fresnel=0,roughness=1),
                            showscale=False,hoverinfo='skip'))
    for i in range(4):
        line3(fig,corners[i],corners[(i+1)%4],color,width=2)
    lower=min(fig.layout.scene.xaxis.range[0],float(corners.min())-1)
    upper=max(fig.layout.scene.xaxis.range[1],float(corners.max())+1)
    fig.update_scenes(xaxis_range=[lower,upper],yaxis_range=[lower,upper],
                      zaxis_range=[lower,upper])

In Chapter 2.4, we described lines and planes through the origin in parametric form. In Chapter 2.5, we described them using linear equations of the form

ax + by + cz = d,

where $d$ was forced to be 0.

Let’s now think about lines and planes in $\mathbb{R}^3$ that are not required to pass through the origin. These are called affine lines and planes.

As we discussed in Chapter 2.1 when we introduced affine lines in $\mathbb{R}^2$ , the idea is to add a fixed vector $\vec p$ to every vector in the set. The direction(s) stay the same, but the starting point changes.

fig=base3()
plane3(fig,(1,-2,1),BLUE,0.18)
plane3(fig,(1,-2,1),ORANGE,0.18,d=3)
vec3(fig,(0,0,3),'<i>p</i>⃗',PINK,offset=(0.4,-0.5,0.3))
fig.add_trace(go.Scatter3d(x=[0],y=[0],z=[0],mode='markers',
                         marker=dict(color='black',size=4),hoverinfo='skip'))
fig.show()

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Translating a plane¶

Recall the plane

P=\operatorname{span}(\vec v_1,\vec v_2),\qquad \vec v_1=\begin{bmatrix}3\\4\\5\end{bmatrix},\qquad \vec v_2=\begin{bmatrix}5\\2\\-1\end{bmatrix}.

It has normal vector and equation

\vec w=\begin{bmatrix}1\\-2\\1\end{bmatrix},\qquad x-2y+z=0.

fig=base3()
plane3(fig,(1,-2,1))
vec3(fig,(3,4,5),'<i>v</i>⃗<sub>1</sub>',BLUE)
vec3(fig,(5,2,-1),'<i>v</i>⃗<sub>2</sub>',ORANGE,offset=(-0.8,0.2,0.4))
fig.add_trace(go.Scatter3d(x=[-3],y=[1],z=[5],mode='text',text=['<i>P</i>'],
                         textfont=dict(family='Palatino',size=26,color=BLUE),hoverinfo='skip'))
fig.show()

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The plane $P=\operatorname{span}(\vec v_1,\vec v_2)$ , with equation $x-2y+z=0$ , before translation.

This plane passes through the origin. Let’s suppose we add the vector $\vec p = \begin{bmatrix} 0 \\ 0 \\ 3 \end{bmatrix}$ to every vector on the plane. The new plane has the vector-parametric form

\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}0\\0\\3\end{bmatrix} +a\begin{bmatrix}3\\4\\5\end{bmatrix} +b\begin{bmatrix}5\\2\\-1\end{bmatrix},\qquad a,b\in\mathbb R.

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Adding $\vec p=\begin{bmatrix}0\\0\\3\end{bmatrix}$ translates $P$ to the parallel plane $x-2y+z=3$ .

How do we express this translated plane as a linear equation? First, note that translation preserves the directions in the plane, so $\vec w=\begin{bmatrix}1\\-2\\1\end{bmatrix}$ is still a normal vector. A vector $\vec x=\begin{bmatrix}x\\y\\z\end{bmatrix}$ is on the translated plane exactly when $\vec x-\vec p$ – in other words, $\vec x$ if we “undo” the translation by the fixed vector $\vec p$ – is on the original plane, $P$ . In other words, $\vec x - \vec p$ is on the translated plane when $\vec x - \vec p$ is orthogonal to the original plane (and new plane)'s normal vector, $\vec w$ . Therefore,

\begin{aligned} \vec w\cdot(\vec x-\vec p)&=0\\ \vec w\cdot\vec x&=\vec w\cdot\vec p\\ \end{aligned}

The equation $\vec w \cdot \vec x = \vec w \cdot \vec p$ is what we will use to find the constant value of $d$ in

ax + by + cz = d.

In our current example,

\vec w \cdot \vec x = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \cdot \begin{bmatrix} x \\ y \\ z \end{bmatrix}

and

\vec w \cdot \vec p = \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix} \cdot \begin{bmatrix} 0 \\ 0 \\ 3 \end{bmatrix} = 1(0) -2(0) +1(3) = 3

so the linear equation for the translated plane is

x - 2y + z = 3.

More generally, a plane has equation $ax+by+cz=d$ , where the normal vector $\begin{bmatrix}a\\b\\c\end{bmatrix}$ is nonzero. If $d=0$ , the equation is homogeneous and the plane passes through the origin. If $d\ne0$ , it is nonhomogeneous and the plane does not pass through the origin.

Activity 1

Consider

Q'=\begin{bmatrix}2\\-1\\4\end{bmatrix} +\operatorname{span}\left(\begin{bmatrix}1\\2\\-1\end{bmatrix}, \begin{bmatrix}2\\-1\\3\end{bmatrix}\right).

Write vector-parametric and scalar-parametric forms, stating the possible parameter values.
Use the normal $\vec n=\begin{bmatrix}1\\-1\\-1\end{bmatrix}$ (check that it is perpendicular to both spanning vectors) to find the constant in $\vec n\cdot\vec x=\vec n\cdot\vec p$ .
Write an equation $ax+by+cz=d$ for the plane. Does it contain the origin?

Solution

The vector-parametric form is

\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}2\\-1\\4\end{bmatrix} +s\begin{bmatrix}1\\2\\-1\end{bmatrix} +t\begin{bmatrix}2\\-1\\3\end{bmatrix},\qquad s,t\in\mathbb R.

The scalar-parametric form is

\begin{aligned}x&=2+s+2t,\\y&=-1+2s-t,\\z&=4-s+3t,\end{aligned}\qquad s,t\in\mathbb R.

The constant is $\vec n\cdot\vec p=2+1-4=-1$ . Hence $\vec n\cdot\vec x=-1$ , or $x-y-z=-1$ . The origin does not satisfy this equation.

Translating a line¶

The same idea works for a line. Recall

\ell=\operatorname{span}(\vec v_1),\qquad \vec v_1=\begin{bmatrix}3\\4\\5\end{bmatrix}.

In Chapter 2.5, we used the two independent normals

\vec n_1=\begin{bmatrix}1\\-2\\1\end{bmatrix},\qquad \vec n_2=\begin{bmatrix}2\\1\\-2\end{bmatrix}.

Both dot products with $\vec v_1$ are zero. The line is therefore the intersection of $x-2y+z=0$ and $2x+y-2z=0$ . Adding $\vec p$ to each vector on $\ell=\operatorname{span}(\vec v_1)$ gives

\begin{bmatrix}x\\y\\z\end{bmatrix} =\begin{bmatrix}0\\0\\3\end{bmatrix} +t\begin{bmatrix}3\\4\\5\end{bmatrix},\qquad t\in\mathbb R.

Again, we haven’t changed the direction of the line, so we can keep the same normal vectors $\vec n_1$ and $\vec n_2$ . Taking their dot products with $\vec p$ gives the new right-hand sides, 3 and -6:

\begin{cases}x-2y+z=3,\\2x+y-2z=-6.\end{cases}

An affine line in $\mathbb R^3$ is the intersection of two planes with independent normal vectors. Translating changes the right-hand sides of their equations while preserving the direction of the line.


fig=base3()
plane3(fig,(1,-2,1),BLUE,0.22,d=3)
plane3(fig,(2,1,-2),ORANGE,0.22,d=-6)
p=np.array([0,0,3]); v=np.array([3,4,5])
line3(fig,p-1.1*v,p+v,PINK,width=7)
vec3(fig,p+v,'<i>v</i>⃗<sub>1</sub>',PINK,start=p,offset=(0.6,0.3,0.5))
fig.add_trace(go.Scatter3d(x=[0],y=[0],z=[3],mode='markers+text',
    marker=dict(color='black',size=4),text=['<i>p</i>⃗'],textposition='top left',
    textfont=dict(family='Palatino',size=16),hoverinfo='skip'))
fig.show()

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The two translated planes intersect in the affine line through $(0,0,3)$ with direction $\vec v_1$ .

Another example¶

Consider the affine line $L$ in scalar-parametric form:

x=1+2t,\qquad y=2-t,\qquad z=-1+3t,\qquad t\in\mathbb R.

Choose the normals $\vec n_1=\begin{bmatrix}0\\3\\1\end{bmatrix}$ and $\vec n_2=\begin{bmatrix}-3\\0\\2\end{bmatrix}$ which are perpendicular to this line’s direction: their dot products with $\begin{bmatrix}2\\-1\\3\end{bmatrix}$ are $-3+3=0$ and $-6+6=0$ . Their dot products with the starting point $\vec p=\begin{bmatrix}1\\2\\-1\end{bmatrix}$ are 5 and -5. Thus,

L=\left\{\vec x\in\mathbb R^3: \vec n_1\cdot\vec x=\vec n_1\cdot\vec p,\quad \vec n_2\cdot\vec x=\vec n_2\cdot\vec p\right\},

or, in scalar equations,

\begin{cases}3y+z=5,\\-3x+2z=-5.\end{cases}

These planes intersect in $L$ . To check, let $z=-1+3t$ in the system; the equations give $y=2-t$ and $x=1+2t$ .

Activity 2

Consider the affine line

L=\left\{\begin{bmatrix}2\\-1\\1\end{bmatrix} +t\begin{bmatrix}1\\2\\-1\end{bmatrix}:t\in\mathbb R\right\}.

Write scalar-parametric equations for $L$ , stating the possible parameter values.
Find two independent vectors perpendicular to the direction of $L$ .
Use these vectors to describe $L$ as the solution set of two linear equations. Does $L$ contain the origin?

Solution

The scalar-parametric equations are

x=2+t,\qquad y=-1+2t,\qquad z=1-t,\qquad t\in\mathbb R.

A perpendicular vector $\begin{bmatrix}a\\b\\c\end{bmatrix}$ must satisfy $a+2b-c=0$ . Two choices are

\vec n_1=\begin{bmatrix}2\\-1\\0\end{bmatrix},\qquad \vec n_2=\begin{bmatrix}1\\0\\1\end{bmatrix}.

They are not scalar multiples, and their dot products with the direction vector are $2-2=0$ and $1-1=0$ .

Taking dot products with $\vec p=\begin{bmatrix}2\\-1\\1\end{bmatrix}$ gives

\vec n_1\cdot\vec p=4+1=5,\qquad \vec n_2\cdot\vec p=2+1=3.

Thus $L$ is the solution set of

\begin{cases}2x-y=5,\\x+z=3.\end{cases}

To check, set $x=2+t$ . The equations then give $y=-1+2t$ and $z=1-t$ , recovering the original line. The origin does not satisfy either equation, so it is not on $L$ .

An affine line or plane need not pass through the origin, but it can. Translation does not automatically make every right-hand side nonzero: each constant is determined by the corresponding dot product.