2.5: Linear equations in ℝ³ - Math 124 Course Notes

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 using spans and parametric equations. Here, we’ll describe them using linear equations:

ax + by + cz = d

Recall, a homogenous linear equation is one whose right-hand side above is 0. We will focus on homogenous equations here, and explore nonzero values of $d$ in Chapter 2.6.

Why does one linear equation describe a plane?¶

In $\mathbb R^2$ , a linear equation of the form $ax+by=c$ , with $a,b$ not both zero, describes a line.

In $\mathbb R^3$ , a single linear equation with at least one nonzero coefficient describes a plane. For now, we’ll focus on homogeneous equations:

ax+by+cz=0.

Why a plane instead of a line? For example, consider

x-2y+z=0,

or equivalently,

z=-x+2y.

We can plug in any $x$ and any $y$ to get an output $z$ . For instance, when $x=1$ and $y=1$ , we get $z=1$ . Allowing every possible pair of $x$ and $y$ gives us a plane.

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The plane $x-2y+z=0$ , or $z=-x+2y$ . Every pair of $x$ and $y$ determines a point on the plane.

But a line only works for very specific pairs of $x$ and $y$ . Recall the line spanned by $\vec v=\begin{bmatrix}3\\4\\5\end{bmatrix}$ from Chapter 2.4. There’s no point on this line that has $x=3$ and $y=1$ . Rather, when $x=3$ , $y$ is forced to be 4, and $z$ is forced to be 5.

So we cannot describe this line by a formula for $z$ in terms of $x$ and $y$ that lets us plug in any $x$ and any $y$ : most pairs of $x$ and $y$ -values do not lie on the line.

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For example, $y=0$ gives us the $xz$ -plane.

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The equation $2x-3y+9z=0$ describes another plane through the origin.

Linear equations for planes¶

Let’s return to the plane from Chapter 2.4:

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}.

Note that we don’t yet have a linear equation describing this plane: we’ve expressed the plane as the span of two vectors. How might we find a linear equation that describes this plane?

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The plane $P$ spanned by $\vec v_1=\begin{bmatrix}3\\4\\5\end{bmatrix}$ and $\vec v_2=\begin{bmatrix}5\\2\\-1\end{bmatrix}$ .

For now, we’ll just tell it to you: an equation for this plane is

x-2y+z=0.

You should verify yourself that both $\vec v_1 = \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix}$ and $\vec v_2 = \begin{bmatrix} 5 \\ 2 \\ -1 \end{bmatrix}$ satisfy the equation above.

How could we have found this equation without guessing? The key is to find a vector perpendicular to the plane. Note that the equation above can be written as

\begin{bmatrix} x \\ y \\ z \end{bmatrix} \cdot \begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}.

All vectors that satisfy $x - 2y + z = 0$ are perpendicular to $\begin{bmatrix} 1 \\ -2 \\ 1 \end{bmatrix}$ !

Normal vectors¶

Remember that in Chapter 2.1, we saw that in the equation of a line in $\mathbb{R}^2$ ,

ax + by = c,

the vector $\begin{bmatrix} a \\ b\end{bmatrix}$ – found by reading the coefficients of $x$ and $y$ above – is orthogonal to the line above.

Let’s try the same thing here. The coefficients on $x$ , $y$ , and $z$ in

x-2y+z=0

imply the normal vector

\vec w=\begin{bmatrix}1\\-2\\1\end{bmatrix}.

fig=base3()
plane3(fig,(1,-2,1))
n=np.array([1,-2,1])
line3(fig,-2*n,2*n,PINK,width=3)
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=(-0.6,0.2,0.4))
vec3(fig,n,'<i>w</i>⃗',PINK,offset=(0.5,-0.5,0.4))
fig.update_layout(scene_camera=dict(eye=dict(x=2,y=-1.1,z=1.1)))
fig.show()

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The vector $\vec w=\begin{bmatrix}1\\-2\\1\end{bmatrix}$ is perpendicular to the plane $P$ .

To show that $\vec w$ is perpendicular to $P$ , we need to show that it’s orthogonal to every vector in $P$ . There are infinitely many such vectors, but we only need two dot products to get started:

\vec v_1\cdot\vec w=3(1)+4(-2)+5(1)=0,

\vec v_2\cdot\vec w=5(1)+2(-2)+(-1)(1)=0.

Remember that every vector in $P$ is a linear combination of $\vec v_1$ and $\vec v_2$ . Since both dot products above are 0, the dot product of $\vec w$ with any of their linear combinations is also 0:

\begin{aligned} (a\vec v_1+b\vec v_2)\cdot\vec w &=a(\vec v_1\cdot\vec w)+b(\vec v_2\cdot\vec w)\\ &=a(0)+b(0)=0. \end{aligned}

That’s why checking the two spanning vectors is enough. It tells us that $\vec w$ is perpendicular to the entire plane.

For our plane through the origin (where $d=0$ ), we can write

P=\left\{\begin{bmatrix}x\\y\\z\end{bmatrix}: \vec w\cdot\begin{bmatrix}x\\y\\z\end{bmatrix}=0\right\}.

The above form is sometimes called the dot-product form equation of a plane.

We’ll write $P^\perp$ for the set of vectors orthogonal to every vector in $P$ . Here it is the line

P^\perp=\operatorname{span}(\vec w).

This extends the perpendicular notation from Chapter 2.1 through Chapter 2.3.

In $\mathbb R^2$ , the vectors perpendicular to a line through the origin form another line.
In $\mathbb R^3$ , the vectors perpendicular to a plane through the origin form a line, while the vectors perpendicular to a line through the origin form a plane.

A normal vector is not unique. Using $3\vec w=\begin{bmatrix}3\\-6\\3\end{bmatrix}$ gives $3x-6y+3z=0$ , which is the original equation multiplied by 3. The plane does not change.

Finding a normal vector for a plane¶

So far, we’ve checked a normal vector after being given an equation. What if we only know the spanning vectors? Let’s find a normal to the same 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}.

Write the unknown normal as $\vec n=\begin{bmatrix}a\\b\\c\end{bmatrix}$ . It must be orthogonal to both spanning vectors, so

3a+4b+5c=0,\qquad 5a+2b-c=0.

The second equation gives $c=5a+2b$ . Substituting into the first gives

3a+4b+5(5a+2b)=0\quad\Longrightarrow\quad28a+14b=0.

Thus $b=-2a$ and $c=5a+2(-2a)=a$ . Choosing $a=1$ gives

\vec n=\begin{bmatrix}1\\-2\\1\end{bmatrix}.

These are exactly the coefficients in the plane’s equation:

x-2y+z=0.

Any nonzero choice of $a$ gives a scalar multiple of this normal and an equivalent equation for the same plane.

Here’s a video reviewing how to find the equation of the plane spanned by two vectors in $\mathbb{R}^3$ .

Activity 1

Find a nonzero normal to the plane spanned by

\vec u=\begin{bmatrix}5\\-7\\3\end{bmatrix},\qquad \vec v=\begin{bmatrix}4\\1\\-2\end{bmatrix}.

Use it to write an equation for the plane.

Solution

Write $\vec n=\begin{bmatrix}a\\b\\c\end{bmatrix}$ . Orthogonality to both spanning vectors gives

5a-7b+3c=0,\qquad 4a+b-2c=0.

The second equation gives $b=-4a+2c$ . Substituting into the first gives $33a-11c=0$ , so $c=3a$ and $b=2a$ .

Choosing $a=1$ gives $\vec n=\begin{bmatrix}1\\2\\3\end{bmatrix}$ . We can check that

\vec n\cdot\vec u=5-14+9=0,\qquad \vec n\cdot\vec v=4+2-6=0.

The plane’s equation is

x+2y+3z=0.

Activity 2

Find a nonzero normal to the plane spanned by

\vec u=\begin{bmatrix}6\\4\\-2\end{bmatrix},\qquad \vec v=\begin{bmatrix}4\\12\\4\end{bmatrix}.

Use it to write an equation for the plane.

Solution

Write the unknown normal as

\vec n=\begin{bmatrix}a\\b\\c\end{bmatrix}.

Its entries must satisfy

6a+4b-2c=0,\qquad 4a+12b+4c=0.

The first equation gives

c=3a+2b.

Substitute this into the second equation:

4a+12b+4(3a+2b)=0\quad\Longrightarrow\quad16a+20b=0.

Choose $a=5$ . Then

b=-4,\qquad c=7,\qquad \vec n=\begin{bmatrix}5\\-4\\7\end{bmatrix}.

Let’s check both dot products:

\vec n\cdot\vec u=30-16-14=0,\qquad \vec n\cdot\vec v=20-48+28=0.

So, the plane spanned by $\vec u = \begin{bmatrix}6\\4\\-2\end{bmatrix}$ and $\vec v = \begin{bmatrix}4\\12\\4\end{bmatrix}$ has the equation

5x-4y+7z=0.

Lines as intersections of planes¶

We now have an understanding of how planes in $\mathbb{R}^3$ can be expressed:

As the span of two linearly independent vectors in $\mathbb{R}^3$ .
As a linear equation,
$ax + by + cz = d$
(so far, we’ve only seen the case where $d=0$ .)

How do we describe lines using linear equations, when a single linear equation in terms of $x$ , $y$ , and $z$ describes a plane? That is what we will now explore. First, some terminology:

A linear system is a collection of linear equations in the same unknowns.
A solution gives values of the unknowns that satisfy every equation simultaneously. We typically express the solutions as vectors.
The solution set is the set of all solutions – we often think of this as a set of vectors.
A system is homogeneous if every right-hand side is zero. Otherwise, it is nonhomogeneous.

For a concrete example in $\mathbb R^2$ , consider

\begin{cases}x+y=7,\\x-y=1.\end{cases}

Adding the equations gives $2x=8$ , so $x=4$ and $y=3$ . Thus, the solution set is

\left\{\begin{bmatrix}4\\3\end{bmatrix}\right\}.

Geometrically, the two lines intersect at the point $(4,3)$ . The system is nonhomogeneous because its right-hand sides are not all zero.

Now let’s return to $\mathbb R^3$ . To describe a line, we need two linear equations and take their common solutions. Consider

\ell=\operatorname{span}\left(\begin{bmatrix}3\\4\\5\end{bmatrix}\right).

How do we find two equations? Pick two independent vectors $\vec n_1$ and $\vec n_2$ in $\ell^\perp$ , the plane of vectors perpendicular to $\ell$ .

fig=base3()
plane3(fig,(3,4,5),BLUE,0.22,bounds=(-5,5))
v=np.array([3,4,5])
line3(fig,-1.1*v,1.3*v,ORANGE,width=8)
vec3(fig,(1,-2,1),'<i>n</i>⃗<sub>1</sub>',PINK,offset=(0.2,-0.6,0.2))
vec3(fig,(2,1,-2),'<i>n</i>⃗<sub>2</sub>',PINK,offset=(0.4,0.2,-0.5))
fig.add_trace(go.Scatter3d(x=[3.9,-4],y=[5.2,4],z=[6.9,-0.8],mode='text',text=['ℓ','ℓ<sup>⊥</sup>'],textfont=dict(family='Palatino',size=25),hoverinfo='skip'))
fig.show()

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The two independent vectors $\vec n_1$ and $\vec n_2$ lie in $\ell^\perp$ (blue), perpendicular to the line $\ell$ (orange).

In our example, a vector $\begin{bmatrix}x\\y\\z\end{bmatrix}$ is in $\ell^\perp$ if

3x+4y+5z=0.

We can choose any values of $x$ and $y$ and solve for $z=-\frac{3x+4y}{5}$ . For instance, choosing $x=1,y=-2$ gives $z=1$ , while choosing $x=2,y=1$ gives $z=-2$ . Thus we can use

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

Both are perpendicular to $\begin{bmatrix}3\\4\\5\end{bmatrix}$ , and they are not multiples of each other. The line is exactly the set of vectors orthogonal to both normals, so it is the solution set of

\begin{cases}x-2y+z=0,\\2x+y-2z=0.\end{cases}

What’s happening geometrically? Each equation describes a plane:

P:\ x-2y+z=0,\qquad P':\ 2x+y-2z=0.

The line consists of the points on both planes. In other words,

\ell=P\cap P'.

The symbol $\cap$ means intersection.

def intersection_plot(normals,names):
    fig=base3()
    for normal,color in zip(normals,(BLUE,ORANGE)):
        plane3(fig,normal,color,0.28,bounds=(-5,7))
    v=np.array([3,4,5])
    line3(fig,-1.1*v,1.3*v,PINK,width=8)
    label_x = (5,-2) if normals[0] == (1,-2,1) else (1,-4)
    for normal,name,color,x,z in zip(normals,names,(BLUE,ORANGE),label_x,(-2,1)):
        y=-(normal[0]*x+normal[2]*z)/normal[1]
        fig.add_trace(go.Scatter3d(x=[x],y=[y],z=[z],mode='text',text=[name],textfont=dict(family='Palatino',size=26,color=color),hoverinfo='skip'))
    fig.add_trace(go.Scatter3d(x=[4.1],y=[5.5],z=[6.9],mode='text',text=['ℓ'],textfont=dict(family='Palatino',size=26,color=PINK),hoverinfo='skip'))
    return fig

fig=intersection_plot([(1,-2,1),(2,1,-2)],['<i>P</i>',"<i>P</i>′"])
fig.show()

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The blue plane $P$ and orange plane $P\prime$ intersect along the pink line $\ell$ .

Our choices of normals were somewhat arbitrary. For example, we could instead use

\vec m_1=\vec n_1+\vec n_2=\begin{bmatrix}3\\-1\\-1\end{bmatrix},\qquad \vec m_2=\vec n_1-\vec n_2=\begin{bmatrix}-1\\-3\\3\end{bmatrix}.

These are also perpendicular to $\ell$ and are not multiples of each other. They give another pair of planes,

Q:\ 3x-y-z=0,\qquad Q':\ -x-3y+3z=0,

whose intersection is the same line: $\ell=Q\cap Q'$ . The two descriptions are shown side by side below.

from plotly.subplots import make_subplots
left=intersection_plot([(1,-2,1),(2,1,-2)],['<i>P</i>',"<i>P</i>′"])
right=intersection_plot([(3,-1,-1),(-1,-3,3)],['<i>Q</i>',"<i>Q</i>′"])
fig=make_subplots(rows=1,cols=2,specs=[[{'type':'scene'},{'type':'scene'}]],horizontal_spacing=0.03,subplot_titles=['ℓ = P ∩ P′','ℓ = Q ∩ Q′'])
for col,panel in enumerate((left,right),1):
    for trace in panel.data:
        fig.add_trace(trace,row=1,col=col)
fig.update_layout(template=left.layout.template,paper_bgcolor='white',font=dict(family='Palatino',size=16),height=480,margin=dict(l=0,r=0,t=45,b=0),showlegend=False)
scene=left.layout.scene.to_plotly_json()
lower=min(left.layout.scene.xaxis.range[0],right.layout.scene.xaxis.range[0])
upper=max(left.layout.scene.xaxis.range[1],right.layout.scene.xaxis.range[1])
for axis in ('xaxis','yaxis','zaxis'):
    scene[axis]['range']=[lower,upper]
fig.update_scenes(scene)
fig.show()

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Different pairs of planes can have the same intersection: $\ell=P\cap P\prime=Q\cap Q\prime$ . Drag either panel to rotate its view.

Activity 3

Find two independent vectors perpendicular to

\vec d=\begin{bmatrix}4\\-7\\6\end{bmatrix}.

Use them to describe $\operatorname{span}(\vec d)$ as the solution set of a linear system. Then find a second pair of independent perpendicular vectors to describe the same line with a different linear system.

Solution

The entries of a normal must satisfy

4a-7b+6c=0.

Setting $c=0,b=4$ gives $a=7$ . Setting $b=0,c=-2$ gives $a=3$ . Hence we can use

\vec n_1=\begin{bmatrix}7\\4\\0\end{bmatrix},\qquad \vec n_2=\begin{bmatrix}3\\0\\-2\end{bmatrix}.

They are not scalar multiples, and

\vec n_1\cdot\vec d=28-28=0,\qquad \vec n_2\cdot\vec d=12-12=0.

One system is

\begin{cases}7x+4y=0,\\3x-2z=0.\end{cases}

Setting $x=4t$ recovers the scalar-parametric form

x=4t,\qquad y=-7t,\qquad z=6t,\qquad t\in\mathbb R.

For a second choice, add and subtract our two normals:

\vec m_1=\vec n_1+\vec n_2=\begin{bmatrix}10\\4\\-2\end{bmatrix},\qquad \vec m_2=\vec n_1-\vec n_2=\begin{bmatrix}4\\4\\2\end{bmatrix}.

Neither has a zero entry. They are not scalar multiples, and

\vec m_1\cdot\vec d=40-28-12=0,\qquad \vec m_2\cdot\vec d=16-28+12=0.

A second system describing the same line is

\begin{cases}10x+4y-2z=0,\\4x+4y+2z=0.\end{cases}

Adding and subtracting these equations recovers twice each equation of the first system, so both systems have the same solution set.

We’ve now described lines and planes through the origin using linear equations (here in Chapter 2.5) as well as in parametric form (in Chapter 2.4). In Chapter 2.6, we’ll translate these objects to study affine lines and planes, which do not need to pass through the origin.