Lines and Planes

MAT 229, Spring 2021

Week 16

Supporting materials

If you wish to get a different perspective on the notes below, try either of the following textbook sections.

Stewart’s Calculus
	Section 12.5: Lines and planes

Boelkins/Austin/Schlicker’s Active Multivariable Calculus
	Section 9.5: Lines and planes in space

Review

Questions

How can we easily determine if two vectors are perpendicular or not?

What is a simple way to find a 3D vector that is orthogonal (perpendicular) to two given vectors?

(Video)

Questions

Consider the parametric equations
	x(t)=3t+5
y(t)=-2t+1

What is the point associated with t=0?

What is the point associated with t=1?

What is the vector with initial point (x(0),y(0)) and terminal point (x(1),y(1))?

Find a Cartesian equation for this parametrically defined curve.

(Video)

Questions

Consider the parametric equations
	x(t)=a t+(Subscript[x, 0])
y(t)=b t+(Subscript[y, 0])

What is the point associated with t=0?

What is the point associated with t=1?

What is the vector with initial point (x(0),y(0)) and terminal point (x(1),y(1))?

Find a Cartesian equation for this parametrically defined curve to show it is a line.

What is its slope?

How does the vector found above compare with the line?

(Video)

Parametric equations of lines in space

Questions

What information is needed to find an equation for a line in the plane?

How many points determine a line in the plane?

How many points determine a line in space?

(Video)

Parametric equations

The parametric equations
	x(t)=a t+(Subscript[x, 0])
y(t)=b t+(Subscript[y, 0])
z(t)=c t+(Subscript[z, 0])
produce a line in space.

As a vector equation:

It's better to think of this as a vector equation:

	⟨x(t),y(t),z(t)⟩
=⟨a t+(Subscript[x, 0]), b t+(Subscript[y, 0]), c t+(Subscript[z, 0])⟩=⟨a, b, c ⟩ t  +  ⟨(Subscript[x, 0]), (Subscript[y, 0]), (Subscript[z, 0])⟩
Furthermore, we can make the analogy of the vector ⟨a, b, c ⟩ as "the slope", and  ⟨(Subscript[x, 0]), (Subscript[y, 0]), (Subscript[z, 0])⟩ as "the intercept" -- the point you’re at when t=0.

Two-dimensional version:

Think about the two-dimensional analogue: in 2d, we think of x as the independent and y as the dependent variable. The two corresponding parametric equations are
	x(t)= t
y(t)=m t+(Subscript[y, 0])
which produce a line in the plane: 

		⟨x(t),y(t)⟩
=⟨ t, m t+(Subscript[y, 0])⟩=⟨1, m ⟩ t  +  ⟨0, (Subscript[y, 0])⟩

Manipulate[
start={0,y0};
end={1,y0+m};
Show[
Plot[m x + y0,{x,-1,2}],
Graphics[Arrow[{start,end}]],
PlotRange->{{-1.5,1.5},{-5,5}}
],
{m,-4,4},
{y0,0,3}
]

Questions

Consider the line given by
	x(t)=2 t+3
y(t)=-t+2
z(t)=3 t

What is the point (x(0),y(0),z(0))?

What is the point (x(1),y(1),z(1))?

What is the vector that points from (x(0),y(0),z(0)) to (x(1),y(1),z(1))?

What relation does this vector have to the line?

(Video)

x[t_]:=2t+3
y[t_]:=-t+2
z[t_]:=3t
start={x[0],y[0],z[0]};
end={x[1],y[1],z[1]};
Manipulate[
Show[
ParametricPlot3D[t (end-start) + start,{t,-2,2}],
Graphics3D[{Red,Arrowheads[0.05],Arrow[Tube[{start,end},0.05]]}],
ListPointPlot3D[{{start,end}}, PlotStyle->PointSize[.05]],
ListPointPlot3D[{{x[time],y[time],z[time]}}, PlotStyle->Black]
],
{time,-2,2}
]

Questions

Given the parametric equations
	x(t)=a t+(Subscript[x, 0])
y(t)=b t+(Subscript[y, 0])
z(t)=c t+(Subscript[z, 0])

What is the point on the line associated with t=0?

What is a vector parallel to the line?

(Video)

⟨x(t),y(t),z(t)⟩
=⟨a, b, c ⟩ t  +  ⟨(Subscript[x, 0]), (Subscript[y, 0]), (Subscript[z, 0])⟩

Questions

Find parametric equations for the line through point (-1,0,1) and parallel to vector ⟨3,4,0⟩.		(Video)

⟨x(t),y(t),z(t)⟩=⟨3,4,0 ⟩ t  +  ⟨-1,0,1⟩=⟨3t-1,4t,1 ⟩

direction={3,4,0};
start={-1,0,1};
end=start+direction;
line[t_]:=t direction+start;
Manipulate[
Show[
ParametricPlot3D[line[t],{t,-2,2}],
Graphics3D[{Red,Arrowheads[0.05],Arrow[Tube[{start,end},0.05]]}],
ListPointPlot3D[{{start,end}}, PlotStyle->PointSize[.05]],
ListPointPlot3D[{{line[time]}}, PlotStyle->Black]
],
{time,-2,2}
]

Find parametric equations for the line through point (2,3,-1) and parallel to the line given parametrically by x=-t,y=2t-3,z=t+1.	(Video)

direction={-1,2,1};
start={2,3,-1};
end=start+direction;
line[t_]:=t direction+start;
Manipulate[
Show[
ParametricPlot3D[t direction + start,{t,-2,2}],
ParametricPlot3D[{-t,2t-3,t+1},{t,-2,2}],
Graphics3D[{Red,Arrowheads[0.05],Arrow[Tube[{start,end},0.05]]}],
ListPointPlot3D[{{start,end}}, PlotStyle->PointSize[.05]],
ListPointPlot3D[{{line[time]}}, PlotStyle->Black],
PlotRange->All
],
{time,-2,2}
]

Find parametric equations for the line through points (0,1,1) and (2,3,5).

start={0,1,1};
end={2,3,5};
direction=end-start;
line[t_]:=t direction+start; (* = <x(t),y(t),z(t)> *)
Manipulate[
Show[
ParametricPlot3D[line[t],{t,-2,2}],
Graphics3D[{Red,Arrowheads[0.05],Arrow[Tube[{start,end},0.05]]}],
ListPointPlot3D[{{start,end}}, PlotStyle->PointSize[.05]],
ListPointPlot3D[{{line[time]}}, PlotStyle->Black]
],
{time,-2,2}
]

Find a different set of parametric equations for the line through points (0,1,1) and (2,3,5).

Just swap start and end points in the commands above! That reverses the motion; and though the point moves differently, it will trace out exactly the same line!

(Video)

Vector equation of a line in space

Definitions

The position vector (Overscript[u, ->])=⟨a,b,c⟩ is the vector with initial point at the origin and terminal point at (a,b,c).

The parametric vector equation
	(Overscript[r, ->])(t)=t (Overscript[u, ->])+(Subscript[(Overscript[r, ->]), 0])
for each value of t is the position vector for a point on the line parallel to vector (Overscript[u, ->]), that passes through the terminal point of position vector (Subscript[(Overscript[r, ->]), 0]).

rr0={1,0,2};
uu={-1,1,1};
r[t_]:=t uu+rr0;
fixedGraphics=Show[Graphics3D[{Arrow[{{0,0,0},rr0}],Arrow[{rr0,rr0+uu}],Text[(Subscript[r, 0]),.5rr0,{1,1}],Text[u,rr0+.5uu,{1,1}]},BaseStyle->(FontSize->18)],ParametricPlot3D[r[t],{t,-2,2}]];
Manipulate[Show[fixedGraphics,Graphics3D[{Blue,Arrow[{{0,0,0},r[t]}],Green,Arrow[{rr0,r[t]}]}]],{{t,0},-2,2}]

Question

What is a vector equation for the line that passes through (1,0,2) and that is parallel to vector ⟨-1,1,1⟩?	(Video)

Planes

Question

What is the minimum information needed to determine a plane?	(Video)

Questions

Given a plane let
	(Overscript[n, ->]) be a vector perpendicular to it,
	(Subscript[P, 0])((Subscript[x, 0]),(Subscript[y, 0]),(Subscript[z, 0])) be a particular point on it.
If P(x,y,z) is any point on the  plane

What is true about the vector (Overscript[(Subscript[P, 0]) P, ⟶])?

What relation does this vector have with vector (Overscript[n, ->])?

If (Overscript[n, ->]) in component form is ⟨a,b,c⟩, write an equation that x,y,z must satisfy to lie on the plane that passes through ((Subscript[x, 0]),(Subscript[y, 0]),(Subscript[z, 0])) and that is perpendicular (normal) to (Overscript[n, ->]).

(Video)

Equation of a plane

The points with coordinates (x,y,z) that satisfy the equation
				a(x-(Subscript[x, 0]))+b(y-(Subscript[y, 0]))+c(z-(Subscript[z, 0]))=0
form a plane that passes through ((Subscript[x, 0]),(Subscript[y, 0]),(Subscript[z, 0])) and that is normal to vector ⟨a,b,c⟩.

Frequently we solve for z, and so think of z(x,y) as the plane, a function of the two "independent" coordinates x and y:
				z(x,y)=[a(x-(Subscript[x, 0]))+b(y-(Subscript[y, 0]))+c(-(Subscript[z, 0]))]/(-c)
Or, maybe better yet, in analogy with the equation of a line in two dimensions
			z=(Subscript[z, 0])-[(a/c)(x-(Subscript[x, 0]))+(b/c)(y-(Subscript[y, 0]))]=(Subscript[z, 0])+(Subscript[m, x])(x-(Subscript[x, 0]))+(Subscript[m, y])(y-(Subscript[y, 0])).
And so we defined “slopes” in both the x and y directions.

Alternatively (certainly better), 
				 ⟨a,b,c⟩· ( ⟨x,y,z⟩  -  ⟨(Subscript[x, 0]),(Subscript[y, 0]),(Subscript[z, 0])⟩ ) = 0 
That is,  the vector ⟨a,b,c⟩ and a vector lying in the plane, given as the difference of the two vectors  ⟨x,y,z⟩  -  ⟨(Subscript[x, 0]),(Subscript[y, 0]),(Subscript[z, 0])⟩ , has a dot product of 0: they're perpendicular.

Questions

Find an equation for the plane that passes through point (1,0,-1) and that is normal to vector ⟨2,3,-4⟩.	(Video)

p0={1,0,-1};
uu={2,3,-4}; 
uhat=uu/Norm[uu]; (* make it a unit vector.... *)
perp1={3,-2,0}; (* This was a cheap and dirty vector perpendicular to uu -- do you see how I did it? *)
perp1hat=perp1/Norm[perp1]; (* make it a unit vector.... *)
(* Now we use that cross-product! *)
perp2hat=Cross[uhat,perp1hat];  (* and this will be a unit vector.... *)r[t_]:=Sin[t]perp1hat+Cos[t]perp2hat+p0
(* r[t] is will trace out a circle in the plane -- just for fun! *)
z[x_,y_]:=(2x+3y-uu.p0)/4 (* The equation of the plane *)

p1=Plot3D[z[x,y],{x,0,2},{y,-1,1}]; (* plot the plane *)
p2=ParametricPlot3D[r[t],{t,0,2Pi}]; (* plot the circle *)
axes=Graphics3D[  (* plot the axes *)
{
Arrow[{p0,p0+uhat}],
Arrow[{p0,p0+perp1hat}],
Arrow[{p0,p0+perp2hat}]
}]
Show[p1,p2,axes,PlotRange->All,AspectRatio->Full]  (* all together now! *)

What is a vector normal to the plane given by 3(x+2)-y+2(z-1)=0? What is a point on this plane?	(Video)

Recall:
The points with coordinates (x,y,z) that satisfy the equation
	3(x-(-2))+(-1)(y-0)+2(z-1)=0
form a plane that passes through (-2,0,1) and that is normal to vector ⟨3,-1,2⟩.

Alternatively (better), 
				 ⟨3,-1,2⟩· ( ⟨x,y,z⟩  -  ⟨-2,0,1⟩ ) = 0

Rewrite the last equation so that all the variables are on one side of the equation and the constants are summed on the other side.	(Video)

What is an equation for the plane that passes through (0,1,0) that is parallel to both ⟨1,1,1⟩ and ⟨0,1,-1⟩?	(Video)

p0={0,1,0};
perp1={1,1,1}; 
perp1hat=perp1/Norm[perp1];
perp2={0,1,-1}; (* This happens to be perpendicular to perp1.  *)
perp2hat=perp2/Norm[perp2];
uu=Cross[perp1hat,perp2hat]; (* this is the normal *)
uhat=uu/Norm[uu];
r[t_]:=Sin[t]perp1hat+Cos[t]perp2hat+p0
z[x_,y_]:=(uhat[[1]]x+uhat[[2]]y-uhat.p0)/(-uhat[[3]])

p1=Plot3D[z[x,y],{x,p0[[1]]-1,p0[[1]]+1},{y,p0[[2]]-1,p0[[2]]+1}]; (* plot the plane *)
p2=ParametricPlot3D[r[t],{t,0,2Pi}]; (* plot the circle *)
axes=Graphics3D[  (* plot the axes *)
{
Arrow[{p0,p0+uhat}],
Arrow[{p0,p0+perp1hat}],
Arrow[{p0,p0+perp2hat}]
}];
Show[p1,p2,axes,PlotRange->All,AspectRatio->2]  (* all together now! *)

Find an equation for the plane that passes through the three points (Subscript[P, 0])(1,1,1), (Subscript[P, 1])(2,0,-1), and (Subscript[P, 2])(0,1,3).	(Video)

p0={1,1,1};
perp1={2,0,-1}-p0; 
perp2={0,1,3}-p0; 
perp1.perp2 (* Not mutually orthogonal! *)
uu=Cross[perp1,perp2]; (* uu is the normal to both *)
uhat=uu/Norm[uu]; (* make it a unit vector *)
perp1hat=perp1/Norm[perp1];(* ditto *)
perp2hat=Cross[uhat,perp1hat]; (* We redefine perp2hat, so that we have an orthogonal coordinate system! *)
r[t_]:=Sin[t]perp1hat+Cos[t]perp2hat+p0
z[x_,y_]:=(uhat[[1]]x+uhat[[2]]y-uhat.p0)/(-uhat[[3]])

p1=Plot3D[z[x,y],{x,p0[[1]]-1,p0[[1]]+1},{y,p0[[2]]-1,p0[[2]]+1}]; (* plot the plane *)
p2=ParametricPlot3D[r[t],{t,0,2Pi}]; (* plot the circle *)
axes=Graphics3D[  (* plot the axes *)
{
Arrow[{p0,p0+uhat}],
Arrow[{p0,p0+perp1hat}],
Arrow[{p0,p0+perp2hat}]
}];
Show[p1,p2,axes,PlotRange->All,AspectRatio->Full]  (* all together now! *)

Find an equation for the plane that is parallel to x+2y+3z=20 and that passes through the point (1,0,-1).	(Video)

Homework

IMath problems on lines and planes.