1.3 Lines and Planes

In the xy-plane, the general form of the equation for a line is ax+by = c.  If b \neq 0 this can be written in the familiar point-slope form: y = -(a/b) x+ c/b where m = -(a/b) is the slope of the line and k = c/d is the y-intercept.  

To get vectors into the picture, examine the line l: 3x-2y = 0.

If we let \vec{n} = [3,-2] and \vec{x} = [x,y] then the equation for the line becomes \vec{n} \cdot \vec{x} = 0.

The vector \vec{n} is perpendicular to the line, i.e., orthogonal to any vector \vec{x} that is parallel to the line.  We say that \vec{n} is a normal vector to the line and that the equation \vec{n} \cdot \vec{x} is the normal form of the equation of the line.  

We can also think of the line in terms of parametric equations.  Suppose a particle is initially at the origin at time t=0 and it moves along the line in such a way that its x-coordinate changes 1 unite per second.  That at t = 1 the particle is at (1, 3/2) and in general if x = t then y = 3/2 t.  We can write this relationship in vector form as

\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r} x \\ y \end{array}\right] = \newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r} t \\ 3/2 t \end{array}\right] = t \newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r}  1 \\ 3/2 \end{array}\right]

What is the significance of the vector \vec{d} = \newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r}  1 \\ 3/2 \end{array}\right]?  

This vector is is a particular vector parallel to the line such that \vec{x} = t \vec{d}.  This form of the line equation is called the vector form.

What happens if we have the equation of the line k: 3x-2y = 5?  So that the line k doesn't go through the origin?  With a slight modification, everything will work.

Notice that the new line k is just l shifted down by 5.  The line n has the same slope as l, 3/2, but a different y-intercept, -5.  So it is clear that the vectors \vec{n} and \vec{d} are the normal vector and direction vector for this line as well.  

From this we see that \vec{n} is orthogonal to every vector that is parallel to k.  To create an example of a vector parallel to k, take P=(1,-1) a point on k.  If X = (x,y) is any point on k, then the vector \vec{PX} = \vec{x}-\vec{p} is parallel to k.  Further, \vec{n} \cdot (\vec{x}-\vec{p}) = 0.  Since we know the properties of the dot product, we can rearrange this equation to \vec{n} \cdot \vec{x} = \vec{n} \cdot \vec{p}.  But \vec{n} \cdot \vec{x} = 3x-2y and \vec{n}\cdot \vec{p} = 5.  

So the normal form equation \vec{n} \cdot \vec{x} = \vec{n} \cdot \vec{p} gives us the general equation of the line k.  Notice that for l, \vec{p} was just the zero vector, so \vec{n} \cdot 0 = 0 gave us the right hand side of the equation for l.

Definition:

The normal form of the equation of a line l in \mathbb{R}^2 is 

\vec{n} \cdot (\vec{x} - \vec{p}) = 0 or \vec{n} \cdot \vec{x} = \vec{n} \cdot \vec{p}

where \vec{p} is a specific point on l and \vec{n} \neq 0 is a normal vector for l.

The general form of the equation of l is ax + by = c where \vec{n} = [a,b] is a normal vector for l.

Continuing with the example, what is the vector form of k?  Note that for each choice of \vec{x}, the vector \vec{x} - \vec{p} must be parallel two and thus a multiple of the direction vector \vec{d}, i.e., \vec{x}-\vec{p} = t \vec{d} for some scalar t.  Rearranging, \vec{x} = \vec{p} + t \vec{d}.  In terms of components we now have:

\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r} x \\ y \end{array}\right] =\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r} 1 \\ -1 \end{array}\right]+ t \newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r}  1 \\ 3/2 \end{array}\right]

or as parametric equations:

x = 1 + t and y = -1 + 3/2 t

The concept of slope is a weird one in \mathbb{R}^3, but replacing it by the normal and direction vectors allows us to generalize nicely.

Definition:

The vector form of the equation of a line l in \mathbb{R}^2 or \mathbb{R}^3 is 

\vec{x} = \vec{p} + t \vec{d}

where \vec{p} is a specific point on l and \vec{d} \neq 0 is a direction vector for l.  

The equations corresponding to the components of the vector form of the equations are called parametric equations of l.

Now a 3 dimensional example, this time starting with a point and a direction vector.

P = (-1,0,1) and \vec{d} = \newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r} 2 \\ 1 \\ -1 \end{array}\right]

The vector equation for this is \vec{x} = \vec{p} + t \vec{d} or 

\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r} x \\ y \\ z \end{array}\right] = \newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r} -1\\ 0 \\ 1 \end{array}\right] + t \newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r} 2 \\ 1 \\ -1 \end{array}\right] 

and so the parametric equations are x = -1 + 2t, y = t, z = 1-t.

Remark:

The vector and parametric equations of a line are not unique, in fact there are infinitely many since we may use any point on the line to construct them.  As all direction vectors are multiples of each other, the direction is in fact unique.  

If we let t = 1, we get another point on the line (1,1,0).  Multiplying \vec{d} by the scalar 2, we get another direction vector [4,2,-2] and thus new vector and parametric equations:

\newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r} x \\ y \\ z \end{array}\right] = \newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r} 1 \\ 1 \\ 0 \end{array}\right] + s \newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r} 4 \\ 2 \\ -2 \end{array}\right]

and x = 1 +4s, y = 1 +2 s, z = -2 s which define the same line.  Each of these equations reduces to t = 1+2s.

Also notice that if we are given another point on the line so that we have P and Q, we no longer need the direction vector, as we can construct one from P and Q, namely \vec{d} = \vec{PQ}.

Intuitively we know that a line is a 1 dimensional object and that \mathbb{R}^2, a plane, is a 2 dimensional object.  We will clarify this concept later (chapters 3 and 6) but for now, notice that the idea of dimension seems to agree with the fact that the vector form of a equation of a line requires one parameter (and as we will see shortly, for a plane, we need two parameters.) 

Also notice, that the number of parameters a line needs (i.e., 1) is independent of where the line 'lives', either in \mathbb{R}^2 or \mathbb{R}^3.

The next question we should ask ourselves is, How does the general form of the equation of a line in \mathbb{R}^2, i.e., ax +by = c, generalize to \mathbb{R}^3?  

The obvious guess is ax+by+cz = d.  In normal form this would be \vec{n} \cdot \vec{x} = \vec{n} \cdot \vec{p} where \vec{n} is a normal vector to the line and \vec{p} corresponds to a point on the line.  First, let's look at the special case of P = (0,0,0) so \vec{p} = 0.  Then the normal form is \vec{n} \cdot \vec{x} = 0.  What does this mean geometrically?  

Geometrically we want all solutions to \vec{n} \cdot \vec{x} = 0 which is all vectors \vec{x} such that \vec{x} is orthogonal to \vec{n}.  

Here's an example where \vec{n} = [0,0,1]

So instead of a line, we get a plane!

Let's make our finding more precise.  Every plain \mathcal{P} in \mathbb{R}^3 is determined by specifying a point \vec{p} on \mathcal{P} and a nonzero vector \vec{n} normal to \mathcal{P}.  Thus, if \vec{x} represents an arbitrary point on \mathcal{P}, we have \vec{n} \cdot (\vec{x} - \vec{p}) = 0 or \vec{n} \cdot \vec{x} = \vec{n} \cdot \vec{p}.  If 

\vec{n} = \newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r} a \\ b \\ c \end{array}\right] and \vec{x} = \newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r} x \\ y \\ z \end{array}\right] the equation becomes

ax + by + cd = d where d = \vec{n} \cdot \vec{p}.  As before these are the normal form and the general form of the plane \mathcal{P}. 

Let \mathcal{P} be the plane defined by having the point P = (1,2,1) and normal vector \vec{n} = \newcommand{\Bold}[1]{\mathbf{#1}}\left[\begin{array}{r}  3 \\ 0 \\ 1 \end{array}\right] .  Then we get the general equation 3x+z = 4.

Geometrically, it is obvious that parallel planes have the same normal vectors.  Thus, their general equations have left hand sides that are multiples of each other.  For example, 9x+3z=3 is parallel to 3x+z=4.

Now we would like to express the equation of a plane in vector or parametric form.  To do so, notice that if we pick one point and one vector, that doesn't give us enough information.

If we specify one point and two direction vectors, then this does determine a plane.  This is equivalent to saying that our parametric equations will have two parameters.  This agrees with the intuition that a line is one dimensional and a plane is two dimensional.

Definition

The vector form of the equation of a plane \mathcal{P} in \mathbb{R}^3 is \vec{x} = \vec{p} + s \vec{u} +t \vec{v} where \vec{p} is a point on the plane and \vec{u} and \vec{v} are direction vectors for \mathcal{P}.  The direction vectors \vec{u} and \vec{v} must be non-zero, parallel to \mathcal{P}, but not parallel to each other.  

The corresponding equations are called the parametric equations of \mathcal{P}.

With the example we had before, we can find two additional points Q = (1,1,1) and R = (0,0,4).  

From these we get the direction vectors \vec{u} = \vec{q}-\vec{p} = [0,1,0] and \vec{v} = \vec{r} - \vec{p} = [-1,-1,3]

which give the parametric equations x = 1-t, y = s-t, z = 1+3t.

Remarks:

A plane is a two-dimensional object and its equation in vector or parametric form requires two parameters.

Given a point P and a vector \vec{n} in \mathbb{R}^3 there are infinitely many lines through P with \vec{n} as a normal vector.  However P and two normal vectors do uniquely determine a line.  Thus a line in \mathbb{R}^3 can be specified by a pair of equations:

a_1 x + b_1 y + c_1 z = d_1 and a_2 x + b_2 y + c_2  z = d_2

one corresponding to each normal vector.  Notice that each equation by itself describes a plane, so in \mathbb{R}^3 we describe lines as the intersections of planes.  

Observe that in \mathbb{R}^2 the general equation described a line and in \mathbb{R}^3 the general equation described a plane.  In higher dimensions the object described by a single equation of this type will be called a hyperplane.  The relationship among the dimension of the object, the number of equations required, and the dimension of the space is given by the "balancing formula"

dimension of object + number of general equations = dimension of the space

i.e., the higher the dimension the object, the fewer equations it needs.  Note that the dimension of the object also agrees with the number of parameters in its vector or parametric form.

If we combine the concept of projection from 1.2 with what we've learned about lines and planes, we can now calculate the distance from a point to a line or  plane.

Example:

Find the distance from the point B = (0,0,1) to the line l through the point A = (-1,0,1) with direction vector \vec{d} = [2,1,-1].

What we need to calculate is the length of \vec{PB} where P is the point on l at the foot of the perpendicular from B.  If \vec{v} = \vec{AB}, then \vec{AP} = proj_{\vec{d}}(\vec{v}) and \vec{PB} = \vec{v} - proj_{\vec{d}}(\vec{v}).

Computing \vec{v} = \vec{b} - \vec{a} = [1,0,0] and

proj_{\vec{d}} (\vec{v}) = \frac{\vec{d} \cdot \vec{v}}{\vec{d} \cdot \vec{d}} \vec{d} = [2/3, 1/3, -1/3] we find the length of \vec{v} - proj_{\vec{d}} (\vec{v}) is \sqrt{1/3}.

Now suppose that we wanted to find the distance between a point an a plane?  How do things change?

Let B = (-1,1,1) and \mathcal{P}: 3x + z = 4.  What is the distance from B to \mathcal{P}?

In this case we need to calculate the length of PB, where P is the point on \mathcal{P} at the foot of the perpendicular from B.  If A is any point on \mathcal{P} and we situate the normal vector \vec{n} = [3,0,1] of \mathcal{P} so that its tail is at A, then we need to find the length of the projection of \vec{AB} onto \vec{n}.  

Note: When we were finding the distance between a line and a point, we didn't have a normal vector, we had a direction vector, which is why we had to compute the length of the difference of two vectors.  

Let A = (1,0,1).  Then the we let \vec{n} = [3,0,1], but starting at (1,0,1), so it is the vector between the points (1,0,1) and (4,0,2).

Computing \vec{v} = \vec{AB} = \vec{b} - \vec{a} = [-2,1,0].  

Then proj_{\vec{n}}(\vec{v}) = (-6/10)[3,0,1] = [-9/5,0,-3/5], which has length 3\sqrt{10}/5.