In this section we’ll study geometrical objects like lines, planes, and vector spaces. We’ll use what we learned about vectors and matrices in the previous chapters to perform geometrical calculations such as projections and distance measurements.
Developing your intuition about the geometrical problems of linear algebra is very important: of all the things you learn in this course, your geometrical intuition will stay with you the longest. Years from now, you may not recall the details of the Gauss–Jordan elimination procedure, but you’ll still remember that the solution to three linear equations in three variables corresponds to the intersection of three planes in R3.
5.1 Lines and planes
Points, lines, and planes are the basic building blocks of geometry. In this section, we’ll explore these geometric objects, the equations that describe them, and their visual representations.
Concepts
• p = (px, py, pz): a point in R3
• ~v = (vx, vy, vz): a vector in R3
• ˆv = k~~vvk: the unit vector in the same direction as the vector ~v
• An infinite line ` is a one-dimensional space defined in one of several possible ways:
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. ` :{po+ t ~v, t∈ R}: a parametric equation of a line with direction vector ~v passing through the point po
. ` :nx−p
ox
vx =y−pvyoy =z−pvzozo: a symmetric equation
• An infinite plane P is a two-dimensional space defined in one of several possible ways:
. P :{Ax + By + Cz = D }: a general equation . P :{po+ s ~v + t ~w, s, t∈ R}: a parametric equation . P : {~n · [(x, y, z) − po] = 0}: a geometric equation of the
plane that contains point po and has normal vector ˆn
• d(a, b): the shortest distance between geometric objects a and b
Points
We can specify a point in R3by its coordinates p = (px, py, pz), which is similar to how we specify vectors. In fact, the two notions are equivalent: we can either talk about the destination point p or the vector ~p that takes us from the origin to the point p. This equivalence lets us add and subtract vectors and points. For example, ~d = q − p denotes the displacement vector that takes the point p to the point q.
We can also specify a point as the intersection of two lines. As an example in R2, let’s define p = (px, py)to be the intersection of the lines x − y = −1 and 3x + y = 9.
We must solve the two equations simultaneously to find the coordi-nates of the point p. We can use the standard techniques for solving equations to find the answer. The intersection point is p = (2, 3). Note that for two lines to intersect at a point, the lines must not be paral-lel.
Example 1 Find where the lines x + 2y = 5 and 3x + 9y = 21 intersect. To find the point of intersection, we solve these equations simultaneously and obtain the point (x, y) that is contained in both lines. The answer is the point p = (1, 2).
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In three dimensions, a point can also be specified as the intersection of three planes. This is precisely what happens when we solve equations of the form:
A1x + B1y + C1z = D1, A2x + B2y + C2z = D2, A3x + B3y + C3z = D3. To solve this system of equations, we must find the point (x, y, z) that sat-isfies all three equations, which means this point is contained in all three planes.
Lines
A line ` is a one-dimensional space that is infinitely long. There are several equivalent ways to specify a line in space.
The parametric equation of a line is ob-tained as follows. Given a direction vector
~vand some point po on the line, we define the line as the following set:
` : {(x, y, z) ∈ R3| (x, y, z) = po+t~v, t∈ R}.
The line consists of all the points (x, y, z) that can be reached starting from the point poand adding any multiple of the direction
vector ~v. We say the line is parametrized by the variable t.
The symmetric equation is an equivalent way to describe a line that does not require an explicit parametrization. Consider the equations that correspond to each coordinate in the parametric equation of a line:
x = pox+ t vx, y = poy+ t vy, z = poz+ t vz.
When we solve for t in these equations and equate the results, we obtain the symmetric equation of a line:
` :
x− pox vx
= y− poy vy
= z− poz vz
.
Note the parameter t does not appear. The symmetric equation spec-ifies the line as the relationships between the x, y, and z coordinates that hold for all points on the line.
You’re probably most familiar with the symmetric equation of lines in R2, which do not involve the variable z. For non-vertical lines
in R2 (vx 6= 0), we can think of y as a function of x and write the equation of the line in the equivalent form:
x− pox vx
= y− poy
vy ⇒ y(x) = mx + b,
where m = vvyx and b = poy− vvyxpox. The equation m = vvyx makes sense intuitively: the slope of a line m corresponds to how much the line “moves” in the y-direction divided by how much the line “moves”
in the x-direction.
Another way to describe a line is to specify two points that are part of the line. The equation of a line that contains the points p and q can be obtained as follows:
` : {~x = p + t (q − p), t ∈ R}, where (q − p) plays the role of the direction vector ~v for this line. Any vector parallel to the line can be used as the direction vector for the line.
Example 2 Find the parametric equation of the line that passes through the points p = (1, 1, 1) and q = (2, 3, 4). What is the sym-metric equation of this line?
Using the direction vector ~v = q − p = (1, 2, 3) and the point p on the line, we can write a parametric equation for the line as {(x, y, z) ∈ R3 | (x, y, z) = (1, 1, 1) + t(1, 2, 3), t ∈ R}. Note that a parametric equation using the direction vector (−1, −2, −3) would be equally valid: {(1, 1, 1) + t(−1, −2, −3), t ∈ R}. The symmetric equation of the line is x−11 = y−12 = z−13 .
Lines as intersections of planes In three dimensions, the intersection of two non-parallel planes forms a line. For exam-ple, the intersection of the xy-plane Pxy : {(x, y, z) ∈ R3 | z = 0} and the xz-plane Pxz :{(x, y, z) ∈ R3 | y = 0} is the x-axis:
{(x, y, z) ∈ R3| (0, 0, 0) + (1, 0, 0)t, t ∈ R}.
For this simple case, we can imagine the two planes (use your hands) and visually establish that they intersect along the x-axis. Wouldn’t it be nice if there was a
general procedure for finding the line of intersection of two planes?
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You already know such a procedure! The line of intersection be-tween the planes A1x + B1y + C1z = D1 and A2x + B2y + C2z = D2
is the solution of the following set of linear equations:
A1x + B1y + C1z = D1, A2x + B2y + C2z = D2.
Example 3 Find the intersection of the planes 0x+0y +1z = 0 and 0x + 1y + 1z = 0. We follow the standard Gauss–Jordan elimination procedure: construct an augmented matrix, perform row operations (denoted ∼), obtain the RREF, and interpret the solution:
0 0 1 0 The first column is a free variable t ∈ R. The solution is the line
which corresponds to the x-axis.
Planes
A plane P in R3 is a two-dimensional space with infinite extent. In general, we specify a plane through a constraint equation that must be satisfied by all points in the plane:
P : {(x, y, z) ∈ R3 | Ax + By + Cz = D}.
The plane P is the set of all points (x, y, z) ∈ R3 that satisfy the equation Ax + By + Cz = D. The equation Ax + By + Cz = D is called the general equation of the plane. This definition represents the algebraic view of planes, which is useful for calculations.
There is an equally useful geometric view of planes. A plane can be specified by a normal vector ~n and some point po
in the plane. The normal vector ~n is per-pendicular to the plane: it sticks out at right angles to the plane like the normal force between surfaces in physics problems.
All points in the plane P can be obtained starting from the point poand moving in a direction orthogonal to the normal vector
~n. The geometric formula of a plane is
P : ~n· [(x, y, z) − po] = 0.
Recall that the dot product of two vectors is zero if and only if these vectors are orthogonal. In the above equation, the expression [(x, y, z)−po]forms an arbitrary vector with one endpoint at po. From all these vectors, we select only those that are perpendicular to ~n and thus we obtain all the points in the plane.
The geometric equation ~n · [(x, y, z) − po] = 0is equivalent to the general equation Ax + By + Cz = D. We can find the parameters A, B, C, and D by calculating the dot product: A = nx, B = ny, C = nz, and D = ~n · po= nxpox+ nypoy+ nypoz.
Observe that scaling the general equation of a plane by a constant factor does not change the plane: the equations Ax + By + Cz = D and αAx + αBy + αCz = αD define the same plane. Similarly the geometric equations ~n · [(x, y, z) − po] = 0and α~n · [(x, y, z) − po] = 0 define the same plane. In each case, the direction of the normal vector
~nis important, but not its length.
We can also give a parametric equation of a plane P . If we know a point poin the plane and two linearly independent vectors ~v and ~w that lie in the plane, then a parametric equation for the plane can be obtained as follows:
P : {(x, y, z) ∈ R3 | (x, y, z) = po+ s ~v + t ~w, s, t∈ R}.
Since a plane is a two-dimensional space, we need two parameters (s and t) to describe the location of arbitrary points in the plane.
Suppose we’re given three points p, q, and r that lie in the plane. How can we find the geometric equation for this plane ~n · [(x, y, z) − po] = 0? We can use the point p as the reference point po, but how do we find the normal vector ~n for the plane? The trick is to use the cross product. First we build two vectors that are parallel to the plane, ~v = q − p and
~
w = r− p, and then compute their cross product to find a vector that is
perpen-dicular to both of them, and hence normal to the plane.
~n = ~v× ~w = (q− p) × (r − p).
We can use the vector ~n to write the geometric equation of the plane
~n· [(x, y, z) − p] = 0. The key property we used is the fact that the cross product of two vectors is perpendicular to both vectors. The cross product is the perfect tool for finding normal vectors.