Chapter 8
Goals
ˆ to be able to recognize the various forms (scalar, parameter, vector) of the equations for
lines in two- and three-space and planes (three-space) and be able to convert between
them
ˆ to understand the properties of lines and planes, including parallelism and perpendicularity and be able to solve distance problems
ˆ to understand the physical properties for the intersections of lines in two- and threespace, lines and planes (three-space) and planes (three-space) and see the connection
with solutions of linear systems of equations
ˆ to understand how to perform elimination on the augmented matrix representing a
linear system to find the solution
Equations of Lines in Two– and Three–Space
In two-space, a line can be given in slope-intercept form y = mx+b or with the scalar equation
Ax + By + C = 0.
Example:
y
y = 2x + 6
12
6
1
x
3
The line y = 2x + 6 has slope m = 2 and y-intercept b = 6. Its scalar equation form would
be 2x − y + 6 = 0.
2
r_0
r
m
We can represent this line with a vector equation. A direction vector parallel to the line is
m
~ = [1, 2]. A position vector that has its tip on the line would be r~0 = [1, 8] (since the point
(1, 8) is on the line). Another position vector ~r = [3, 12] also touches the line (since (3, 12)
is on the line). Then, if we let ~s be the vector joining (1, 8) to (3, 12), we would have that
~r = r~0 +~s. But since ~s is parallel to m,
~ we have that ~r = r~0 +tm.
~ In this particular case, t = 2
– can you see that ? If we were to let t vary, we could get any other point on the line, so we
~ t ∈ R or [x, y] = [x0 , y0 ] + t[m1 , m2 ].
have that the vector equation of the line is ~r = r~0 + tm,
Notice that this is not unique – we can use any point (x0 , y0 ) on the line and any direction
vector m
~ parallel to it.
If we separate the components, we have the parametric equations of the line, x = x0 + tm1 ,
y = y0 + tm2 , t ∈ R.
So for our example above, x = 1 + t and y = 8 + 2t.
Example:
Is the point (−1, 4) on the line ? How about the point (2, 9) ?
If (−1, 4) is on the line, then we must have that −1 = 1 + t and 4 = 8 + 2t for the same
value of t. So, yes, (−1, 4) is on the line since t = −2 satisfies both equations.
For the other point, 2 = 1 + t requires t = 1 but 9 = 8 + 2t requires that t = 1/2 (ie not the
same value) so the point (2, 9) is not on the line.
Example:
Find the vector equation of the line passing through the points P1 = (−1, 2) and P2 = (2, −4).
We need a direction vector, so we use m
~ = P1~P2 = [3, −6]. We can use either point, so take
r~0 = [−1, 2] and thus ~r = r~0 + tm
~ is [x, y] = [−1, 2] + t[3, −6] (and then P2 corresponds to
t = 1 ).
Two lines L1 and L2 are parallel if they have the same slope and are coincident if they are
the same line. For lines in vector equation form, they will be parallel if the direction vectors
are parallel.
So the lines in our two examples above are not parallel, whereas the lines
3
L1 ~r = (2 + 3t)ı̂ + (1 − 2t)̂ and L2 ~r = (6 − 6t)ı̂ + (4 + 4t)̂ are parallel.
(The notation has changed – can you see what you need to see to see that the lines are parallel ?)
Let’s go back to our example y = 2x + 6. We know that m
~ = [1, 2] is a direction vector
for the line, so any vector ~n that is perpendicular to m
~ (ie m
~ · ~n = 0 ) will be normal or
perpendicular to the line, like ~n = [2, −1]. But notice the scalar equation 2x − y + 6 = 0. In
other words, given the scalar equation Ax + By + C = 0 for a line, the vector ~n = [A, B] is
normal to it.
What do we do in three-space ?
First of all, a scalar equation of three variables Ax + By + Cz + D = 0 describes a plane
in three-space, not a line, so we do not have a scalar equation, nor do we have a slopeintercept form. But we do have a vector equation and parametric equations. The line passing through P0 = (x0 , y0 , z0 ) with direction vector m
~ = [m1 , m2 , m3 ] is ~r = r~0 + tm,
~ t∈R
or [x, y, z] = [x0 , y0 , z0 ]+t[m1 , m2 , m3 ] or x = x0 +tm1 , y = y0 +tm2 and z = z0 +tm3 for t ∈ R.
Example:
The line passing through P1 = (1, 0, 5) and P2 = (4, 2, 1) has m
~ = [3, 2, −4], so
~r = [4, 2, 1] + t[3, 2, −4] (and then P1 corresponds to t = −1 ).
Does (10, 5, −7) lie on the line ?
4 + 3t = 10 =⇒ t = 2
2 + 2t = 5 =⇒ t = 3/2
1 − 4t = −7 =⇒ t = 2 so no, it does not.
All of the vectors normal to a line in two-space are parallel, but this is not the case in threespace.
y
n_2
n_1
x
4
z
n_2
n_1
y
x
Equations of Planes
Example:
Consider the scalar equation x + y − z − 2 = 0. if we set two of the coordinates equal to 0, we
can see where the graph of the object would cross the axes. If y = z = 0, x + (0) − (0) − 2 =
0 =⇒ x = 2, so the x-intercept is 2. If x = z = 0, (0) + y − (0) − 2 = 0 =⇒ y = 2, so the
y-intercept is 2. If x = y = 0, (0) + (0) − z − 2 = 0 =⇒ z = −2, so the z-intercept is −2.
z
(2,0,0)
x
y
(0,2,0)
(0,0,-2)
This object is certainly not a line. The graph of this equation is a plane, a flat, twodimensional surface in three-space that extends off to infinity in every direction.
Other points on the plane are P = (2, 2, 2), Q = (0, 7, 5), R = (6, 3, 7) and S = (4, 0, 2).
There are infinitely many points on the plane – can you see that these points satisfy the
equation ?
However, the point (10, 5, 7) is not on the plane as (10) + (5) − (7) − 2 6= 0 (ie the point does
not satisfy the equation).
Using the points on the plane, we can find direction vectors that are parallel to the plane
~ = [−2, −3, −5].
(and not to each other). For example, P~Q = [−2, 5, 3] and RS
Suppose we have two non-parallel direction vectors ~a and ~b on a plane.
5
P
ta
a
sb
P_0
b
Then, the vector from any point P0 = (x0 , y0 , z0 ) to any other point P = (x, y, z) on the
plane can be represented by a linear combination of the direction vectors, ie P~0 P = t~a + s~b,
where t, s ∈ R. But since P~0 P = [x, y, z] − [x0 , y0 , z0 ], we can rearrange the vector equation
to write [x, y, z] = [x0 , y0 , z0 ]+t~a +s~b = [x0 , y0 , z0 ]+t[a1 , a2 , a3 ]+s[b1 , b2 , b3 ] or ~r = r~0 +t~a +s~b
and we have the vector equation of a plane.
So for our example above, we can take P0 = (2, 2, 2), ~a = [−2, 5, 3] and ~b = [−2, −3, −5], so
[x, y, z] = [2, 2, 2] + t[−2, 5, 3] + s[−2, −3, −5]. Notice that if t = −1/4 and s = 1/4, we have
[x, y, z] = [2, 0, 0], the position vector of the x-intercept.
The parametric equations of a plane would be x = x0 + ta1 + sb1 , y = y0 + ta2 + sb2 and
z = z0 + ta3 + sb3 where t, s ∈ R.
So for our example, x = 2 − 2t − 2s, y = 2 + 5t − 3s and z = 2 + 3t − 5s. Picking values for
t and s will give us other points on the plane, like if t = 1 and s = −2, we’ll have (4, 13, 15).
Can you see that this does satisfy the original scalar equation ?
Example:
Let’s find the equation of the plane that contains the line [x, y, z] = [2, 1, 4] + t[2, 3, 4] and is
parallel to the line [x, y, z] = [7, 4, 2] + s[−1, 0, 6].
Since the first line lies on the plane, we can use (2, 1, 4) as the point on the plane and [2, 3, 4]
as a direction vector. Since the plane is parallel to the second line and [−1, 0, 6] is not parallel
to [2, 3, 4] (not a scalar multiple of it), we can use [−1, 0, 6] as the second direction vector. A
vector equation for the plane would be [x, y, z] = [2, 1, 4] + p[2, 3, 4] + q[−1, 0, 6] for p, q ∈ R.
Or, in parametric form, x = 2 + 2p − q, y = 1 + 3p and z = 4 + 4p + 6q.
Properties of Planes
Suppose we know a point P0 = (x0 , y0 , z0 ) on a plane and that the vector ~n = [A, B, C] =
Aı̂ + B̂ + C k̂ is normal (perpendicular or orthogonal) to the plane.
6
n = [A,B,C]
P = (x,y,z)
P_0 = (x_0, y_0, z_0)
If we take any point P = (x, y, z) on the plane, then the vector P~0 P = [x − x0 , y − y0 , z − z0 ]
is parallel to the plane and perpendicular to ~n. Thus P~0 P · ~n = 0.
So [x − x0 , y − y0 , z − z0 ] · [A, B, C] = 0
or A(x − x0 ) + B(y − y0 ) + C(z − z0 ) = 0
or Ax + By + Cz + (−Ax0 − By0 − Cz0 ) = 0
or Ax + By + Cz + D = 0 where D = −Ax0 − By0 − Cz0 is a constant. This is the
scalar equation of the plane.
Example:
The plane containing the point P0 = (4, −2, 3) and having normal vector ~n = [1, −2, 1] has
scalar equation (1)x+(−2)y +(1)z +(−(1)(4)−(−2)(−2)−(1)(3)) = 0 or x−2y +z −11 = 0,
which is also written as x − 2y + z = 11.
Example:
Find the scalar equation of the plane containing the points P = (2, 1, 4), Q = (4, 0, 3) and
R = (3, 4, −2).
The vectors P~Q = [2, −1, −1] and P~R = [1, 3, −6] are vectors on the plane. Then P~Q × P~R
will be normal to the plane, so
~n = [2, −1, −1] × [1, 3, −6]
= [(−1)(−6) − (−1)(3), (−1)(1) − (2)(−6), (2)(3) − (−1)(1)]
= [9, 11, 7].
The scalar equation has the form 9x + 11y + 7z + D = 0. We can plug any of the three points
in to find D. Let’s use Q = (4, 0, −3), then 9(4) + 11(0) + 7(3) + D = 0 =⇒ D = −57 and
the scalar eqaution of the plane is 9x + 11y + 7z = 57.
Intersections of Lines in Two– and Three–Space
For two lines L1 and L2 in two-space, there are three possibilities.
7
y
L1
L2
x
(i) The lines intersect at a single point.
y
L1
L2
x
(ii) The lines are coincident (and hence intersect at infinitely many points).
y
L1
L2
x
(iii) The lines are parallel and distinct (and hence do not intersect).
There is a direct connection to linear systems of two equations in two unknowns.
(a) Consider
2x + 3y = 4
3x + 4y = 9
we can rewrite the system as an augmented matrix
8
2 3 4
3 4 9
where the first row represents the first equation and the second row represents the second
equation. The first column represents x and the second column represents y. The vertical
bar represents the equalities and the last column represents the constants on the right-hand
sides of the equations. This is simply a shorthand notation for the system of equations
where we keep track of the coefficients. We perform elementary row operations like multiplying or dividing a row by a nonzero scalar constant, interchanging rows or adding or
subtracting multiples of one row from another. The goal is to change the matrix into its
reduced row echelon form (RREF), where the first nonzero entry in each row is a 1, called a
leading 1 or pivot. Each leading 1 is strictly to the right of any leading 1 in a row above it and
all other elements in a column that contains a leading 1 are zero (so both above and below the
leading 1). Any zero rows must appear at the bottom of the matrix. When we have reduced
row echelon form, we will be able to read the solution of the linear system right out of the
matrix. The procedure to reduce a matrix to this form is called Gauss-Jordan elimination.
2 3 4
R1 /2 (create leading 1 in first row)
3 4 9
1 3/2 2
R2 − 3R1 (clear column under leading 1)
3
4 9
1
3/2 2
R2 × −2 (make leading 1 in second row)
0 −1/2 3
1 3/2
2
R1 − (3/2)R2 (clear column above leading 1)
0
1 −6
1 0 11
this is RREF, the solution is x = 11 and y = −6.
0 1 −6
These lines intersect at a single point, which is a unique solution.
(b) Consider
2x + 3y = 4
6x + 9y = 12
as a
2
6
1
6
1
0
matrix
3 4
9 12
3/2
9
3/2
0
R1 /2
2
R2 − 6R1
12
2
(RREF)
0
we have a row of zeros and the solution is x+(3/2)y = 2, which means that there are infinitely
9
many solutions (these lines are coincident). Let y = t be a parameter, then x + (3/2)t = 2
or x = 2 − (3/2)t. So the parametric equations of the line of solutions are x = 2 − (3/2)t
and y = t and the vector equation would be [x, y] = [2, 0] + t[−3/2, 1].
(c) Consider
2x + 3y = 4
6x + 9y = 16
as a
2
6
1
6
1
0
matrix
3 4
9 16
3/2
9
3/2
0
R1 /2
2
R2 − 6R1
16
2
4
this system is inconsistent. The last row says 0x + 0y = 4, which is impossible, so there is
no solution here (these lines are parallel).
If we have two lines L1 and L2 in three-space, there are four possibilities.
z
L2
y
L1
x
(i) The lines intersect at a single point (unique solution).
z
L1
L2
y
x
10
(ii) The lines are coincident (there are infinitely many solutions).
z
L1
L2
y
x
(iii) The lines are parallel and distinct (no solution).
z
L1
y
x
L2
(iv) The lines are skew – they are distinct, not parallel and do not intersect (no solution).
Example:
The lines L1 [x, y, z] = [2, 3, 4] + t[1, 1, 3] and L2 [x, y, z] = [1, 0, −1] + s[1, 1, 3] are parallel
(same direction vector). Are they coincident or distinct ?
If they are coincident, every point on L1 would be on L2 , so all we have to do is check if
(2, 3, 4) is on L2 . The parametric equations for L2 are x = 1 + s, y = s and z = −1 + 3s. If
we substitute the point (2, 3, 4) in, 2 = 1 + s =⇒ s = 1
3 = s =⇒ s = 3
and 4 = −1 + 3s =⇒ s = 5/3 and so the point is not on L2 and the lines are distinct.
Example:
Find the point of intersection of L1 ~r = [2, 1, 5] + t[3, 1, −2] and L2 ~r = [0, 3, 1] + s[5, −1, 2]
These lines are not parallel because their direction vectors are not parallel.
We write the lines in parametric form x = 2 + 3t, y = 1 + t and z = 5 − 2t; and x = 5s,
y = 3 − s and z = 1 + 2s. We look for a point of intersection by equating the coordinates
2 + 3t = 5s, 1 + t = 3 − s and 5 − 2t = 1 + 2s. This system has a unique solution s = t = 1.
Plugging these values for the parameters into the equations for the lines yields the point of
intersection (5, 2, 3) (which is a unique solution).
11
Example:
Are the lines L1 [x, y, z] = [2, 1, 5] + t[3, 1, −2] and L2 [x, y, z] = [1, 3, 2] + s[5, −1, 2] skew ?
The lines are not parallel, so we check for a point of intersection.
2 + 3t = 1 + 5s =⇒ 3t − 5s = −1
1 + t = 3 − s =⇒ t + s = 2
and 5 − 2t = 2 + 2s =⇒ 2t + 2s = 3, which contradicts the previous equation, so this system
is inconsistent and there is no point of intersection. ∴ the lines are skew.
The shortest distance between two skew lines L1 and L2 would be the length of the common
~
perpendicular |AB|.
B
A
P2
P1
L2
L1
Suppose we have points P1 and P2 on the lines, then if ~n is any vector in the direction of
~ we would have that proj~n P1~P2 = AB.
~ But |AB|
~ is the length we want, so
AB,
~
~ = |proj~n P1~P2 | = |P1 P2 · ~n| . We can find a normal vector ~n that is perpendicular to
|AB|
|~n|
both lines by taking the cross product of the direction vectors of the lines, ie ~n = m
~1 × m
~ 2.
Example:
For our skew lines above, m
~ 1 = [3, 1, −2] and m
~ 2 = [5, −1, 2].
So ~n = m
~1 × m
~2
= [3, 1, −2] × [5, −1, 2]
= [(1)(2) − (−2)(−1), (−2)(5) − (3)(2), (3)(−1) − (1)(5)]
= [0, −16, −8].
We can divide by −8 and use ~n = [0, 2, 1] for convenience.
Use the points on the lines P1 = (2, 1, 5) and P2 = (1, 3, 2) to get P1~P2 = [−1, 2, −3].
~
~ = |P1 P2 · ~n|
So |AB|
|~n|
|[−1, 2, −3] · [0, 2, 1]|
= p
(0)2 + (2)2 + (1)2
|(−1)(0) + (2)(2) + (−3)(1)|
√
=
5
√
= 1/ 5 ≈ 0.45 (so the lines are close).
12
Intersections of Lines and Planes
If we have a line and a plane in three-space, there are three possibilities.
(i) The line intersects the plane at a single point (unique solution).
(ii) The line lies on the plane (so they intersect at infinitely many points and there are infinitely many solutions).
(iii) The line is parallel to the plane (but not on it), so the line does not intersect the plane
(no solution).
Example:
Dose the line x = 2 + t, y = 6 − t and z = 3 + 2t intersect the plane 2x + 3y + 5z = 55 ? If
yes, at what point ?
We plug the parametric equations of the line into the scalar equation of the plane to see if
there is a solution.
2(2 + t) + 3(6 − t) + 5(3 + 2t) = 55
4 + 2t + 18 − 3t + 15 + 10t = 55
9t = 18 =⇒ t = 2
Since there is a single solution for t, the line and plane intersect at the single point (4, 4, 7).
Example:
Is the line ~r = [2, 3, 1] + t[1, 5, −2] parallel to the plane 3x + y + 4z = 13 ? If yes, is the line
13
in the plane or not ?
The direction vector of the line is m
~ = [1, 5, −2] = ı̂ + 5̂ − 2k̂ and a normal vector of the
plane is ~n = [3, 1, 4] = 3ı̂ + ̂ + 4k̂.
Let’s check m
~ · ~n = [1, 5, −2] · [3, 1, 4] = (1)(3) + (5)(1) + (−2)(4) = 0.
Since m
~ · ~n = 0, m
~ ⊥ ~n and hence the line is parallel to the plane.
Let’s check for points of intersection.
3(2 + t) + (3 + 5t) + 4(1 − 2t) = 13
6 + 3t + 3 + 5t + 4 − 8t = 13
13 + 0t = 13
0t = 0
Since all values of t satisfy this equation, there are infinitely many points of intersection and
hence the line does lie on the plane.
The line would be parallel to the plane 3x + y + 4z = 10 as well, but it would not lie on it.
3(2 + t) + (3 + 5t) + 4(1 − 2t) = 10
13 + 0t = 10
0t = −3
Which has no solution.
If a line is parallel to a plane, but not on it, how far apart are they ? Suppose we have a
point P on the line and a point Q on the plane.
P
n
d
Q
If ~n is a normal vector to the plane, then the projection of P~Q onto ~n would be the perpen|P~Q · ~n|
dicular distance d from P to the plane. So we have d =
(which is similar to what
|~n|
we did for the skew lines).
Example:
How far is the line ~r = [2, 3, 1] + t[1, 5, −2] from the plane 3x + y + 4z = 10 ?
Take P = (2, 3, 1) and Q = (2, 0, 1), then P~Q = [0, −3, 0] and we know ~n = [3, 1, 4].
|[0, −3, 0] · [3, 1, 4]|
So d = p
(3)2 + (1)2 + (4)2
|(0)(3) + (−3)(1) + (0)(4)|
√
=
26
√
= 3/ 26 ≈ 0.59 (so pretty close).
14
Intersections of Planes
If we have two planes in three-space, there are three possibilities.
(i) They intersect in a line (infinitely many solutions with one parameter).
(ii) They are coincident (infinitely many solutions with two parameters).
(iii) They are parallel and distinct (no solution).
A system of two equations in three unknowns is either inconsistent (has no solution) or has
infinitely many solutions – a unique solution is not possible (two planes cannot intersect at
a single point).
Example:
Consider the planes 2x + 2y + z = 7 and x + y + z = 6.
The normals for the planes are not parallel, so the planes are not parallel and hence they
must intersect.
Let’s solve for the line of intersection. We’ll rewrite the system as a matrix.
1 1 1 6
R2 − 2R1
2 2 1 7
15
1 1
1
6
R2 × −1
0 0 −1 −5
1 1 1 6
R1 − R2
0 0 1 5
1 1 0 1
(RREF)
0 0 1 5
The solution is x + y = 1, z = 5.
Let y = t, then x = 1 − t, y = t and z = 5 are the parametric equations of the line.
The vector equation is [x, y, z] = [1, 0, 5] + t[−1, 1, 0].
Example:
Consider the planes x − 2y + 3z = 10 and 2x − 4y + 6z = 12.
The normals are parallel, so the planes are parallel. But are they coincident or distinct ?
1 −2 3 10
R2 − 2R1
2 −4 6 12
1 −2 3 10
0
0 0 −8
This is an inconsistent system, so there is no solution and the planes are distinct.
The planes x − 2y + 3z = 10 and 2x − 4y + 6z = 20 are coincident – can you see that ?
For three planes, there are six possibilities. There are three where the linear system is
consistent (has solutions).
(figure of 3 planes intersecting at point)
(i) The planes intersect at a single point (unique solution).
(figure of 3 planes intersecting in line)
(ii) The planes intersect in a line (infinitely many solutions with one parameter).
(figure of 3 coincident planes)
(iii) The planes are coincident (infinitely many solutions with two parameters).
And there are three where the system is inconsistent (no solutions).
(figure of 3 parallel planes)
(iv) The planes are parallel (and at least two are distinct).
16
(figure of 2 parallel planes, one not)
(v) Two of the planes are parallel (and distinct), but the third is not parallel.
(figure of planes intersecting in pairs)
(vi) The planes intersect in pairs.
The normals in the situations would have to be:
(i) not parallel, nor coplanar
(ii) coplanar, but not parallel
(iii) parallel
(iv) parallel
(v) two are parallel
(vi) coplanar, but not parallel.
Example:
Consider the planes x + 2y + 3z = 10, 2x + 3y − z = 4 and 3x + 4y + 5z = 12.


1 2
3 10
 2 3 −1 4  R2 − 2R1 , R3 − 3R1
3 4
5 12


1
2
3
10
 0 −1 −7 −16  R2 × −1
0 −2 −4 −18


1
2
3
10
 0
16  R3 + 2R2
1
7
0 −2 −4 −18


1 2 3 10
 0 1 7 16  R3 /10
0 0 10 14


1 2 3 10
 0 1 7 16  R1 − 3R3 , R2 − 7R3
0 0 1 7/5


1 2 0 29/5
 0 1 0 31/5  R1 − 2R2
0 0 1 7/5


1 0 0 −33/5
 0 1 0
31/5  (RREF)
0 0 1
7/5
These planes intersect at a single point (−33/5, 31/5, 7/5).
Example:
Consider the planes x + y − 2z = 4, 2x + 2y − 4z = 6 and 3x + 5y + 2z = 10.
17

1
 2
3

1
 0
0

1 −2 4
2 −4 6  R2 − 2R1 , R3 − 3R1
5
2 10

1 −2
4
0
0 −2 
2
8 −2
This is an inconsistent system, so there is no solution.
Can you see that two of the planes are parallel but distinct ?
Example:
Consider the planes x − 5y + 2z = 10, x + 7y − 2z = −6 and 8x + 5y + z = 20.


1 −5
2 10
 1
7 −2 −6  R2 − R1 , R3 − 8R1
8
5
1 20


1 −5
2
10
 0 12 −4 −16  R2 /12
0 45 −15 −60


1 −5
2
10
 0
1 −1/3 −4/3  R3 − 45R2
0 45 −15 −60


1 −5
2
10
 0
1 −1/3 −4/3  R1 + 5R2
0
0
0
0


1 0
1/3 10/3
 0 1 −1/3 −4/3  (RREF)
0 0
0
0
So we have x + (1/3)z = 10/3 and y − (1/3)z = −4/3.
Let z = t be the parameter, then we have x = 10/3 − (1/3)t, y = −4/3 + (1/3)t and z = t
or ~r = [10/3, −4/3, 0] + t[−1/3, 1/3, 1] and the planes intersect in a line.