0 + b y 0 + c z 0 − ( ax + b y + c z)| =
0 + b y 0 + c z 0 − (− d)| =
0 + b y 0 + c z 0 + d |
a 2 + b 2 + c 2
a 2 + b 2 + c 2
a 2 + b 2 + c 2
If n points away from the side of P where the point Q is located, then 90◦ < θ < 180◦ and
so cos θ < 0. The distance D is then |cos θ| r , and thus repeating the same argument as
above still gives the same result.
QED
Example 1.25. Find the distance D from (2, 4, −5) to the plane from Example 1.24.
Solution: Recall that the plane is given by 5 x − 3 y + z − 10 = 0. So
|5(2) − 3(4) + 1(−5) − 10|
|−17|
17
D =
=
=
≈ 2.87
52 + (−3)2 + 12
35
35
38
CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
Line of intersection of two planes
Note that two planes are parallel if they have normal vectors that
are parallel, and the planes are perpendicular if their normal vectors
are perpendicular. If two planes do intersect, they do so in a line (see
L
Figure 1.5.9). Suppose that two planes P and P with normal vectors
1
2
n and n , respectively, intersect in a line L. Since n
, then
1
2
1 × n2 ⊥ n1
n
is parallel to the plane P . Likewise, n
means that
1 × n2
1
1 × n2 ⊥ n2
Figure 1.5.9
n
is also parallel to P . Thus, n
is parallel to the intersection
1 × n2
2
1 × n2
of P and P , i.e. n
is parallel to L. Thus, we can write L in the following vector form:
1
2
1 × n2
L : r + t(n
) , for
1 × n2
− ∞ < t < ∞
(1.28)
where r is any vector pointing to a point belonging to both planes. To find a point in both
planes, find a common solution ( x, y, z) to the two normal form equations of the planes. This
can often be made easier by setting one of the coordinate variables to zero, which leaves you
to solve two equations in just two unknowns.
Example 1.26. Find the line of intersection L of the planes 5 x − 3 y + z − 10 = 0 and 2 x + 4 y −
z + 3 = 0.
Solution: The plane 5 x − 3 y + z − 10 = 0 has normal vector n1 = (5,−3,1) and the plane 2 x +
4 y − z + 3 = 0 has normal vector n
and n are not scalar multiples, then
2 = (2, 4, −1). Since n1
2
the two planes are not parallel and hence will intersect. A point ( x, y, z) on both planes will
satisfy the following system of two equations in three unknowns:
5 x − 3 y + z − 10 = 0
2 x + 4 y − z + 3 = 0
Set x = 0 (why is that a good choice?). Then the above equations are reduced to:
−3 y + z − 10 = 0
4 y − z + 3 = 0
The second equation gives z = 4 y + 3, substituting that into the first equation gives y = 7.
Then z = 31, and so the point (0,7,31) is on L. Since n1 × n2 = (−1,7,26), then L is given by:
r + t(n
)
1 × n2 = (0, 7, 31) + t(−1, 7, 26),
for − ∞ < t < ∞
or in parametric form:
x = − t,
y = 7 + 7 t,
z = 31 + 26 t, for − ∞ < t < ∞
1.5 Lines and Planes
39
Exercises
A
For Exercises 1-4, write the line L through the point P and parallel to the vector v in the
following forms: (a) vector, (b) parametric, and (c) symmetric.
1. P = (2,3,−2), v = (5,4,−3)
2. P = (3,−1,2), v = (2,8,1)
3. P = (2,1,3), v = (1,0,1)
4. P = (0,0,0), v = (7,2,−10)
For Exercises 5-6, write the line L through the points P and P in parametric form.
1
2
5. P 1 = (1,−2,−3), P 2 = (3,5,5)
6. P 1 = (4,1,5), P 2 = (−2,1,3)
For Exercises 7-8, find the distance d from the point P to the line L.
7. P = (1,−1,−1), L : x = −2 − 2 t, y = 4 t, z = 7 + t
8. P = (0,0,0), L : x = 3 + 2 t, y = 4 + 3 t, z = 5 + 4 t
For Exercises 9-10, find the point of intersection (if any) of the given lines.
9. x = 7 + 3 s, y = −4 − 3 s, z = −7 − 5 s and x = 1 + 6 t, y = 2 + t, z = 3 − 2 t x − 6
x − 11
y − 14
z + 9
10.
= y + 3 = z and
=
=
4
3
−6
2
For Exercises 11-12, write the normal form of the plane P containing the point Q and per-
pendicular to the vector n.
11. Q = (5,1,−2), n = (4,−4,3)
12. Q = (6,−2,0), n = (2,6,4)
For Exercises 13-14, write the normal form of the plane containing the given points.
13. (1, 0, 3), (1, 2, −1), (6,1,6)
14. (−3,1,−3), (4,−4,3), (0,0,1)
15. Write the normal form of the plane containing the lines from Exercise 9.
16. Write the normal form of the plane containing the lines from Exercise 10.
For Exercises 17-18, find the distance D from the point Q to the plane P.
17. Q = (4,1,2), P : 3 x − y − 5 z + 8 = 0
18. Q = (0,2,0), P : −5 x + 2 y − 7 z + 1 = 0
For Exercises 19-20, find the line of intersection (if any) of the given planes.
19. x + 3 y + 2 z − 6 = 0, 2 x − y + z + 2 = 0
20. 3 x + y − 5 z = 0, x + 2 y + z + 4 = 0
B
x − 6
21. Find the point(s) of intersection (if any) of the line
= y + 3 = z with the plane
4
x + 3 y + 2 z − 6 = 0. ( Hint: Put the equations of the line into the equation of the plane. )
40
CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
1.6 Surfaces
In the previous section we discussed planes in Euclidean space. A plane is an example of
a surface, which we will define informally8 as the solution set of the equation F( x, y, z) = 0
in R3, for some real-valued function F. For example, a plane given by ax + b y + cz + d = 0
is the solution set of F( x, y, z) = 0 for the function F( x, y, z) = ax + b y + cz + d. Surfaces are 2-dimensional. The plane is the simplest surface, since it is “flat”. In this section we will
look at some surfaces that are more complex, the most important of which are the sphere
and the cylinder.
Definition 1.9. A sphere S is the set of all points ( x, y, z) in R3 which are a fixed distance r
(called the radius) from a fixed point P
, y , z ) (called the center of the sphere):
0 = ( x 0
0
0
S = {( x, y, z) : ( x − x )2
)2
)2
0
+ ( y − y 0 + ( z − z 0 = r 2 }
(1.29)
Using vector notation, this can be written in the equivalent form:
S = {x : x − x0 = r }
(1.30)
where x = ( x, y, z) and x
, y , z ) are vectors.
0 = ( x 0
0
0
Figure 1.6.1 illustrates the vectorial approach to spheres.
z
z
x − x
x = r
0
= r
x
x
x − x0
y
( x , y , z )
0
0
0
0
x0
y
0
x
x
(a) radius r, center (0, 0, 0)
(b) radius r, center ( x 0, y 0, z 0)
Figure 1.6.1
Spheres in R3
Note in Figure 1.6.1(a) that the intersection of the sphere with the x y-plane is a circle
of radius r (i.e. a great circle, given by x 2 + y 2 = r 2 as a subset of R2). Similarly for the
intersections with the xz-plane and the yz-plane. In general, a plane intersects a sphere
either at a single point or in a circle.
8See O’NEILL for a deeper and more rigorous discussion of surfaces.
1.6 Surfaces
41
Example 1.27. Find the intersection of the sphere x 2 + y 2 + z 2 = 169 with the plane z = 12.
z
Solution: The sphere is centered at the origin and has radius
13 = 169, so it does intersect the plane z = 12. Putting
z = 12
z = 12 into the equation of the sphere gives
y
x 2 + y 2 + 122 = 169
0
x 2 + y 2 = 169 − 144 = 25 = 52
which is a circle of radius 5 centered at (0, 0, 12), parallel to
x
the x y-plane (see Figure 1.6.2).
Figure 1.6.2
If the equation in formula (1.29) is multiplied out, we get an equation of the form:
x 2 + y 2 + z 2 + ax + b y + cz + d = 0
(1.31)
for some constants a, b, c and d. Conversely, an equation of this form may describe a sphere,
which can be determined by completing the square for the x, y and z variables.
Example 1.28. Is 2 x 2 + 2 y 2 + 2 z 2 − 8 x + 4 y − 16 z + 10 = 0 the equation of a sphere?
Solution: Dividing both sides of the equation by 2 gives
x 2 + y 2 + z 2 − 4 x + 2 y − 8 z + 5 = 0
( x 2 − 4 x + 4) + ( y 2 + 2 y + 1) + ( z 2 − 8 z + 16) + 5 − 4 − 1 − 16 = 0
( x − 2)2 + ( y + 1)2 + ( z − 4)2 = 16
which is a sphere of radius 4 centered at (2, −1,4).
Example 1.29. Find the points(s) of intersection (if any) of the sphere from Example 1.28
and the line x = 3 + t, y = 1 + 2 t, z = 3 − t.
Solution: Put the equations of the line into the equation of the sphere, which was ( x − 2)2 +
( y + 1)2 + ( z − 4)2 = 16, and solve for t:
(3 + t − 2)2 + (1 + 2 t + 1)2 + (3 − t − 4)2 = 16
( t + 1)2 + (2 t + 2)2 + (− t − 1)2 = 16
6 t 2 + 12 t − 10 = 0
4
The quadratic formula gives the solutions t = −1 ±
. Putting those two values into the
6
equations of the line gives the following two points of intersection:
4
8
4
4
8
4
2 +
, −1 +
, 4 −
and
2 −
, −1 −
, 4 +
6
6
6
6
6
6
42
CHAPTER 1. VECTORS IN EUCLIDEAN SPACE
If two spheres intersect, they do so either at a single point or in a circle.
Example 1.30. Find the intersection (if any) of the spheres x 2 + y 2 + z 2 = 25 and x 2 + y 2 +( z−
2)2 = 16.
Solution: For any point ( x, y, z) on both spheres, we see that
x 2 + y 2 + z 2 = 25
⇒
x 2 + y 2 = 25 − z 2, and
x 2 + y 2 + ( z − 2)2 = 16
⇒
x 2 + y 2 = 16 − ( z − 2)2, so
16 − ( z − 2)2 = 25 − z 2
⇒
4 z − 4 = 9
⇒
z = 13/4
⇒
x 2 + y 2 = 25 − (13/4)2 = 231/16
∴ The intersection is the circle x 2 + y 2 = 231 of radius 231
).
16
4
≈ 3.8 centered at (0,0, 13
4
The cylinders that we will consider are right circular cylinders. These are cylinders ob-
tained by moving a line L along a circle C in R3 in a way so that L is always perpendicular
to the plane containing C. We will only consider the cases where the plane containing C is
parallel to one of the three coordinate planes (see Figure 1.6.3).
z
z
r
z
r
y
y
r
y
0
0
0
x
x
x
(a) x 2 + y 2 = r 2, any z
(b) x 2 + z 2 = r 2, any y
(c) y 2 + z 2 = r 2, any x
Figure 1.6.3
Cylinders in R3
For example, the equation of a cylinder whose base circle C lies in the x y-plane and is
centered at ( a, b, 0) and has radius r is
( x − a)2 + ( y − b)2 = r 2,
(1.32)
where the value of the z coordinate is unrestricted. Similar equations can be written when
the base circle lies in one of the other coordinate planes. A plane intersects a right circular
cylinder in a circle, ellipse, or one or two lines, depending on whether that plane is parallel,
oblique9, or perpendicular, respectively, to the plane containing C. The intersection of a
surface with a plane is called the trace of the surface.
9i.e. at an angle strictly between 0◦ and 90◦.
1.6 Surfaces
43
The equations of spheres and cylinders are examples of second-degree equations in R3, i.e.
equations of the form
Ax 2 + B y 2 + Cz 2 + Dxy + Exz + F yz + Gx + H y + I z + J = 0
(1.33)
for some constants A, B, . . . , J. If the above equation is not that of a sphere, cylinder, plane,
line or point, then the resulting surface is called a quadric surface.
z
One type of quadric surface is the ellipsoid, given
c
by an equation of the form:
x 2
y 2
z 2
y
+
+
= 1
(1.34)
a 2
b 2
c 2
0
b
a
In the case where a = b = c, this is just a sphere.
In general, an ellipsoid is egg-shaped (think of an
x
ellipse rotated around its major axis). Its traces in