2.1 Three Dimensional Curves and Surfaces

2.1.1 Parametric Equation of a Line

Any line in two- or three-dimensional space can be uniquely specified by apoint on the line and a vector parallel to the line. The line then is the lineparallel to the vector v = (a, b, c) passing through the point P0(x0, y0, z0).In particular, if we view the vector as having its initial point at P0, then wecan move away from P0 along the line by adding multiples of v. In other

v

Ha,b,cL

L

P0Hx0,y0,z0L

0.0

0.5

1.0

x

0.0

0.5

1.0y

0.0

0.5

1.0

1.5

2.0

z

Figure 1: Parametric equation of a line.

words, we can move tv along the line. In other words, the position of pointQ(x, y, z) on the line is given by

(x, y, z) = (x0, y0, z0) + t(a, b, c) = (x0 + at, y0 + bt, z0 + ct) . (1)

This gives us the parametric equations of a line in 3-space for a linepassing through P (x0, y0, z0) and parallel to v = ai + bj + ck :

x = x0 + at , y = y0 + bt , z = z0 + ct . (2)

1

Here t is the parameter that determines how far along the line you havemoved.

Example: Find the parametric equation of the line passing through P (1, 3, 2),parallel to 2i− j + 3k.

Solution: The line is given by parametric equations

x = 1 + 2t , y = 3− t , z = 2 + 3t .

Note that these lines are infinite. If we want only a line segment we mustrestrict the parameter. Thus, the line given in the example can be restrictedto the line segment joining P (1, 3, 2) to Q(3, 2, 5) if we restrict t to the interval0 ≤ t ≤ 1.Finally, two lines that are not parallel and do not cross are called skew lines.

2.1.1.1 Representing the parametric equation using vectors

If instead of an initial point P (x0, y0, z0), we define a position vector r0 =(x0, y0, z0) to give an initial position on a line and r = (x, y, z) for any pointon the line parallel to v = (a, b, c) through P , then equation (1) can berepresented by the vector equation of a line as

r = r0 + tv . (3)

This is shown in Figure 2. Note that all these vectors are defined with initialpoints at the origin. This is entirely equivalent to the equations in (2), butthe notation is obviously briefer, and the interpretation is purely in termsof vectors. If you need to think about this more intuitively, think of r asordinary position vectors in space, r as a “velocity” vector and t as “time”.This then looks like one of the usual equations for linear motion in vectorform. However, we can use this more generally for any parameter t andappropriate vector v. Let’s return to the previous example:

Example: Find the vector equation of the line passing through P (1, 3, 2),parallel to 2i− j + 3k..

Solution: The initial position is given by the vector

r0 = (1, 3, 2) ,

2

v

r

r0

v t

L

P0Hx0,y0L

0.0 0.2 0.4 0.6 0.8 1.0 1.2 x0.0

0.5

1.0

1.5

2.0

y

Figure 2: Vector equation of a line r = r0 + vt.

and the vector equation is

r = (1, 3, 2)+t(2,−1, 3) = (1+2t, 3−t, 2+3t) = (1+2t)i+(3−t)j+(2+3t)k .

It is important to be comfortable with all these ways of writing parametricequations of a line.

2.1.2 Planes

The first type of planes we might think of are the coordinate planes. Thexy-plane, for example, is the set of a x and y with z = 0. By extension, wecan imagine planes parallel to the coordinate planes. For example, the planex = a is the set of all y and z such that x = a. These types of planes areshown in Figure 3.

3

y

z

x

x=a

z

x

y

y=b

x

y

z

z=c

Figure 3: Planes parallel to the coordinate axes.

2.1.2.1 Planes specified by a point and normal vector

Any plane in 3-space can be uniquely determined by giving a point on theplane and a vector perpendicular to the plane, called a normal vector. Sup-pose we want to find an equation of the plane passing through P (x0, y0, z0)and perpendicular to n = (a, b, c). Let us define the vector r0 = (x0, y0, z0)pointing to P , and r = (x, y, z) pointing to another point Q. Since n isperpendicular to the plane, n · r0 = n · r = 0, and therefore

n · (r− r0) = 0 , (4)

which in component form is written

(a, b, c) · (x− x0, y − y0, z − z0) = 0 , (5)

which results in the equation

a(x− x0) + b(y − y0) + c(z − z0) = 0 . (6)

This is known as the point-normal form of the equation of a plane. Equa-tion (4) is the vector form of this equation.

Example: Find the equation of the plane passing though (1, 2,−1) and withnormal vector n = (2, 3, 1).

Solution: We use equation (6) to give us

2(x− 1) + 3(y − 2) + 1(z + 1) = 0 ,

4

which can be simplified to

2x+ 3y + z − 7 = 0 .

In fact, this final form can be generalised to

ax+ by + cz + d = 0 , (7)

which is the equation with graph that is a plane with n = (a, b, c) as a normal.This is called the general form of the equation of a plane. d is determinedby knowing a point that lies in the plane and substituting into the equationto find d.Other ways to determine a plane are a point and two vectors parallel to theplane, or three points in the plane. Note that these allow us to find a normalvector and so the equation of the plane. The cross-product of two vectors inthe plane will give a vector normal to the plane. Three points P1, P2 and P3

allow us to define two vectors−−→P1P2 and

−−→P1P3, which again allow us to find

the normal vector.

2.1.2.2 Intersecting planes

Two distinct planes have an acute angle of intersection, 0 ≤ θ ≤ π/2. Theangle is the same as either the angle between n1 and n2 or between n1 and−n2 depending on the direction of the normal vectors. However, in eithercase the angle is given by

cos θ =|n1 · n2|||n1||||n2||

. (8)

The absolute value ensures that regardless of the sign of the normal vectorswe always have an acute angle.

Example: Find the angle between the planes

2x− y + 2z = 3 , 2x+ 4y − 4z = 0 .

Solution: The normal vectors are given by n1 = (2,−1, 2) and n2 =(2, 4,−4), and the angle is found from

cos θ =|4− 4− 8|√

9√

36=| − 8|(3)(6)

=4

9,

5

Θ

Θ

n1 n2

-1.0

-0.5

0.0

0.5

1.0

x

-1.0 -0.5 0.0 0.5 1.0

y

-5

0

5

z

Figure 4: Intersecting planes.

which gives the angle

θ = cos−1 4

9= 1.11024 rad = 63.6122o .

2.1.2.3 Distance problems involving planes

There are three distance problems we will be concerned with: the distancebetween a point and a plane; the distance between two parallel planes; findthe distance between two skew lines.

Theorem: The distance D between a point P (x0, y0, z0) and the plane ax+by + cz + d = 0 is

D =|ax0 + by0 + cz0 + d|√

a2 + b2 + c2. (9)

Proof: Let Q(x1, y1, z1) be a point in the plane, and n = (a, b, c) the normalvector with its initial point is at Q. Consider Figure 5. The distance D is

equal to the length of the orthogonal projection of−−→QP0 onto n. Therefore,

recalling

projbv =v · b||b||2

b , (10)

6

QHx1,y1,z1L

P0Hx0,y0,z0L

projnQP0D

Figure 5: Projection onto normal vector: distance from plane.

we get

D = ||projn−−→QP0|| =

∣∣∣∣∣∣∣∣∣∣−−→QP0 · n||n||2

n

∣∣∣∣∣∣∣∣∣∣ =|−−→QP0 · n|||n||

. (11)

However, we can write

−−→QP0 = (x1 − x0, y1 − y0, z1 − z0)

⇒−−→QP0 · na(x1 − x0) + b(y1 − y0) + c(z1 − z0) ,

(12)

and||n|| =

√a2 + b2 + c2 , (13)

which together give us

D = |a(x0 − x1) + b(y0 − y1) + c(z0 − z1)√a2 + b2 + c2 .

(14)

Moreover, since Q lies in the plane, it satisfies the equation

ax1 + by1 + cz1 + d = 0 , (15)

which allows us to find d,

d = −ax1 − by1 − cz1 , (16)

and therefore we find the result.

7

Example: Find the distance between the point P (4, 4, 2) and the plane

x− 2y + 4z + 2 = 0 .

Solution: Using (11) we get

D =|(1)(4) + (−2)(4) + (4)(2) + 2|√

42 + 42 + 22=|6|6

= 1 . (17)

To compute the distance between two parallel planes, compute the distancebetween one plane and any point in the other plane. To find the distancebetween skew lines, define two parallel planes each of which contains oneof the skew lines. Then the distance between the planes gives the distancebetween the skew lines.

Example: Find the distance between the skew lines

L1 : x = 1 + 4t y = 5− 4t , z = −1 + 5t ,

L1 : x = 2 + 8t y = 4− 3t , z = 5 + t .

Solution: Let P1 and P2 be parallel planes containing L1 and L2 respectively.We can find a point on each line and hence in each plane by setting t = 0,giving Q1(1, 5,−1) and Q2(2, 4, 5), see Figure 6. Let’s use Q1 and find the

D

L1

L2

Q1

Q2

Figure 6: Distance between skew lines.

equation of the plane P2. Since the planes are parallel, the vectors usedto define the parametric equations of the lines u1 = (4,−4, 5) and u2 =(8,−3, 1) are both parallel to P2. Hence

n = u1 × u2 =

∣∣∣∣∣∣i j k4 −4 58 −3 1

∣∣∣∣∣∣ = 11i + 36j + 20k ,

8

is normal to P1 and P2. With this normal vector and the point Q2, we findthe equation of P2:

11(c− 2) + 36(y − 4) + 20(z − 5) = 0 ,

which can be written in the general form

11x+ 36y + 20z − 266 = 0 .

Therefore the distance from P2 to the point Q1(1, 5,−1) is

D =|(11)(1) + (36)(5) + (20)(−1)− 266|√

112 + 362 + 202=

95√1817

,

which in turn is the distance from L1 to L2, since they lie in the parallelplanes.

2.1.3 Quadric Surfaces

The generalisation to the general quadratic equation ?? which from whichconic sections were derived in 2-space is the second-degree equation in x,y and z,

Ax2 +By2 + Cz2 +Dxy + Exz + Fyz +Gx+Hy + Ik + J = 0 . (18)

The graphs of this family of equations are called the quadric surfaces.There are six common types of quadrics,

1. Ellipsoid:x2

a2+y2

b2+z2

c2= 1

2. Hyperboloid of One Sheet:x2

a2+y2

b2− z2

c2= 1

3. Hyperboloid of Two Sheets:z2

c2− x2

a2− y2

b2= 1

4. Elliptic Cone: z2 =x2

a2+y2

b2

5. Elliptic Paraboloid: z =x2

a2+y2

b2

6. Hyperbolic Paraboloid: z =y2

b2− x2

a2

(19)

where we assume that a, b, c > 0. These are shown in Figure 7. They havethe following traces

9

1. Ellipsoid: Traces in the coordinate planes are ellipses.

2. Hyperboloid of One Sheet: The trace in the xy-plane are ellipses, andtraces in the xz- and yz-planes are hyperbolas.

3. Hyperboloid of Two Sheets: There is no trace in the xy-plane, althoughthe traces in planes parallel to the xy-plane are ellipses provided thereis a trace, and traces in the xz- and yz-planes are hyperbolas.

4. Elliptic Cone: The trace in the xy-plane is a point and in planes parallelto the xy-plane the traces are ellipses, and traces in the xz- and yz-planes are pairs of intersecting lines.

5. Elliptic Paraboloid: The trace in the xy-plane is a point and in planesparallel to and above the xy-plane the traces are ellipses, and traces inthe xz- and yz-planes are parabolas.

6. Hyperbolic Paraboloid: The trace in the xy-plane is a pair of intersectinglines, and traces in planes parallel to the xy-plane are hyperbolas, whichopen in the y-direction when above the xy-plane and open in the x-direction when below the xy-plane. Traces in the xz- and yz-planes areparabolas.

These of course can also appear in other orientations along different coordi-nate axes, or indeed with cross-product terms which would result in other ori-entations. If the elliptic cross-section of an elliptic cone or elliptic paraboloidis circular the are called a circular cone and a circular paraboloid respec-tively. Of course an ellipsoid with all the cross-sections circular is a sphere,i.e. a = b = c.You are not expected to accurately draw any of these surfaces. If asked fora sketch, draw the traces on the planes, and join these to give a rough ideaof the shape. As with the conic sections we can translate a quadric surfaceby moving away from the origin to (a, b, c), which will result in the change(x, y, z) → (x − a, y − b, z − c) in the equations for the surfaces. However,this will not be required for this course.

10

1.

-2-1

01

2

x

-2

-1

0

1

2y

-1.0

-0.5

0.0

0.5

1.0

z

2.

-2

-1

0

1

2

x

-2-1

01

2

y

-4

-2

0

2

4

z

3.

-2-1

01

2

x

-2-1

01

2

y

-5

0

5

z

4.

-2

-1

0

1

2x

-2

-1

0

1

2

y

-1

0

1

z

5.

-2

-1

0

1

2

x

-2

-1

0

1

2

y

0

1

2

3

z

6.

-2

-1

0

1

2x

-2

-1

0

1

2

y

-1

0

1

2

z

Figure 7: Quadric surfaces.

11

2.1.4 Cylindrical and Spherical Coordinates

We have already met polar coordinates in 2-space. Now we introduce twocoordinate systems that are often useful when rectangular coordinates areawkward in 3-space.

Cylindrical coordinates (ρ, θ, z): In terms of cylindrical coordinates therectangular coordinates are written

x = ρ cos θ , y = ρ sin θ , z , (20)

where 0 ≤ ρ < ∞, 0 ≤ θ < 2π. This is shown in Figure 8 If we wish to go

ΡΘ

z

0.00.5

1.01.5

2.0

x

0.0 0.5 1.0 1.5 2.0

y

0.0

0.5

1.0

1.5

2.0

z

Figure 8: Cylindrical coordinates (ρ, θ, φ).

from rectangular to cylindrical coordinates directly, we can use the relations

ρ =√x2 + y2 , tan θ =

y

x, z = z . (21)

Spherical coordinates: In terms of spherical coordinates the rectangularcoordinates are written

x = ρ sinφ cos θ , y = ρ sinφ sin θ , z = ρ cosφ , (22)

where 0 ≤ ρ <∞, 0 ≤ θ < 2π and 0 ≤ φ ≤ π. Figure 9 shows spherical coor-dinates. If we wish to go from rectangular to spherical coordinates directly,

12

ΡΦ

Θ

0.00.5

1.01.5

2.0

x

0.00.5

1.01.5

2.0

y

0.0

0.5

1.0

1.5

2.0

z

Figure 9: Spherical coordinates.

we can use the relations

ρ =√x2 + y2 + z2 , tan θ =

y

x, cosφ =

z√x2 + y2 + z2

. (23)

Example: Change x2 + y2 + z2 = 9 to spherical polar coordinates.

Solution: The coordinates for this surface (a sphere) are for ρ = 3 and so

x = 3 sinφ cos θ , y = 3 sinφ sin θ , z = 3 cosφ , (24)

where θ and φ are the parameters.

13