Chapter 12 - Vectors and the Geometry of Space

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Calculus III Chapter 12 - Vectors and the Geometry of Space

1. Three-Dimensional Coordinate System

Third-semester calculus is the study of functions of more than one variable. Much of what we do will be to generalize concepts from your previous two semesters. However, multivariable calculus will soon take on a life of its own and there are many instances where the picture is signiﬁcantly more complicated than in single variable calculus. Nevertheless, you will often be able to rely on intuition you have built up in previous courses.

Initially, we will focus on building up the geometry of three-dimensional space. Much of what we do will extend beyond this, but pictures are harder to come by.

The three-dimensional coordinate system is formed by three coordinate axes labeled the x- axis, y-axis, and z-axis, all pairwise perpendicular. The origin, denoted O, is the intersection of all three axes. In order to standardize direction, we use the right-hand rule.

Right-hand rule: Extend your right hand, with ﬁngers curled and thumb pointed up. Your ﬁngers align with the positive direction of the x-axis, your arm with the positive direction of the y-axis, and your thumb with the positive direction of the z-axis.

These axes form three planes, labeled the xy-plane, xz-plane, and yz-plane. These planes in turn divide space into eight octants. We will not concern ourselves with numbering the eight octants, except for the ﬁrst octant which is bounded by the positive part of each plane.

A point P in three-dimensional space can be represented by an ordered triple (a, b, c), in- dicating the distance from the origin O in the x, y, and z direction, respectively. Hence, O is represented by (0, 0, 0). This gives a one-to-one bijection between points in space and ordered triples in the Cartesian product R × R × R = {(x, y, z): x, y, z ∈ R}. Given a point P (a, b, c), we can drop a perpendicular to the xy-plane1. This perpendicular intersects the xy-plane at the point Q(a, b, 0). We call this point the projection of P onto the xy-plane. One can project similarly onto the xz- and yz-planes.

1What this means precisely will be deﬁned later. Roughly, it means that we draw a line from P through the xy-plane such that it is perpendicular to every line that lies on that plane. 1 Given a function f : Rn → R (i.e., a function of n-variables), the graph of f is the set of points P such that f(P ) = 0. When n = 3, we call this graph a surface.

Example 1. (1) The surface represented by z − 3 is a plane parallel to the xy-plane through the point (0, 0, 3). (2) The surface represented by x2 + y2 − 1 is a cylinder of radius 1, perpendicular to the xy-plane.

The distance between two points P1(x1, y1, z1) and P2(x2, y2, z2), denoted |P1P2|, is the length of the line segment whose endpoints are P1 and P2. To ﬁnd this distance, we form a box whose opposite endpoints are P1 and P2. Set two other points on this box, A(x2, y1, z1) and

B(x2, y2, z1). Applying the Pythagorean Theorem (twice) gives

2 2 2 |P1P2| = |P1B| + |BP2| (PT) 2 2 2 = |P1A| + |AB| + |BP2| (PT) 2 2 2 = |x2 − x1| + |y2 − y1| + |z2 − z1| (distance formula on real line). Hence, we have now derived the following.

Distance Formula in Three Dimensions: The distance |P1P2| between the points

P1(x1, y1, z1) and P2(x2, y2, z2) is

p 2 2 2 |P1P2| = (x2 − x1) + (y2 − y1) + (z2 − z1) .

A sphere of radius r > 0 centered at C(h, k, `) is the set of points whose distance from (h, k, `) is r. Thus, it follows from the distance formula that the equation of a sphere is

(x − h)2 + (y − k)2 + (z − `)2 = r2.

2 2. Vectors

A vector is, informally (for now), a quantity that has magnitude and direction. We will formalize this eventually, for now we think of a vector v as a ray, having an initial point A −→ (the tail) and a terminal point B (the tip), denoted v = AB. The length of v is the distance |AB|. We say that two vectors u and v are equivalent if they have the same magnitude and direction, even if they are in diﬀerent positions, and write u = v. The zero vector, denoted by 0, is the unique vector with length zero and no direction.

Deﬁnition of Vector Addition: If u and v are vectors positioned so that the initial point of v is the terminal point of u, then the sum of u and v is the vector from the initial point of u to the terminal point of v. −→ −−→ −→ Hence, if u = AB and v = BC, then u+v = AC. Note that vector addition is commutative, as can be seen by the parallelogram law:

Deﬁnition of Scalar Multiplication: If c is a scalar and v is a vector, then the scalar multiple cv is the vector whose length is |c| times the length of v and whose direction is the same as v if c > 0 and opposite if c < 0. If c = 0, then cv = 0.

Two vectors are said to be parallel if they are scalar multiples of one another. We call −v = (−1)v the negative of v. By the diﬀerence of vectors u and v we mean

u − v = u + (−1)v.

If we fix a coordinate system, we can denote a vector by its terminal point P , fixing its initial point to be the origin (since position doesn’t matter). Of course, a vector may have many representations, i.e., different initial and terminal points. We call this one the position vector of the point P . The coordinates for this point are called the components of the

vector. In two-dimensions, we write a = ha1, a2i, and in three-dimensions, we write a = 3 ha1, a2, a3i. We denote by Vn the set of all n-dimensional vectors. So if a ∈ Vn, then −→ a = ha1, a2, . . . , ani. Given any representation AB of the vector a, where A(x1, y1, z1) and

B(x2, y2, z2), the position vector has coordinates

a = hx2 − x1, y2 − y1, z2 − z1i.

The length (or magnitude) of the vector v is the length of any representation and is denoted p 2 2 by |v|. Hence, the length of the two-dimensional vector a = ha1, a2i is |a| = a1 + a2. The p 2 2 2 length of the three-dimensional vector a = ha1, a2, a3i is |a| = a1 + a2 + a3. Given components, it is straightforward to ﬁnd the sum, diﬀerence, and scalar multiple of vectors. Let a = ha1, a2i, v = hb1, b2i, and c a scalar. Then

a + b = ha1 + b1, a2 + b2i, a − b = ha1 − b1, a2 − b2i, ca = hca1, ca2i.

This extends in the obvious way to three or more dimensions. Note that, |ca| = |c||a|. This can be veriﬁed directly, q p 2 2 2 2 2 |ca| = (ca1) + (ca2) = c (a1 + a2) = |c||a|.

Caution: It is not true that |a + b| = |a| + |b|.

Properties of Vectors For all a, b, c ∈ Vn and all scalars c, d,

(i) a + b = b + a (v) c(a + b) = ca + cb (ii) (a + b) + c = a + (b + c) (vi) (c + d)a = ca + da (iii) a + 0 = a (vii) c(da) = (cd)a (iv) a + (−a) = a + (−1)a = 0 (viii) 1a = a

Another way to represent vectors is as sums (linear combinations) of the standard basis vectors. In V2, these are the vectors i = h1, 0i and j = h0, 1i. In V3, these are the vectors

i = h1, 0, 0i, j = h0, 1, 0i, k = h0, 0, 1i.

If a = ha1, a2, a3i, then a = a1i + a2j + a3k. Of course, it is important to know whether you are working in V2 or V3. A unit vector is a vector whose length is 1. Hence, the standard basis vectors are, by deﬁnition, unit vectors. In general, if a 6= 0, the the unit vector in the same direction as a is 1 u = a. |a| We can check this easily. Let c = 1/|a|. Then |u| = |ca| = |c||a| = |a|. 4 Application. A 100-lb weight hangs from two wires as shown in Figure 19 on page 804.

Find the tensions (forces) T1 and T2 in both wires and the magnitudes of the tensions. This is left mostly as a reading exercise. We will walk through some of the initial steps.

We regard T1 and T2 as vectors in V2. Draw a line through the common initial point parallel to the “top” of the triangle and drop perpendiculars from the terminal points of both vectors to this line. This gives two right triangles with hypotenuses |T1| and |T2|, respectively. Hence, using right triangle trigonometry, we have

◦ ◦ T1 = −|T1| cos 50 i + |T1| sin 50 j ◦ ◦ T2 = |T2| cos 32 i + |T2| sin 32 j. The resultant force experience by an object is the vector sum of the forces acting on it. In

this case, the resultant T1 + T2 of the tensions counterbalances the weight w = −100j and

so we have T1 + T2 = −w = 100j. This equation must hold in each component. Hence we have the system of equations,

◦ ◦ −|T1| cos 50 + |T2| cos 32 = 0 ◦ ◦ |T1| sin 50 + |T2| sin 32 = 100. We can solve this system using elementary means2. We obtain

|T1| ≈ 85.64lb and |T2| = 64.9lb.

2Students who have taken Linear Algebra should be able to solve this very quickly. 5 3. The Dot Product

Our discussion of vectors so far included no multiplication-type operation. One way to multiply vectors is using the dot product that is related to the angle between two vectors.

Let a, b ∈ Vn and write a = ha1, . . . , ani, b = hb1, . . . , bni. The dot product of a and b, 3 denoted by a · b, is a scalar (not an element of Vn) given by

a · b = a1b1 + ··· + anbn.

Example 2. Let a = 2i − 3j + k and b = i − 5k. Then

a · b = h2, −3, 1i · h1, 0, −5i = (2)(1) + (−3)(0) + (1)(−5) = −3.

The following properties are easy to check directly for V2 and V3.

Properties of the Dot Product For all a, b, c ∈ Vn and all scalars c, (i) a · a = |a|2 (iii) a · (b + c) = (a · c) + (b · c) (v) 0 · a = 0 (ii) a · b = b · a (iv) (ca) · b = c(a · b) = a · (cb) −→ −−→ Theorem 3. Let A, B be points such that the angle θ between OA and OB satisﬁes 0 ≤ θ ≤ π. If a and b are the position vectors for A and B, respectively, then

a · b = |a| · |b| cos θ.

Proof. Apply the Law of Cosines to the triangle ∆OAB,

|AB|2 = |OA|2 + |OB|2 − 2|OA||OB| cos θ.

In terms of the position vectors, this is

(1) |a − b|2 = |a|2 + |b|2 − 2|a||b| cos θ.

Using the properties of the dot product, we have

(2) |a − b|2 = (a − b) · (a − b) = |a|2 − 2(a · b) + |b|2.

Putting (1) and (2) together we arrive at the desired result.

Example 4. Suppose a = h2, 1i and b = h1, 3i. Then √ √ 5 = a · b = |a||b| cos θ = 5 10 cos θ.

Hence, cos θ = √1 , and so θ = π/4. 2

3 Formally, we would say that · is a function Vn × Vn → R. 6 Two nonzero vectors are perpendicular (or orthogonal) if the angle between them is π/2, in which case cos(π/2) = 0 and so a · b = 0. In general, we say two vectors are orthogonal if and only if a · b = 0.

Example 5. The vectors 3i + 2j − 5k and i + 6j + 3k are orthogonal.

The direction angles of a nonzero vector a ∈ V3 are the angles α, β, γ ∈ [0, π] that a makes with the positive x-, y−, and z− axes, respectively. The cosines of these angles are called the direction cosines of the vector a. We have a · i a cos α = = 1 . |a||i| |a| Similarly, a a cos β = 2 and cos γ = 3 . |a| |a| Hence, cos2 α + cos2 β + cos2 γ = 1 and a = |a|hcos α, cos β, cos γi. It now follows that 1 a = hcos α, cos β, cos γi. |a| −→ −→ Let PQ and PR be representations of two vectors a and b, respectively. We drop a perpen- −→ −→ dicular from R to PQ and call the foot of the perpendicular S. The vector PS is called the vector projection of b onto a, denoted proja b. The scalar projection of b onto a is deﬁned as the signed magnitude of proja b, denoted compa b (for the component of b along a). Let θ be the angle between a and b. Then using right triangle trigonometry we have a · b a · b comp b = |b| cos θ = |b| = . a |a||b| |a|

The vector projection is now the vector of magnitude compa b in the direction of a, so a · b a a · b proj b = = a. a |a| |a| |a|2 Example 6. Find the scalar and vector projections of b = h2, 3i onto a = h1, 4i. a · b 14 a · b 14 comp b = = proj b = = . a |a| 3 a |a|2 9 Application. Suppose a wagon is pulled a distance of 100m along a horizontal path by a constant force of 70 N. The handle of the wagon is held at an angle of 35◦ above the horizontal. Find the work done by the force.

When a constant force F moves an object through a distance d along the line of motion of the object, the work done by the force is deﬁned as W = F d. Now suppose the constant 7 −→ force is a vector F = PR. If the force moves the object from P to Q, then the displacement −→ vector is D = PQ and the work done by this force is deﬁned to be the component of the force along D and the distance moved:

W = (|F| cos θ)|D| = F · D.

In the example, we have

W = F · D = |F||D| cos 35◦ = (70)(100) cos 35◦ ≈ 5734N · m = 5734J.

8 4. The Cross Product

The cross product of two vectors a = ha1, a2, a3i and b = hb1, b2, b3i is deﬁned as the vector c = hc1, c2, c3i that is orthogonal to both a and b. Thus, a · c = 0 and b · c = 0. Writing this out we have

a1c1 + a2c2 + a3c3 = 0

b1c1 + b2c2 + b3c3 = 0.

Solving this system for the ci we obtain

a × b = hc1, c2, c3i = ha2b3 − a3b2, a3b1 − a1b3, a1b2 − a2b1i.

This deﬁnition, though correct, can be diﬃcult to remember. To simplify, we use determi- nants.

The determinant of order 2 is deﬁned as

a b

= ad − bc. c d The determinant of order 3 is deﬁne as

a1 a2 a3 b b b b b b 2 3 1 3 1 2 b1 b2 b3 = a1 − a2 + a3 . c2 c3 c1 c3 c1 c2 c1 c2 c3 In terms of the standard basis vectors, the cross product can then be expressed as

i j k

a × b = a1 a2 a3 .

b1 b2 b3

Here’s another method for computing the determinant, called the butterﬂy method. We rewrite the ﬁrst and second column to the right of the matrix. We draw diagonals going down (these are the positive summands) and then draw diagonals going up (these are the negative summands). The result is the determinant of the matrix.

Example 7. Find the cross product of a = 3i + 2j − 5k and b = i + 6j + 3k. We want the determinant of the matrix,

i j k

3 2 −5 .

1 6 3 9 We will use the butterﬂy method to compute the determinant. The diagonals going up are colored below: ijk i j 32 -5 32 = 6i − 5j + 18k. 1 63 16

The diagonals going down are colored below:

i jk ij 32 -5 32 = −30i + 9j + 2k. 163 1 6

Thus, the cross product (determinant) is

(6i − 5j + 18k) − (−30i + 9j + 2k) = 36i − 14j + 16k.

Suppose a, b, and a × b are represented by vectors with the same endpoint. Then a × b points in a direction perpendicular to the plane through a and b. The direction is determined by the right-hand rule. Extend your right hand, thumb extended, with your ﬁngers curled in the direction of the angle from a to b. Then your thumb points in the direction of a × b.

Theorem 8. Let a and b be vectors such that the angle θ between them satisﬁes 0 ≤ θ ≤ π. Then |a × b| = |a||b| sin θ.

Proof. A tedious computation using the original deﬁnition of the cross product (see page 817 in the text) shows that

|a × b|2 = |a|2|b|2 − (a · b)2. 10 Now

|a × b|2 = |a|2|b|2 − (a · b)2 = |a|2|b|2 − |a|2|b|2 cos2 θ = |a|2|b|2(1 − cos2 θ) = |a|2|b|2 sin2 θ. √ 2 The result now follows because sin θ ≥ 0 for all 0 ≤ θ ≤ π, so sin θ = sin θ.

Corollary 9. Two nonzero vectors a and b are parallel if and only if a × b = 0.

Proof. Two vectors are parallel if and only if θ = 0 or π if and only if sin θ = 0.

Let a and b represent two vectors with the same initial point and let θ, 0 ≤ θ ≤ π be the angle between them. Form a parallelogram with base |a| and altitude |b| sin θ. Thus, the area of the parallelogram is A = |a|(|b| sin θ) = |a × b|. Thus, the magnitude of the cross product of a and b is the area of the parallelogram determined by these vectors.

Properties of the Dot Product For all a, b, c ∈ V3 and all scalars c,

(i) a × b = −b × a (iv) (a + b) × c = (a × c) + (b × c) (ii) (ca) × b = c(a × b) = a × (cb) (v) a · (b × c) = (a × b) · c (iii) a × (b + c) = (a × c) + (b × c) (vi) a × (b × c) = (a · c)b − (a · b)c

The product a · (b × c) is called the scalar triple product of the vectors a, b, and c. This can be written as the determinant

a1 a2 a3

a · (b × c) = b1 b2 b3 .

c1 c2 c3 Consider the parallelepiped determined by these vectors. The area of the base parallelogram is A = |b × c|. If θ is the angle between a and b × c, then the height of the parallelepiped is h = |a|| cos θ|. Thus, the volume of the parallelepiped is

V = Ah = |b × c||a|| cos θ| = |a · (b × c)|.

Note that a · (b × c) = 0 implies that the vectors are all coplanar. 11 Application. A bolt is tightened by applying a 40-N force to a 0.25-m wrench. Let θ be the angle between the position and force vectors, r and F, respectively, and suppose θ measures 75◦. Let τ be the torque (relative to the origin) deﬁned to be the cross product of the position and force vectors, so τ = r × F. Hence, in this example, we have

|τ| = |r × F| = |r||F| sin 75◦ = (0.25)(40) sin 75◦ ≈ 9.66N · m.

12 5. Equations of lines and planes

For points in the Cartesian (xy-)plane, a line is determined by a point and slope. Really what the slope is telling us is the direction of the line. In three dimensions, we need the same information, but we express it in terms of vectors.

Let L be a line in three-dimensional space and let v be a vector parallel to L. Let P0(x0, y0, z0) be a given point on L and let P (x, y, z) be an arbitrary point on L. Let r0 and r denote the position vectors associated to P0 and P , respectively. Denote by a the vector with −−→ representation P0P , so by the Triangle Law, r = r0 + a. Also, a is parallel to v so a = tv for some scalar t. Thus, the vector equation for the line L (in vector form) is

r = r0 + tv.

The values of the parameter t correspond to points on the line L, where r0 is the point with t = 0. To indicate only a portion of the line, we could restrict the values of t. For example, setting t ≥ 0 would give one “half” of the line.

If we write ha, b, ci for the direction vector, the values a, b, and c are called the direction numbers of L. The parametric equations for the line L through the point (x0, y0, z0) and parallel to ha, b, ci are

x = x0 + at y = y0 + bt z = z0 + ct.

Note that we could also solve each of these equations for t to obtain x − x y − y z − z t = 0 t = 0 t = 0 . a b c This gives the symmetric equations of L, x − x y − y z − z 0 = 0 = 0 . a b c Example 10. Find the vector, parametric, and symmetric equations for the line through the points A(1, 2, 5) and B(3, 1, 0). At what point does this line intersect the xy-plane? Write the equation of the line segment from A to B. −→ The vector v with representation AB is parallel to the line and

v = h3 − 1, 1 − 2, 0 − 5i = h2, −1, −5i.

Thus, taking r0 = h1, 2, 5i, we have that the vector equation for the line is

r = r0 + tv = (1 + 2t)i + (2 − t)j + (5 − 5t)k. 13 The parametric equations are

x = 1 + 2t y = 2 − t z = 5 − 5t.

The symmetric equations are then, x − 1 y − 2 z − 5 = = . 2 −1 −5

The line intersects the xy-plane when z = 0. Plugging this in the symmetric equations gives, x − 1 y − 2 = = 1. 2 −1 Thus, this occurs when x = 3 and y = 1.

Recalling the vector equation r = r0 + tv, we note that A occurs (technically, its position vector) when t = 0. On the other hand, B occurs when t = 1. Hence, the equation for the line segment AB is

r = r0 + tv, 0 ≤ t ≤ 1.

Let r0 and r1 be points on the line L. Then the direction vector is v = r1 − r0. Thus, in this case, the vector equation is

r = r0 + tv = r0 + t(r1 − r0) = (1 − t)r0 + tr1.

The line segment from r0 to r1 corresponds to the values 0 ≤ t ≤ 1.

Example 11. Consider the lines L1 and L2 with symmetric equations given below x y − 1 z − 2 x − 1 y − 3 z L : = = L : = = . 1 1 −1 3 2 2 −2 7 These lines are not parallel because their respective direction vectors, h1, −1, 3i and h2, −2, 7i, are not scalar multiples of one another.

One could then ask whether these lines intersect. To check this, we rewrite in parametric form, careful to use a diﬀerent parameter for each:

L1 : x = t y = 1 − t z = 2 + 3t

L2 : x = 1 + 2s y = 3 − 2s z = 7s.

We are asking whether there are values for t and s so that the respective equations match. The equations for x tell us that t = 1+2s. Then y = 1−t = 1−(1+2s) = −2s 6= 3−2s. Thus, there is no point of intersection and thus these lines do not intersect. Lines in three-space that do not intersect and are not parallel are called skew lines. 14 A plane in space is determined by a point P0(x0, y0, z0) and a vector n that is orthogonal to the plane, called a normal vector to the plane. The normal vector determines the direction of the plane, while the point determines its position. Let P (x, y, z) be an arbitrary point and let r0 and r be the position vectors of P0 and P , respectively. The normal vector is −−→ orthogonal to every vector in the plane, including P0P , and so we have the vector equation of the plane

n · (r − r0) = 0.

Write n = ha, b, ci, r = hx, y, zi, and r0 = hx0, y0, z0i. We can then rewrite the vector equation as the scalar equation of the plane

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

If we set d = −(ax0 + by0 + cz0), then the linear equation of the plane is

ax + by + cz + d = 0.

Example 12. The plane through the point (5, 3, 5) and with normal vector 2i + j − k is

2(x − 5) + 1(y − 3) − 1(z − 5) = 0.

This can be rewritten as a linear equation,

2x + y − z − 8 = 0.

Example 13. Find the equation of the plane through the points P (3, 0, −1), Q(−2, −2, 3), and R(7, 1, −4). −→ −→ The vectors a and b corresponding to PQ and PR, respectively, are

a = h−5, −2, 4i b = h4, 1, −3i.

Both of these vectors lie in the plane, thus, their cross product is orthogonal to the plane. We take this as the normal vector,

i j k

n = a × b = −5 −2 4 = 2i + j + 3k.

4 1 −3 Thus, using P as our point, the equation of the plane is

2(x − 3) + 1(y − 0) + 3(z + 1) = 0. or 2x + y + 3z = 3. 15 The normal vector determines the direction of the plane, thus, given two planes, the angle between their respective normal vectors determines the angle between the planes, if they intersect. The planes are parallel if and only if the normal vectors are scalar multiples of one another.

Example 14. (1) Consider the planes with equations 9x − 3y + 6z = 2 and 2y = 6x + 4z.

The normal vectors for these planes are n1 = h9, −3, 6i and n2 = h6, −2, 4i. Note that 2 n2 = 3 n1, and so the planes are parallel. (2) Consider the planes with equations x − y + 3z = 1 and 3x + y − z = 2. The normal vectors for these planes are n1 = h1, −1, 3i and n2 = h3, 1, −1i. Then n · n −1 cos θ = 1 2 = . |n1||n2| 11 Thus, θ = arccos(−1/11) ≈ 95.22◦.

Let P1(x1, y1, z1) be a point and ax + by + cz + d = 0 the equation of a plane. Given any −−→ point P0(x0, y0, z0) on the plane, we let b = hx1 − x0, y1 − y0, z1 − z0i denote P0P1. The

distance D from P1 to the plane is the length of the scalar projection of b onto the normal vector n = ha, b, ci of the plane. Hence, |n · b| D = | comp b| = n |n| |a(x − x ) + b(y − y ) + c(z − z )| = 1 0 √ 1 0 1 0 a2 + b2 + c2 |(ax + by + cz ) − (ax + by + cz )| = 1 1 √ 1 0 0 0 . a2 + b2 + c2

As P0 lies on the plane, its coordinates satisfy the equation of the plane. That is, ax0 +by0 + cz0 = −d. Hence we have |ax + by + cz + d| D = 1√ 1 1 . a2 + b2 + c2 Example 15. Find the distance between the parallel planes with equations 6z = 4y − 2x and 9z = 1 − 3x + 6y.

The point P1(0, 0, 0) lies on the ﬁrst plane while n = h−3, 6, −9i is the normal vector of the second plane. Hence, using the above equation, |ax + by + cz + d| 1 D = 1√ 1 1 = √ . a2 + b2 + c2 126

16 6. Cylinders and quadric surfaces

Quadric surfaces are the analogs in three dimensions of conic sections in the plane. They are represented by a degree two equation in the three variables x, y, and z. A somewhat tedious argument shows that they can be written in one of two forms:

Ax2 + By2 + Cz2 + J = 0 or Ax2 + By2 + Iz = 0.

The observant reader will recognize the ﬁrst one as the equation of a sphere, centered at the origin, when A = B = C = 1. The sphere is one example of a quadric surface.

There are six families of quadric surfaces, detailed on page 837 of your text. You should familiarize yourself with these:

How do we draw these? For that matter, how do we have any idea what these should look like? The key idea is to take traces.

Example 16. Let’s sketch the surface 4x2 + 9y2 + 9z2 = 36.

Set z = 0, then the equation reduces to 4x2 + 9y2 = 36, or x2 y2 + = 1. 9 4 17 We say the trace of the surface at z = 0 is an ellipse is the equation of an ellipse. So, we draw an ellipse in the xy-plane.

Now set z = 1, then the equation becomes 4x2 + 9y2 + 9 = 36, or x2 y2 25 + = . 9 4 36 Again, we have an ellipse in the plane z = 1. Repeating this, we obtain the ellipsoid shown below: