Lines and Planes

1. Find an equation describing the plane which passes through the points (2, 2, 1), (3, 1, 0), and (0,−2, 1).

2. Find an equation describing the plane which goes through the point (1, 3, 5) and is perpendicular tothe vector 〈2, 1,−3〉.

3. Let L be the line which passes through the points (1,−2, 3) and (4,−5, 6).

(a) Find a parametric vector equation for L.

(b) Find parametric (scalar) equations for L.

(c) Find symmetric equations for L.

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4. Let L1 be the line with parametric vector equation ~r1(t) = 〈7, 1, 3〉 + t〈1, 0,−1〉 and L2 be the linedescribed parametrically by x = 5, y = 1+3t, z = t. How many planes are there which contain L2 andare parallel to L1? Find an equation describing one such plane.

5. Find the distance from the point (0, 1, 1) to the plane 2x + 3y + 4z = 15.

6. Find the distance from the point (1, 3,−2) to the line x3 = y − 1 = z + 2.

7. True or false: The line x = 2t, y = 1 + 3t, z = 2 + 4t is parallel to the plane x− 2y + z = 7.

8. True or false: Let S be a plane normal to the vector ~n, and let P and Q be points not on the plane S.If ~n ·

−−→PQ = 0, then P and Q lie on the same side of S.

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Quadric Surfaces

Six basic types of quadric surfaces:

• ellipsoid

• cone

• elliptic paraboloid

• hyperboloid of one sheet

• hyperboloid of two sheets

• hyperbolic paraboloid

(A) (B) (C)

(D) (E) (F)

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Quick reminder:

• x2

a2+

y2

b2= r describes . . .

– an ellipse if r > 0.

– a point if r = 0 (we consider this a “degenerate” ellipse).

– nothing if r < 0.

• x2

a2− y2

b2= r describes . . .

– a hyperbola if r 6= 0.

– a pair of lines if r = 0 (we consider this a “degenerate” hyperbola).

• y = ax2 + b describes a parabola.

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1. For each surface, describe the traces of the surface in x = k, y = k, and z = k. Then pick the termfrom the list above which seems to most accurately describe the surface (we haven’t learned any ofthese terms yet, but you should be able to make a good educated guess), and pick the correct pictureof the surface.

(a)x2

9− y2

16= z.

• Traces in x = k:

• Traces in y = k:

• Traces in z = k:

(b)x2

4+

y2

25+

z2

9= 1.

• Traces in x = k:

• Traces in y = k:

• Traces in z = k:

(c)x2

4+

y2

9=

z

2.

• Traces in x = k:

• Traces in y = k:

• Traces in z = k:

(d)z2

4− x2 − y2

4= 1.

• Traces in x = k:

• Traces in y = k:

• Traces in z = k:

(e) x2 +y2

9=

z2

16.

• Traces in x = k:

• Traces in y = k:

• Traces in z = k:

3

(f)x2

9+ y2 − z2

16= 1.

• Traces in x = k:

• Traces in y = k:

• Traces in z = k:

2. Sketch the surface 9y2 + 4z2 = 36. What type of quadric surface is it?

3. Sketch the surface y2 + 2y + z2 = x2. What type of quadric surface is it?

4. What type of quadric surface is 4x2 − y2 + z2 + 9 = 0?

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Lines and Planes

1. Find an equation describing the plane which passes through the points (2, 2, 1), (3, 1, 0), and (0,−2, 1).

Solution. Let’s give the three points names, say P = (2, 2, 1), Q = (3, 1, 0), and R = (0,−2, 1). Apoint S(x, y, z) is in the plane if (and only if) the three vectors

−−→PQ,

−→PR, and

−→PS are coplanar. As we

saw before, this is the same as saying that the scalar triple product−→PS · (

−−→PQ×

−→PR) is equal to 0.

Now, we can just write this all out:−−→PQ = 〈1,−1,−1〉,

−→PR = 〈−2,−4, 0〉, so

−−→PQ×

−→PR = 〈−4, 2,−6〉.

−→PS = 〈x− 2, y− 2, z − 1〉, so the equation

−→PS · (

−−→PQ×

−→PR) = 0 can be rewritten as −4(x− 2) + 2(y−

2)−6(z−1) = 0, or 4x−2y +6z = 10. We can make this equation slightly simpler by dividing throughby 2 to get 2x− y + 3z = 5 .

2. Find an equation describing the plane which goes through the point (1, 3, 5) and is perpendicular to thevector 〈2, 1,−3〉.

Solution. Let’s give the point (1, 3, 5) a name; we’ll call it P . A point S(x, y, z) is on the plane ifthe vector

−→PS is perpendicular to the vector 〈2, 1,−3〉. That is, 〈x − 1, y − 3, z − 5〉 · 〈2, 1,−3〉 = 0.

Simplifying, 2(x− 1) + 1(y − 3)− 3(z − 5) = 0, which can also be written as 2x + y − 3z = −10 .

3. Let L be the line which passes through the points (1,−2, 3) and (4,−5, 6).

(a) Find a parametric vector equation for L.

Solution. Let’s call the points P (1,−2, 3) and Q(4,−5, 6). The line we are looking for passesthrough P and is parallel to

−−→PQ = 〈3,−3, 3〉, so we can describe it using the parametric vector

equation ~r(t) = 〈1,−2, 3〉+ t〈3,−3, 3〉.

(b) Find parametric (scalar) equations for L.

Solution. From (a), ~r(t) = 〈1 + 3t,−2− 3t, 3 + 3t〉, so we could also write the line as x = 1 + 3t,y = −2− 3t, and z = 3 + 3t.

(c) Find symmetric equations for L.

Solution. To find symmetric equations, we solve the parametric scalar equations we found in(b) for t: t = x−1

3 , t = −2−y3 , and t = z−3

3 . Then, we set these equal to each other: x−13 =

−2−y3 = z−3

3 . Of course, we could multiply these all by 3 to get the slightly simpler lookingx− 1 = −2− y = z − 3 .

4. Let L1 be the line with parametric vector equation ~r1(t) = 〈7, 1, 3〉 + t〈1, 0,−1〉 and L2 be the linedescribed parametrically by x = 5, y = 1 + 3t, z = t. How many planes are there which contain L2 andare parallel to L1? Find an equation describing one such plane.

Solution. Let’s first rewrite L2 using a parametric vector equation ~r2(t) = 〈5, 1, 0〉+ t〈0, 3, 1〉.

From the two parametric vector equations, we can see that:

• L1 goes through the point P (7, 1, 3) and is parallel to the vector ~u = 〈1, 0,−1〉.

• L2 goes through Q(5, 1, 0) and is parallel to the vector ~v = 〈0, 3, 1〉.

Therefore, we are looking for a plane which:

1

• contains Q and is parallel to ~v (because it contains L2)

• is parallel to ~u (because it is parallel to L1)

Since such a plane plane is parallel to both ~u and ~v, it must be orthogonal to ~u×~v. We compute that~u× ~v = 〈3,−1, 3〉.

There is only one plane which is orthogonal to 〈3,−1, 3〉 and contains the point (5, 1, 0). (We saw

this idea already in #2.) Its equation is 3x− y + 3z = 14 .

Note: If L1 and L2 had been parallel, there would have been infinitely many planes which containedL2 and were parallel to L1: any plane containing L2 would automatically be parallel to L1.

5. Find the distance from the point (0, 1, 1) to the plane 2x + 3y + 4z = 15.

Solution. We have a point P (0, 1, 1) and a plane, and we want to find the distance between the two.Here is one method. Suppose Q is a point in the plane and ~n is a normal vector for the plane.

P

nÓÖ

Q

Then, the distance from P to the plane is simply the length of the projection vector proj~n−−→QP :

P

projnÓÖ QPÓÖÖÖÖÖ

Q

In this particular case, Q = (0, 5, 0) is a point on the plane, and ~n = 〈2, 3, 4〉 is a normal vector for

the plane. Then,−−→QP = 〈0,−4, 1〉, and proj~n

−−→QP = ~n·

−−→QP

~n·~n ~n = − 829 〈2, 3, 4〉. The length of this vector is

8√29

.

6. Find the distance from the point (1, 3,−2) to the line x3 = y − 1 = z + 2.

Solution. We have a point P (1, 3,−2) and a line, and we want to find the distance between the two:

P

2

Here’s one way to do that. Find a point Q on the line and a vector ~v parallel to the line; then, thedistance is the height of the parallelogram determined by

−−→QP and ~v:

Q

P

vÓ

The height of the parallelogram is equal to the area of the parallelogram divided by the length of the

base, or |−−→QP×~v||~v| .

For the given line, it will be easier to find a point on the line and a vector parallel to the line if werewrite it using a parametric vector equation. To do this, let’s set t equal to x

3 = y− 1 = z + 2. Then,x = 3t, y = 1 + t, and z = −2 + t, so we can describe the line by the parametric vector equation〈0, 1,−2〉+ t〈3, 1, 1〉. From this, we can see that Q(0, 1,−2) is a point on the line and ~v = 〈3, 1, 1〉 is avector parallel to the line.

Now, we just compute |−−→QP×~v||~v| :

−−→QP = 〈1, 2, 0〉, so

−−→QP × ~v = 〈2,−1,−5〉 and

|−−→QP × ~v||~v|

=

√3011

.

7. True or false: The line x = 2t, y = 1 + 3t, z = 2 + 4t is parallel to the plane x− 2y + z = 7.

Solution. True.

A normal vector for the plane is ~n = 〈1,−2, 1〉. The line x = 2t, y = 1 + 3t, z = 2 + 4t is parallel tothe vector 〈2, 3, 4〉, and this vector is orthogonal to ~n, so this vector must be parallel to the plane.

Another way to see that the line and plane are parallel is to try to compute the intersection. If (x, y, z)is in both the line and plane, then the four equations x = 2t, y = 1 + 3t, z = 2 + 4t, and x− 2y + z = 7must all be satisfied. Plugging the first three equations into the fourth, 2t− 2(1 + 3t) + (2 + 4t) = 7,which simplifies to 0 = 7, so there are no solutions to all 4 equations. This means that the line andplane do not intersect, so they must be parallel.

8. True or false: Let S be a plane normal to the vector ~n, and let P and Q be points not on the plane S.If ~n ·

−−→PQ = 0, then P and Q lie on the same side of S.

Solution. True. The fact that ~n ·−−→PQ = 0 means that ~n is orthogonal to

−−→PQ, so

−−→PQ is parallel to the

plane.

3

Quadric Surfaces

Six basic types of quadric surfaces:

• ellipsoid

• cone

• elliptic paraboloid

• hyperboloid of one sheet

• hyperboloid of two sheets

• hyperbolic paraboloid

(A) (B) (C)

(D) (E) (F)

1. For each surface, describe the traces of the surface in x = k, y = k, and z = k. Then pick the termfrom the list above which seems to most accurately describe the surface (we haven’t learned any ofthese terms yet, but you should be able to make a good educated guess), and pick the correct pictureof the surface.

(a)x2

9− y2

16= z.

1

• Traces in x = k: parabolas

• Traces in y = k: parabolas

• Traces in z = k: hyperbolas (possibly a pair of lines)

Solution. The trace in x = k of the surface is z = −y2

16 + k2

9 , which is a downward-openingparabola.

The trace in y = k of the surface is z = x2

9 −k2

16 , which is an upward-opening parabola.

The trace in z = k of the surface is x2

9 −y2

16 = k, which is a hyperbola if k 6= 0 and a pair of lines(a degenerate hyperbola) if k = 0.

This surface is called a hyperbolic paraboloid , and it looks like picture (E) . It is also sometimescalled a saddle.

(b)x2

4+

y2

25+

z2

9= 1.

• Traces in x = k: ellipses (possibly a point) or nothing

• Traces in y = k: ellipses (possibly a point) or nothing

• Traces in z = k: ellipses (possibly a point) or nothing

Solution. The trace in x = k of the surface is y2

25 + z2

9 = 1− k2

4 . Since 1− k2

4 can be positive, zero,or negative (depending on what k is), this intersection can be an ellipse, a point (a degenerateellipse), or nothing.

The trace in y = k of the surface is x2

4 + z2

9 = 1− k2

25 , which is an ellipse, a point, or nothing.

The trace in z = k of the surface is x2

4 + y2

25 = 1− k2

9 , which is an ellipse, a point, or nothing.

The picture which matches this description is (A) , and this surface is an ellipsoid.

(c)x2

4+

y2

9=

z

2.

• Traces in x = k: parabolas

• Traces in y = k: parabolas

• Traces in z = k: ellipses (possibly a point) or nothing

Solution. The trace in x = k of this surface is z = 2y2

9 −k2

2 , which is a parabola.

The trace in y = k of this surface is z = x2

2 −2k2

9 , which is a parabola.

The trace in z = k of this surface is x2

4 + y2

9 = k2 , which is an ellipse when k > 0. If k = 0, it is a

point (a degenerate ellipse), and if k < 0, it is nothing.

This matches picture (C) , and this surface is called an elliptic paraboloid .

(d)z2

4− x2 − y2

4= 1.

2

• Traces in x = k: hyperbolas (never a pair of lines)

• Traces in y = k: hyperbolas (never a pair of lines)

• Traces in z = k: ellipses (possibly a point) or nothing

Solution. The trace in x = k of this surface is z2

4 −y2

4 = 1 + k2, which is a hyperbola. Since1 + k2 is always positive, this is never a pair of lines.

The trace in y = k of this surface is z2

4 − x2 = 1 + k2

4 , which is a hyperbola (and never a pair oflines).

The trace in z = k of this surface is x2 + y2

4 = k2

4 − 1. Since k2

4 − 1 can be positive, zero, ornegative, this intersection can be an ellipse, a point, or nothing.

The pictures (B), (D), and (F) all show surfaces whose traces in x = k and y = k are hyperbolasand whose traces in z = k are ellipses. Of these three graphs, (D) is the only one in which sometraces in z = k are nothing, so the correct picture is (D) . This surface is an example of a

hyperboloid of two sheets .

(e) x2 +y2

9=

z2

16.

• Traces in x = k: hyperbolas (possibly a pair of lines)

• Traces in y = k: hyperbolas (possibly a pair of lines)

• Traces in z = k: ellipses (possibly a point)

Solution. The trace in x = k of this surface is z2

16 −y2

9 = k2. Since k2 can be either non-zero orzero, this is either a hyperbola or a pair of lines. (It is only a pair of lines when k = 0; otherwise,it is a hyperbola.)

The trace in y = k of this surface is z2

16 − x2 = k2

9 , which is again either a hyperbola or a pair oflines (a pair of lines only when k = 0).

The trace in z = k of this surface is x2 + y2

9 = k2

16 . Since k2

16 is either positive or 0, this intersectionis either an ellipse or a point (but never nothing, and it is only a point when k = 0).

This matches picture (F) and is an example of an cone .

(f)x2

9+ y2 − z2

16= 1.

• Traces in x = k: hyperbolas (possibly a pair of lines)

• Traces in y = k: hyperbolas (possibly a pair of lines)

• Traces in z = k: ellipses (never a point)

Solution. The trace in x = k of this surface is y2 − z2

16 = 1 − k2

9 . Since 1 − k2

9 can be non-zeroor zero, this is either a hyperbola or a pair of lines.

The trace in y = k of this surface is x2

9 −z2

16 = 1 − k2, which is either a hyperbola or a pair oflines.

3

The trace in z = k of this surface is x2

9 + y2 = 1 + k2

16 . Since 1 + k2

16 is always positive, this isalways an ellipse (and never a point or nothing).

This description matches picture (B) , and this surface is a hyperboloid of one sheet .

2. Sketch the surface 9y2 + 4z2 = 36. What type of quadric surface is it?

Solution. Notice that the equation describing this surface does not involve x at all. Each trace of thesurface in x = k is the ellipse 9y2 + 4z2 = 36, which can also be written as y2

4 + z2

9 = 1.

Therefore, the surface looks like this:

This is not any of the quadrics we have seen so far; it is a (quadric) cylinder .

3. Sketch the surface y2 + 2y + z2 = x2. What type of quadric surface is it?

Solution. We have a y2 term as well as a y term, and we can make the equation simpler by completingthe square. In this case, we accomplish that by adding 1 to both sides of the equation:

y2 + 2y + 1 + z2 = x2 + 1(y + 1)2 + z2 = x2 + 1

−x2 + (y + 1)2 + z2 = 1

To sketch, let’s look at the traces:

• The trace in x = k of the surface is (y + 1)2 + z2 = 1 + k2, which describes a circle (centered aty = −1, z = 0 and having radius

√1 + k2).

• The trace in y = k of the surface is −x2 + z2 = 1 − (k + 1)2, which is a hyperbola (or a pair oflines).

• The trace in z = k of the surface −x2 +(y +1)2 = 1−k2, which is a hyperbola (or a pair of lines).

Since we know the traces in x = k are circles centered at y = −1, z = 0, we can think of the surfaceas being obtained by rotating some sort of curve about the line y = −1, z = 0 (this line is parallel tothe x-axis). To figure out what that curve is, we can look at the trace of the surface in either y = −1or z = 0.

Let’s do the trace in z = 0: this is (y + 1)2 − x2 = 1, which is the following hyperbola:

4

-3 -2 -1 1 2 3x

-4

-3

-2

-1

1

2

y

If we draw this in space (in the plane z = 0, since we’re talking about the trace in z = 0), it looks likethe left figure below. Therefore, the surface looks like the right figure.

xy

z

This is a hyperboloid of one sheet .

4. What type of quadric surface is 4x2 − y2 + z2 + 9 = 0?

Solution. Let’s look at the traces:

• The trace in x = k of this surface is y2 − z2 = 4k2 + 9, which is always a hyperbola.

• The trace in y = k of this surface is 4x2 + z2 = k2 − 9, which is an ellipse, a point, or nothing,depending on k.

• The trace in z = k of this surface is y2 − 4x2 = k2 + 9, which is always a hyperbola.

Since the traces in x = k and z = k are hyperbolas, we know the surface is either a hyperboloid or acone. The fact that the traces in y = k can be nothing tells us that the surface must be a hyperboloidof two sheets.

The surface looks like this:

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6