Math 200: Exam 1 Review Sheet

Exam Information

Exam 1 will be given on Tuesday, July 15th from 7-7:50 pm. The exam willcover sections 11.1, 11.2, 11.3, 11.4, 11.5, and 11.6.

The exam is closed book, closed notes, and without calculator.

The exam is located in Randell 114.

Exam Preparation Suggestions

To prepare for this exam, I recommend the following:

• Review your lecture notes, the suggested practice problems, and theappropriate chapter contents of a calculus textbook. Specifically, makesure you have mastered the ‘‘Expected Skills’’ which are listed at thetop of each set of practice problems.

• Depending on your learning style, it may be a good idea to form anoutline of the important topics discussed in each chapter - includingone or two sample examples of the types of problems that you expectto be asked.

• Re-work the suggested practice problems! Try to identify those problemsand topics on which you have difficulty and then review the notes orchapter related to that material.

• Form a study group with your peers. After reviewing the material,make a mock exam and trade with a friend. Try to solve the problemswithout the aid of a textbook or notes. Again, use this to help youlearn which material is still unclear so that you can effectively use anyadditional study time working on these topics.

• Make sure that you get a good night’s sleep before test day, and eata hearty breakfast in the morning, especially one rich in protein andfiber. DON’T CRAM OR PULL AN ALL-NIGHTER! DON’T TESTON AN EMPTY STOMACH!

1

Exam Taking Tips & Advice

• Arrive early to the exam location. If you arrive late, then you get lesstime, not more time.

• Bring a watch; classrooms and lecture halls in Drexel tend to lack clocks.Cell phones should be turned off and packed away before the exambegins.

• Use a pencil instead of a pen, and bring more than one with you. Crossout or erase any work that you do not want graded.

• As soon as you receive your exam, write your name and section numberon the front page. Exams without names will result in a loss of 5 points.

• Quickly skim through all of the problems on the exam before doing anyone problem. Then do the problems that you think are easiest first.

• Pace yourself. Know when you should stop working on a problem andmove on to another one. It is not worth leaving a 10-point problemblank because you struggled for ten minutes on a 5-point problem.

• There is no partial credit for true/false, always/sometimes/never, andmultiple choice questions. You can show as little or as much work as youwant, and the work does not have to be very organized. You should tryto eliminate incorrect answers and, if necessary, you can guess from theremaining choices. Never leave a true/false, always/sometimes/never,or multiple choice question blank.

• For all of the other questions you must show all of your work. Be surethat your work is organized and legible and that you do not skip a lotof steps. Answers (even correct ones) will not receive a lot of creditwithout the necessary work to back them up. Partial credit will beawarded based on how much correct work that you show.

• If you have the time, you should check your answers. (Just be sureto manage your time - see ‘‘Pace yourself’’ above.) It doesn’t pay toleave twenty minutes early without checking your answers. A lot ofsilly mistakes can be avoided by going through your work again.

2

Exam 1 Practice Problems

Practice problems are similar, both in difficulty and in scope, to the type ofproblems you will see on the exam. Problems marked with a ? are ‘‘for yourentertainment’’ and are not essential.

EXPECTED SKILLS:

• Be able to determine the location of a point in space using rectangularcoordinates.

• Be able to find the distance between and the midpoint of the linesegment connecting two points in space.

• Know the standard equation of a sphere and be able to find the centerand radius of a sphere.

• Be able to sketch cylindrical surfaces.

• Be able to perform arithmetic operations on vectors and understandthe geometric consequences of the operations.

• Know how to compute the magnitude of a vector and normalize avector.

• Be able to use vectors in the context of geometry and force problems.

• Know how to compute the dot product of two vectors.

• Be able to use the dot product to find the angle between two vectorsand, in particular, be able to determine if two vectors are orthogonal.

• Know how to compute the direction cosines of a vector.

• Be able to decompose vectors into orthogonal components and knowhow to compute the orthogonal projection of one vector onto another.

• Be able to use the dot product to find the work W done by a constantforce applied at a constant angle with the direction of motion to anobject.

• Know how to calculate the cross product of two vectors in 3-space.

3

• Be able to use a cross product to find a vector perpendicular to twogiven vectors.

• Know how to use a cross product to find areas of parallelograms andtriangles.

• Be able to use a cross product together with a dot product to computevolumes of parallelepipeds.

• Be able to use the cross product to find the torque done by applying aconstant force to an object so that it rotates about an axis.

• Be able to find the parametric equations of a line that satisfies certainconditions by finding a point on the line and a vector parallel to theline.

• Know how to determine whether two lines in space are parallel, skew orintersecting, and, if the lines intersect, be able to determine the pointof intersection.

• Know how to determine where a line intersects a surface.

• Be able to find the equation of a plane that satisfies certain conditionsby finding a point on the plane and a vector normal to the plane.

• Know how to find the parametric equations of the line of intersectionof two (non-parallel) planes.

• Be able to find the (acute) angle of intersection between two planes.

• Be able to solve distance problems involving points, lines, and planes.

OLD EXAMS WITH SOLUTIONS:

• http://www.math.drexel.edu/classes/math200/200935/ (go to the‘‘Course Material’’ section)

• http://www.math.drexel.edu/classes/math200/201135/resources/

Exam1-2010.pdf

Solutions: http://www.math.drexel.edu/classes/math200/201135/resources/Exam1-2010-Solutions.pdf

4

• http://www.math.drexel.edu/~dp399/resources/Math-200/Exam1.pdf

Solutions: http://www.math.drexel.edu/~dp399/resources/Math-200/Exam1Sol.pdf

• http://www.math.drexel.edu/classes/math200/201235/resources/

Exam1a-200-201235.pdf

• http://www.math.drexel.edu/classes/math200/201235/resources/

Exam1b-200-201235.pdf

• http://www.math.drexel.edu/classes/math200/201335/resources/

Math200_Exam1_Draft1_201335.pdf

Solutions: http://www.math.drexel.edu/classes/math200/201335/resources/Exam1-Solutions.pdf

PRACTICE PROBLEMS:

1. For each of the following, compute u×v and verify that it is orthogonalto both u and v.

(a) u = 〈3, –4, 1〉; v = 〈2, –2, 3〉(b) u = 〈2, –2, 6〉; v = 〈–1, 2, –1〉(c) u = 2i+ 3k; v = i− j

2. (a) Using appropriate properties of the cross product (Not Determi-nants), compute (i− j)× (j− i).

(b) Verify that your answer to part (a) is correct by using determinants.

3. Compute two unit vectors which are normal to the plane which isdetermined by the points A(1, 2, 3), B(6, 4, 7), and C(1, 5, 2).

4. Compute the area of the triangle with vertices A(1, 2, 3), B(6, 4, 7), andC(1, 5, 2).

5. Compute ||u× v|| if ||u|| = 2, ||v|| = 5, and the angle between u andv is 30◦.

5

6. The following questions deal with finding the distance from a point toa line:

(a) Given three points A,B, and P in 3-space as shown in the picturebelow, explain how you could use the cross product to calculatethe distance d between the point P and the line which contains Aand B.

[Hint: Consider the vectors−→AP and

−→AB.]

(b) Use your method from part (a) to compute the distance fromthe point P (5, 3, 0) to the line containing A(1, 0, 1) and B(2, 3, 1).Verify your answer with Quiz 2 Practice Problem #10(b).

7. Consider the parallelepiped with adjacent edges u = 〈1, 2, 3〉, v =〈3, 4, 0〉, and w = 〈–1, 3, –2〉.

(a) Compute the volume of the parallelepiped.

(b) Determine the area of the face determined by v and w.

(c) Compute the angle between u and the plane containing the facedetermined by v and w.

For problems 8-11, compute parametric equations of the line whichsatisfies the given conditions.

8. The line which passes through the point (1, 0, –1) and is parallel tov = 〈1, –2, 0〉.

9. The line which passes through points A(3, –6, 6) and B(2, 0, 7).

6

10. The line which passes through the point (–1, 2, 4) and is parallel to

L1 :

x = 3− 4ty = 3 + 2tz = t

.

11. The line which passes through the point (–2, 1, 4) and is parallel to boththe xy-plane and the xz-plane.

12. Is the line which passes through the points A1(1, 2, 3) and B1(5, 8, 9) par-allel to the line which passes through pointsA2(–2, 5, 3) andB2(4, 14, 12)?

13. Find the coordinates of the points at which the line L1 :

x = 3− 6ty = 3 + 3tz = t

intersects the given plane:

(a) The xy-plane,

(b) The xz-plane.

(c) The yz-plane.

14. Find the coordinates of the points in 3-space where the line L1 :x = ty = 1 + tz = 1− t

intersects the sphere x2 + y2 + z2 = 29.

For the problems 15-18, determine whether the given lines inter-sect, are parallel, or are skew. If the lines intersect, find their pointof intersection.

15. L1 : x = 2 + 3t, y = 1− 2t, z = 4 + 5t

L2 : x = 3− 6t, y = –2 + 4t, z = –1− 10t

16. L1 : x = 1, y = t, z = 2− tL2 : x = 2 + 3t, y = 4− 3t, z = t

17. L1 : x = 1− 2t, y = 14 + t, z = 5− tL2 : x = t, y = 4 + 3t, z = 3 + t

7

18. L1 : x = 2 + 5t, y = 4− t, z = t+ 1

L2 : x = 3 + 6t, y = 1− t, z = t

19. Verify that the following lines are parallel. Then compute the distancebetween them. [Hint: See Practice Problem #6 above or Quiz 2 PracticeProblem #10.]

L1 : x = 5 + 3t, y = 3 + 9t, z = 0

L2 : x = 1 + t, y = 3t, z = 1

20. Two bugs are walking along lines in 3-space. At time t, bug 1’s position

is the point (x, y, z) on the line L1 :

x = 1 + 2ty = 3 + 5tz = 5 + 2t

and bug 2’s position

is the point (x, y, z) on the line L2 :

x = ty = 11− tz = 4 + t

.

(a) Compute the distance between the bugs’ initial positions.

(b) At which point in space will the bugs’ paths intersect? (Note: thepaths may not intersect at the same moment in time.)

21. Consider the point P (5, 3, 0) and the line L which contains pointsA(1, 0, 1) and B(2, 3, 1). This problem will show you another way tofind the distance d between the point P and the line L.

(a) Compute an equation of line L.

(b) Find a function f(t) which gives the distance from the point P toan arbitrary point on the line.

8

(c) The distance from the point P to line L is the shortest distance.Calculate the value of t that minimizes the distance from the pointP to line L; that is, calculate the value of t which minimizes f(t)from part (b).

(d) Compute the distance from the point P (5, 3, 0) to line L by cal-culating the distance from this point P to the point on your linewhich corresponds to your value of t from part (c). Verify youranswer with Quiz 2 Practice #10(b).

22. For each of the following, find an equation of the plane indicated bythe figure.

9

For Problems 23-27, determine whether the following are parallel,perpendicular, or neither.

23. Plane P1 : 5x− 3y + 4z = –1 and plane P2 : 2x− 2y − 4z = 9.

24. Plane P1 : 3x− 2y + z = –3 and plane P2 : 5x+ y − 6z = 8.

25. Plane P1 : 3x− 2y + z = –3 and plane P2 : –6x+ 4y − 2z = 1.

26. Plane P1 : 5x− 3y + 4z = –1 and line−→` (t) = 〈2 + 2t, 3− 2t, 5− 4t〉.

27. Plane P1 : 5x− 3y + 4z = –1 and line−→` (t) =

⟨2 + 5

2t, 3− 3

2t, 5 + 2t

⟩.

28. Give an example of a plane P and a line L which are neither parallelnor perpendicular to each other.

For problems 29-35, find an equation of the plane which satisfiesthe given conditions.

29. The plane which passes through the point P (1, 2, 3) and which has anormal vector of n = 4i− 2j+ 6k.

30. The plane which passes through P (–2, 0, 1) and is perpendicular to the

line−→` (t) = 〈1, 2, 3〉+ t〈3, –2, 2〉.

31. The plane which passes through points A(1, 2, 3), B(2, –1, 5), andC(–1, 3, 3).

32. The plane which passes through A(1, 2, 3) and is parallel to the plane3x− 5y + z = 2.

33. The plane which passes through A(–2, 1, 5) and is perpendular to theline of intersection of P1 : 3x+ 2y − z = 5 and P2 : –y + z = 7.

34. The plane which contains the point A(–2, –1, 3) and which contains theline L : x = 1 + t, y = 3− 2t, z = 4t.

35. The plane that contains the line L : x = 3+t, y = 1−2t, z = 4+t, and isparallel to the line of intersection of P1 : y+z = 1 and P2 : 2x−y+z = 0.

10

36. Consider the planes P1 : x+ y + z = 7 and P2 : 2x+ 4z = 6.

(a) Compute an equation of the line of intersection of P1 and P2.

(b) Compute an equation of the plane which passes through the pointA(1, 2, 3) and contains the line of intersection of P1 and P2.

37. Find the acute angle of intersection of P1 : 3x − 2y + 5z = 0 andP2 : –x− y + 2z = 3.

38. Find the acute angle of intersection of P1 : 3x − 2y − 5z = 0 andP2 : –x− y + 2z = 3.

39. Consider the plane which passes through the point Q and whose normalvectors are parallel to n. Let P be another point in space, as illustratedbelow.

(a) Show that the distance between the point P and the given plane

is d =|QP •n|||n||

.

(b) Use this method to compute the distance between the pointP (2, –1, 4) and the plane x+ 2y + 3z = 5.

11

40. Consider the planes P1 : 2x− 4y + 5z = –2 and P2 : x− 2y +5

2z = 5.

(a) Verify that P1 and P2 are parallell.

(b) Compute the distance between P1 and P2. [Hint: See the previousproblem.]

? Let a, b, and c be vectors in 3-space.

(a) Prove or disprove that the cross product is associative.

(b) Prove the following formula for the vector triple product:

a× (b× c) = (a • c)b− (a •b)c.

(c) Prove that

a× (b× c) + b× (c× a) + c× (a× b) = 0.

? Does there exist a one-to-one correspondence f of

(a) 2-space with itself such that for any distinct points A, B, the lines←→AB

and←−−−−−→f(A)f(B) are perpendicular?

(b) 3-space with itself such that for any distinct points A, B, the lines←→AB

and←−−−−−→f(A)f(B) are perpendicular?

? Let P be a plane in space and let v be a vector. The vector projection of vonto the plane P, projPv, can be defined informally as follows. Suppose thesun is shining so that its rays are normal to the plane P. Then projPv is theshadow of v onto P. If P is the plane x+ 2y + 6z = 6 and v = i+ j+ k, findprojPv.

12

SOLUTIONS

1. (a)

u× v =

∣∣∣∣∣∣i j k3 –4 12 –2 3

∣∣∣∣∣∣=

∣∣∣∣ –4 1–2 3

∣∣∣∣ i− ∣∣∣∣ 3 12 3

∣∣∣∣ j+ ∣∣∣∣ 3 –42 –2

∣∣∣∣k= [(–4)(3)− (1)(–2)]i− [(3)(3)− (1)(2)]j+ [(3)(–2)− (–4)(2)]k

= [–12− (–2)]i− (9− 2)j+ [–6− (–8)]k = –10i− 7j+ 2k

Check:

(u× v) •u = (–10)(3) + (–7)(–4) + (2)(1) = –30 + 28 + 2 = 0,

(u× v) •v = (–10)(2) + (–7)(–2) + (2)(3) = –20 + 14 + 6 = 0.

Indeed u× v is orthogonal to both u and v.

(b)

u× v =

∣∣∣∣∣∣i j k2 –2 6–1 2 –1

∣∣∣∣∣∣=

∣∣∣∣ –2 62 –1

∣∣∣∣ i− ∣∣∣∣ 2 6–1 –1

∣∣∣∣ j+ ∣∣∣∣ 2 –2–1 2

∣∣∣∣k= [(–2)(–1)− (2)(6)]i− [(2)(–1)− (6)(–1)]j+ [(2)(2)− (–2)(–1)]k

= (2− 12)i− [–2− (–6)]j+ (4− 2)k = –10i− 4j+ 2k

Check:

(u× v) •u = (–10)(2) + (–4)(–2) + (2)(6) = –20 + 8 + 12 = 0,

13

(u× v) •v = (–10)(–1) + (–4)(2) + (2)(–1) = 10− 8− 2 = 0.

Indeed u× v is orthogonal to both u and v.

(c)

u× v =

∣∣∣∣∣∣i j k2 0 31 –1 0

∣∣∣∣∣∣=

∣∣∣∣ 0 3–1 0

∣∣∣∣ i− ∣∣∣∣ 2 31 0

∣∣∣∣ j+ ∣∣∣∣ 2 01 –1

∣∣∣∣k= [(0)(0)− (3)(–1)]i− [(2)(0)− (3)(1)]j+ [(2)(–1)− (0)(1)]k

= [0− (–3)]i− (0− 3)j+ (–2− 0)k = 3i+ 3j− 2k

Check:

(u× v) •u = (3)(2) + (3)(0) + (–2)(3) = 6 + 0− 6 = 0,

(u× v) •v = (3)(1) + (3)(–1) + (–2)(0) = –3− 3 + 0 = 0.

Indeed u× v is orthogonal to both u and v.

2. (a) Note that i− j = –(j− i) so that i− j is parallel to j− i by virtueof being a constant multiple of it. So (i− j)× (j− i) as the crossproduct of two parallel vectors is always 0. One can also deducethis by FOIL:

(i− j)× (j− i) = i× j+ i× (–i) + (j)× j+ (–j)× (–i)

= i× j− (i× i)− (j× j) + j× i

= i× j− 0− 0− (i× j) = 0

14

(b)

(i− j)× (j− i) =

∣∣∣∣∣∣i j k1 –1 0–1 1 0

∣∣∣∣∣∣=

∣∣∣∣ –1 01 0

∣∣∣∣ i− ∣∣∣∣ 1 0–1 0

∣∣∣∣ j+ ∣∣∣∣ 1 –1–1 1

∣∣∣∣k= [(–1)(0)− (0)(1)]i− [(1)(0)− (0)(–1)]j+ [(1)(–1)− (–1)(1)]k

= (0− 0)i− (0− 0)j+ [–1− (–1)]k

= 0i+ 0j+ 0k = 0

3. Vectors−→AB = 〈6−1, 4−2, 7−3〉 = 〈5, 2, 4〉 and

−→AC = 〈1−1, 5−2, 2−

3〉 = 〈0, 3, –1〉 lie in this plane, so a normal to the plane is provided bythe cross product:

−→AB ×

−→AC =

∣∣∣∣∣∣i j k5 2 40 3 –1

∣∣∣∣∣∣=

∣∣∣∣ 2 43 –1

∣∣∣∣ i− ∣∣∣∣ 5 40 –1

∣∣∣∣ j+ ∣∣∣∣ 5 20 3

∣∣∣∣k= [(2)(–1)− (4)(3)]i− [(5)(–1)− (4)(0)]j+ [(5)(3)− (2)(0)]k

= (–2− 12)i− (–5− 0)j+ (15− 0)k = –14i+ 5j+ 15k

To get two unit vectors normal to the plane, we take

±−→AB ×

−→AC∣∣∣∣∣∣−→AB ×−→AC∣∣∣∣∣∣ = ± 〈–14, 5, 15〉√

(–14)2 + 52 + 152= ± 〈–14, 5, 15〉√

196 + 25 + 225= ±

⟨–

14√446

,5√446

,15√446

⟩.

4. These are the same points from the previous problem. Recall that thearea of the triangle is half of the magnitude of the cross product, whichis 1

2

√446, based on our work above.

5. ||u× v|| = ||u|| ||v|| sin θu,v| = (2)(5) sin 30◦ = 10(1/2) = 5.

15

6. (a) d =∣∣∣∣∣∣−→AP ∣∣∣∣∣∣ sin∠BAP =

∣∣∣∣∣∣−→AP ∣∣∣∣∣∣ ∣∣∣∣∣∣−→AB∣∣∣∣∣∣ sin∠BAP∣∣∣∣∣∣−→AB∣∣∣∣∣∣ =

∣∣∣∣∣∣−→AP ×−→AB∣∣∣∣∣∣∣∣∣∣∣∣−→AB∣∣∣∣∣∣ .

(b) We see that−→AP = 〈1 − 5, 0 − 3, 1 − 0〉 = 〈–4, –3, 1〉,

−→AB =

〈2− 1, 3− 0, 1− 1〉 = 〈1, 3, 0〉,

−→AP ×

−→AB =

∣∣∣∣∣∣i j k–4 –3 11 3 0

∣∣∣∣∣∣=

∣∣∣∣ –3 13 0

∣∣∣∣ i− ∣∣∣∣ –4 11 0

∣∣∣∣ j+ ∣∣∣∣ –4 –31 3

∣∣∣∣k= [(–3)(0)− (1)(3)]i− [(–4)(0)− (1)(1)]j+ [(–4)(3)− (–3)(1)]k

= (0− 3)i− (0− 1)j+ [–12− (–3)]k = –3i+ j− 9k,

and

d =

∣∣∣∣∣∣−→AP ×−→AB∣∣∣∣∣∣∣∣∣∣∣∣−→AB∣∣∣∣∣∣ =

√(–3)2 + 12 + (–9)2√

12 + 32 + 02=

√9 + 1 + 81√1 + 9 + 0

=

√91

10.

7. (a) The volume of the parallelepiped is the absolute value of the scalartriple product

|u • (v ×w)| =

∣∣∣∣∣∣∣∣∣∣∣∣1 2 33 4 0–1 3 –2

∣∣∣∣∣∣∣∣∣∣∣∣

=

∣∣∣∣∣∣∣∣ 4 03 –2

∣∣∣∣ 1− ∣∣∣∣ 3 0–1 –2

∣∣∣∣ 2 + ∣∣∣∣ 3 4–1 3

∣∣∣∣ 3∣∣∣∣= |[(4)(–2)− (0)(3)]1− [(3)(–2)− (0)(–1)]2 + [(3)(3)− (4)(–1)]3|= |(–8− 0)− (–6− 0)2 + [9− (–4)]3|= |–8 + (6)2 + (13)3|= |–8 + 12 + 39|= |43| = 43

16

(b) The face determined by v and w is a parallelogram, and so thearea of that face is the length of the cross product v × w. Yet,remember that the scalar triple product u • (v ×w) is similar tothe cross product v × w in that the components of u replace i,j, and k in the calculations of them. Inspecting our computationabove, we thus have v ×w = –8i+ 6j+ 13k, which has length√

(–8)2 + 62 + 132 =√64 + 36 + 169 =

√269.

(c) Start by using the dot product to find the angle between u andv ×w :

θu,v×w = cos–1(

u • (v ×w)

||u|| ||v ×w||

)= cos–1

(43√

12 + 22 + 32√269

)= cos–1

(43

√1 + 4 + 9

√269

)= cos–1

(43√

14√269

)Since u • (v × w) > 0, the angle between u and v × w is acute,and so u and v × w lie on the same side of the plane. Yet, bydefinition, v ×w is normal (at a right angle) to the plane, and sou must lie between the plane and v×w. The desired angle is thus

π

2− cos–1

(43√

14√269

).

8. This is simply r(t) = 〈1, 0, –1〉+〈1, –2, 0〉t, which is the same as x = 1+t,y = –2t, z = –1.

9. Taking A to be the initial point, the direction of the line is given by the

vector−→AB = 〈2− 3, 0− (–6), 7− 6〉 = 〈–1, 6, 1〉, and so the equation of

the line is r(t) = 〈3, –6, 6〉+ 〈–1, 6, 1〉t, which is the same as x = 3− t,y = –6 + 6t, z = 6 + t.

10. Remember that the coefficients of the parameter t in the parametricequation of the line is what gives you the components of a vector

17

parallel to that line. So for L1, the vector 〈–4, 2, 1〉 is parallel to theline and thus also to the line that we are looking for. Using the factthat the line we want passes through (–1, 2, 4), we get the equationr(t) = 〈–1, 2, 4〉+〈–4, 2, 1〉t, which is the same as x = –1−4t, y = 2+2t,and z = 4 + t.

11. Recall that the xy-plane has the equation z = 0 and the xz-plane hasthe equation y = 0. Thus a vector parallel to both the xy-plane andthe xz-plane is i = 〈1, 0, 0〉. (Any vector of the form 〈k, 0, 0〉 where kis a real number suffices as the x-component is free to be whatever wewant.) So the equation of the line is r(t) = 〈–2, 1, 4〉+ 〈1, 0, 0〉t, whichis the same as x = –2 + t, y = 1, z = 4.

12. The line which passes through A1 and B1 is parallel to the vector−−−→A1B1 = 〈5−1, 8−2, 9−3〉 = 〈4, 6, 6〉 while the line which passes through

A2 and B2 is parallel to the vector−−−→A2B2 = 〈4− (–2), 14− 5, 12− 3〉 =

〈6, 9, 9〉 = 32〈4, 6, 6〉. Since

−−−→A1B1 and

−−−→A2B2 are scalar multiples of each

other, it follows that these two vectors, and hence the two lines, areparallel.

13. (a) We need z = 0. So t = 0. Plugging t = 0 into the parametricequations for x and y yield x = 3− 6(0) = 3 and y = 3+ 3(0) = 3.Hence L intersects the xz-plane at the point (3, 3, 0).

(b) We need y = 0. So 3 + 3t = 0 ⇒ t = –1. Plugging t = –1 intothe parametric equations for x and z yield x = 3− 6(–1) = 9 andz = –1. Hence L intersects the xz-plane at the point (9, 0, –1).

(c) We need x = 0. So 3−6t = 0⇒ t = 1/2. Plugging t = 1/2 into theparametric equations for y and z yield y = 3 + 3(1/2) = 9/2 andz = 1/2. Hence L intersects the xz-plane at the point (0, 9/2, 1/2).

14. Substitute the parametric equations of L into the equation of the sphere:

t2 + (1 + t)2 + (1− t)2 = 29⇒ t2 + (1 + 2t+ t2) + (1− 2t+ t2) = 29

⇒ 3t2 + 2 = 29⇒ 3t2 = 27

⇒ t2 = 9⇒ t = ±3.

18

This yields two points of intersection x = 3, y = 1+3 = 4, z = 1−3 = –2for the point (3, 4, –2) and x = –3, y = 1− 3 = –2, z = 1− (–3) = 4 forthe point (–3, –2, 4).

15. A vector parallel to L1 is v1 = 〈3, –2, 5〉 and a vector parallel to L2 isv2 = 〈–6, 4, –10〉 = –2〈3, –2, 5〉. Since v1 and v2 are scalar multiples ofeach other, it follows that v1 and v2, and hence L1 and L2 are parallel.

16. A vector parallel to L1 is v1 = 〈0, 1, –1〉 and a vector parallel to L2 isv2 = 〈3, –3, 1〉, which is clearly not a scalar multiple of v1. (Note thatyou would have to multiply v2 by 0 in order for the first componentsto match, but neither v1 nor v2 is 0.)

However, if L1 and L2 do intersect, then because the x-coordinate ofall points on L1 is a constant 1, we need the x-coordinate of the pointof intersection on L2 to also be 1. So 2 + 3t0 = 1 ⇒ t0 = –1/3 sothat y = 4 − 3(–1/3) = 5 and z = –1/3. In other words, the pointof intersection must be (1, 5, –1/3). In order for the y-coordinates tomatch, we thus need t = 5. Yet z = 2− 5 = –3 6= –1/3, a contradiction.So the two lines do not intersect; L1 and L2 are skew.

17. A vector parallel to L1 is v1 = 〈–2, 1, –1〉 and a vector parallel to L2 isv2 = 〈1, 3, 1〉, which is clearly not a scalar multiple of v1. (Note thatyou would have to multiply v2 by –2 in order for the first componentsto match, but 1, the y-component of v1, is not negative two times 3,the y-component of v2.)

Now suppose that L1 and L2 intersect. Then there are parameters tand s such that plugging t into the parametric equations for L1 andplugging s into the parametric equations for L2 give the same points.This would yield the following system of equations:

1− 2t = s

14 + t = 4 + 3s

5− t = 3 + s

19

Let us consider the first and third equations only for now, as they arethe two simplest ones. If 1− 2t = s then substitution into the equation5 − t = 3 + s yields 5 − t = 3 + 1 − 2t ⇒ t = 4 − 5 = –1. Hences = 1− 2(–1) = 3. We now need to verify this solution with the secondequation. We have 14+ t = 14− 1 = 13 whereas 4+3s = 4+3(3) = 13.Since they match, we see that the lines do in fact intersect. The pointof intersection can be found by plugging either parameter into theirrespective parametric equations. If we use the equation for L2 we findthat x = s = 3, y = 4 + 3s = 13, and z = 3 + s = 3 + 3 = 6. ThereforeL1 and L2 intersect at the point (3, 13, 6).

18. A vector parallel to L1 is v1 = 〈5, –1, 1〉 and a vector parallel to L2 isv2 = 〈6, –, 1〉, which is clearly not a scalar multiple of v1. (Note that y-and z-components of v1 and v2 match while their x-components differ.)

Now suppose that L1 and L2 intersect. Then there are parameters tand s such that plugging t into the parametric equations for L1 andplugging s into the parametric equations for L2 give the same points.This would yield the following system of equations:

2 + 5t = 3 + 6s

4− t = 1− st+ 1 = s

Let us consider the second and third equations only for now, as they arethe two simplest ones. If t+ 1 = s then substitution into the equation4− t = 1− s yields 4− t = 1− t− 1⇒ 4 = 0, which is absurd. HenceL1 and L2 cannot intersect anywhere; L1 and L2 are skew.

19. This is an exercise left to the reader. The lines are parallel because〈3, 9, 0〉 = 3〈1, 3, 0〉. Pick a point from both L1 and L2 and then a thirdpoint however you please. (For simplicity, I recommend sticking topoints where the parameter is either 0 or 1.) Then use either one of theformulas derived in the problems mentioned in the hint. In any case,you should get d =

√91/10.

20

20. (a) The bugs’ initial positions are attained by plugging t = 0 into theequations for L1 and L2. So bug 1’s initial position is (1, 3, 5) andbug 2’s initial position is (0, 11, 4). The distance between them is√

(0− 1)2 + (11− 3)2 + (4− 5)2 =√(–1)2 + 82 + (–1)2 =

√1 + 64 + 1 =√

66.

(b) Letting t1 be bug 1’s time and t2 be bug 2’s time, we want

1 + 2t1 = t2

3 + 5t1 = 11− t25 + 2t1 = 4 + t2

Substituting the first equation into the second one gives 3 + 5t1 =11 − (1 + 2t1) ⇒ 3 + 5t1 = 10 − 2t1 ⇒ 7t1 = 7 ⇒ t1 = 1.Hence 1 + 2(1) = t2 ⇒ t2 = 3. We check these with the thirdequation just to make sure that their paths do in fact intersect:5 + 2(1) = 7 = 4 + 3, which is what we want. So the point ofintersection is (3, 11− 3, 4 + 3) = (3, 8, 7).

21. (a) Taking A to be the initial point, the direction of the line is given

by the vector−→AB = 〈2 − 1, 3 − 0, 1 − 1〉 = 〈1, 3, 0〉, and so the

equation of the line is r(t) = 〈1, 0, 1〉+ 〈1, 3, 0〉t, which is the sameas x = 1 + t, y = 3t, z = 1.

(b) For any given t, the distance between the point the point P andthe point (1 + t, 3t, 1) on the line is

f(t) =√(1 + t− 5)2 + (3t− 3)2 + (1− 0)2 =

√(t− 4)2 + (3t− 3)2 + 1.

(c) This is a calc 1 problem. The value of t that minimizes f(t) alsominimizes [f(t)]2. We will work with the latter quantity insteadto eliminate the square root and make differentiation easier. So ifg(t) = (t− 4)2+(3t− 3)2+1, then g′(t) = 2(t− 4)+2(3t− 3) · 3 =2(t− 4) + 6(3t− 3). The minimum occurs at a critical point, sowe set the derivative equal to 0. Thus

21

2(t−4)+6(3t−3) = 0⇒ t−4+3(3t−3) = 0⇒ t+9t−9 = 4⇒ 10t = 13⇒ t =13

10,

is the only critical point of g. Just to ensure that this is in fact aminimum, we calculate the second derivative g′′(t) = 2+6·3 > 0 forall t, which implies that g has an absolute minimum at t = 13/10,as desired.

(d) Based on the last three parts, the distance from P to L is

f

(13

10

)=

√(13

10− 4

)2

+

(313

10− 3

)2

+ 1

=

√(–27

10

)2

+

(39

10− 30

10

)2

+ 1

=

√729

100+

(9

10

)2

+100

100

=

√829

100+

81

100

=

√910

100=

√91

10,

an answer that is very familiar to us by now.

22. (a) Note that the points A(2, 0, 0), B(0, 3, 0), and C(0, 0, 4) lie on the

plane, and so the plane contains the vectors−→AB = 〈0 − 2, 3 −

0, 0 − 0〉 = 〈–2, 3, 0〉 and−→AC = 〈0 − 2, 0 − 0, 4 − 0〉 = 〈–2, 0, 4〉.

Therefore, a normal perpendicular to the plane is given by

22

−→AB ×

−→AC =

∣∣∣∣∣∣i j k–2 3 0–2 0 4

∣∣∣∣∣∣=

∣∣∣∣ 3 00 4

∣∣∣∣ i− ∣∣∣∣ –2 0–2 4

∣∣∣∣ j+ ∣∣∣∣ –2 3–2 0

∣∣∣∣k= [(3)(4)− (0)(0)]i− [(–2)(4)− (0)(–2)]j+ [(–2)(0)− (3)(–2)]k

= (12− 0)i− (–8− 0)j+ [0− (–6)]k = 12i+ 8j+ 6k,

and so the plane’s equation is 12(x−2)+8(y−0)+6(z−0) = 0⇒12x− 24+8y+6z = 0⇒ 12x+8y+6z = 24⇒ 6x+4y+3z = 12.

(b) Note that the plane goes straight up and down in the z-directionso that the plane is also a cylindrical surface. It thus suffices tofind the equation of the line that is the intersection of the planewith the xy-plane. This line goes through the points (2, 0) and(0, 3) in the xy-plane. The equation of this line, and thus also ourplane, is

y − 0 =3− 0

0− 2(x− 2)⇒ y = –

3

2(x− 2)⇒ y = –

3

2x+ 3.

This equation can be put in general form as follows by multiplyingthe equation by 2 and adding the x to both sides to get 3x+2y = 6.

23. (Perpendicular) Vectors normal to P1 and P2 are n1 = 〈5, –3, 4〉 andn2 = 〈2, –2, –4〉, respectively. Note that

n1, •n2 = (5)(2) + (–3)(–2) + (4)(–4) = 10 + 6− 16 = 0

so that n1 and n2 are orthogonal. Consequently P1 and P2 are perpen-dicular.

24. (Neither) Vectors normal to P1 and P2 are n1 = 〈3, –2, 1〉 and n2 =〈5, 1, –6〉, respectively. We see that n1 and n2 are not parallel. Indeed,

23

the y-component of n1 is –2 times the y-component of n2, but the othertwo components do not satisfy this. Furthermore,

n1, •n2 = (3)(5) + (–2)(1) + (1)(–6) = 15− 2− 6 = 7 6= 0

so that n1 and n2 are not orthogonal. Consequently P1 and P2 are notperpendicular.

25. (Parallel) Vectors normal to P1 and P2 are n1 = 〈3, –2, 1〉 and n2 =〈–6, 4, –2〉 = –2〈3, –2, 1〉, respectively. Since n1 and n2 are scalar multi-ples of each other, n1 and n2, and consequently P1 and P2, are parallel.

26. (Parallel) The vector n = 〈5, –3, 4〉 is normal to P1, and the vectorv = 〈2, –2, –4〉 is parallel to `. We found these vectors to be orthogonalin problem 2. Since any vector in P1 is orthogonal to n, it follows thatv, and consequently `, is parallel to P1.

27. (Perpendicular) The vector n = 〈5, –3, 4〉 is normal to P1, and thevector v = 〈5/2, –3/2, 2〉 is parallel to `. Note that v = 1

2n so that v

and n are parallel. Hence v, like n is normal to P1, and so P1 and ` areperpendicular.

28. Suppose the equation of L has the form−→` (t) =

−→` 0 + t−→v and that P

has −→n as a normal vector. Then all possible answers are those for which−→v 6‖ −→n (i.e., −→v 6= c−→n for some scalar c) and −→v 6⊥ −→n (i.e., −→v •−→n 6= 0).The first condition ensures that L and P are not perpendicular. Thesecond conditions ensures that L and P are not parallel. We foundexamples of such vectors in problem 3. Let −→v = 〈3, –2, 1〉 and −→n =

〈5, 1, –6.〉. Since−→`0 is free to be anything, we will set it equal to

−→0 for

space conservation. So let L have equation−→` (t) = t〈3, –2, 1〉 and P

have equation 5x+ y − 6z = 0.

29. Using the point-normal form for the equation of a plane, we have thatits equation is 4(x− 1)− 2(y − 2) + 6(z − 3) = 0.

30. Since the plane and the line are perpendicular, the vector 〈3, –2, 2〉,which is parallel to

−→` , is normal to the plane. The plane’s equation is

thus 3(x+ 2)− 2y + 2(z − 1) = 0.

24

31. The plane contains the vectors−→AB = 〈2−1, –1−2, 5−3〉 = 〈1, –3, 2〉 and

−→AC = 〈–1−1, 3−2, 3−3〉 = 〈–2, 1, 0〉. Therefore, a normal perpendicularto the plane is given by

−→AB ×

−→AC =

∣∣∣∣∣∣i j k1 –3 2–2 1 0

∣∣∣∣∣∣=

∣∣∣∣ –3 21 0

∣∣∣∣ i− ∣∣∣∣ 1 2–2 0

∣∣∣∣ j+ ∣∣∣∣ 1 –3–2 1

∣∣∣∣k= [(–3)(0)− (2)(1)]i− [(1)(0)− (2)(–2)]j+ [(1)(1)− (–3)(–2)]k

= (0− 2)i− [0− (–4)]j+ (1− 6)k = –2i− 4j− 5k,

and so the plane’s equation is –2(x− 1)− 4(y − 2)− 5(z − 3) = 0.

32. Parallel planes have the same normal vectors, and so our plane’s equationis 3(x− 1)− 5(y − 2) + (z − 3) = 0.

33. Vectors normal to P1 and P2 are n1 = 〈3, 2, –1〉 and n2 = 〈0, –1, 1〉,respectively. It follows that any vector parallel to the line of intersectionof P1 and P2 is orthogonal to both n1 and n2. Such a vector can byfound via the cross product:

n1 × n2 =

∣∣∣∣∣∣i j k3 2 –10 –1 1

∣∣∣∣∣∣=

∣∣∣∣ 2 –1–1 1

∣∣∣∣ i− ∣∣∣∣ 3 –10 1

∣∣∣∣ j+ ∣∣∣∣ 3 20 –1

∣∣∣∣k= [(2)(1)− (–1)(–1)]i− [(3)(1)− (–1)(0)]j+ [(3)(–1)− (2)(0)]k

= (2− 1)i− (3− 0)j+ (–3− 0)k = i− 3j− 3k,

Since the line of intersection is perpendicular to our plane, n1 × n2 isnormal to it. The plane’s equation is therefore (x + 2) − 3(y − 1) −3(z − 5) = 0.

25

34. Since this plane contains L, it goes through any two points on it.In particular, if we let t = 0, then we get the point B(1 + 0, 3 −2(0), 4(0)) = B(1, 3, 0), and if we let t = 1, then we get the pointC(1 + 1, 3− 2(1), 4(1)) = C(2, 1, 4). This is now similar to problem 10.

Our plane contains the vectors−→AB = 〈1−(–2), 3−(–1), 0−3〉 = 〈3, 4, –3〉

and−→AC = 〈2 − (–2), 1 − (–1), 4 − 3〉 = 〈4, 2, 1〉. Therefore, a normal

perpendicular to the plane is given by

−→AB ×

−→AC =

∣∣∣∣∣∣i j k3 4 –34 2 1

∣∣∣∣∣∣=

∣∣∣∣ 4 –32 1

∣∣∣∣ i− ∣∣∣∣ 3 –34 1

∣∣∣∣ j+ ∣∣∣∣ 3 44 2

∣∣∣∣k= [(4)(1)− (–3)(2)]i− [(3)(1)− (–3)(4)]j+ [(3)(2)− (4)(4)]k

= [4− (–6)]i− [3− (–12)]j+ (6− 16)k = 10i− 15j− 10k,

and so the plane’s equation is 10(x+ 2)− 15(y + 1)− 10(z − 3) = 0,which can be simplified by dividing both sides of the equation by 5:2(x+ 2)− 3(y + 1)− 2(z − 3) = 0.

35. Since L lies on our plane, all points on L are also points on our plane.In particular, if we let t = 0, then we get the point (3, 1, 4). Also, sinceL is parallel to the vector v = 〈1, –2, 1〉, v is parallel to our plane.

Next, vectors normal to P1 and P2 are n1 = 〈0, 1, 1〉 and n2 = 〈2, –1, 1〉,respectively. It follows that any vector parallel to the line of intersectionof P1 and P2 is orthogonal to both n1 and n2. Such a vector can byfound via the cross product:

26

n1 × n2 =

∣∣∣∣∣∣i j k0 1 12 –1 1

∣∣∣∣∣∣=

∣∣∣∣ 1 1–1 1

∣∣∣∣ i− ∣∣∣∣ 0 12 1

∣∣∣∣ j+ ∣∣∣∣ 0 12 –1

∣∣∣∣k= [(1)(1)− (1)(–1)]i− [(0)(1)− (1)(2)]j+ [(0)(–1)− (1)(2)]k

= [1− (–1)]i− (0− 2)j+ (0− 2)k = 2i+ 2j− 2k.

To make calculations easier, divide n1 × n2 by 2 to get the vector〈1, 1, –1, which is parallel to the intersection of P1 and P2 and thereforeparallel to our plane. Hence, the vector n orthogonal to our plane mustalso be orthogonal to 〈1, –2, 1〉 and 〈1, 1, –1〉. We attain n by againtaking the cross product:

v ×(1

2n1 × n2

)=

∣∣∣∣∣∣i j k1 –2 11 1 –1

∣∣∣∣∣∣=

∣∣∣∣ –2 11 –1

∣∣∣∣ i− ∣∣∣∣ 0 12 1

∣∣∣∣ j+ ∣∣∣∣ 1 –21 1

∣∣∣∣k= [(–2)(–1)− (1)(1)]i− [(0)(1)− (1)(2)]j+ [(1)(1)− (–2)(1)]k

= (2− 1)i− (0− 2)j+ [1− (–2)]k = i+ 2j+ 3k.

Since our plane goes through the point (3, 1, 4) and has normal 〈1, 2, 3〉,it follows that our plane’s equation is (x− 3) + 2(y − 1) + 3(z − 4) = 0.

36. Let us first simplify the equation of P2 by dividing it by 2 to getx+ 2z = 3.

(a) Note that the point (1, 5, 1) lies on both P1 and P2. Vectors normalto P1 and P2 are n1 = 〈1, 1, 1〉 and n2 = 〈0, 1, 2〉, respectively. Itfollows that any vector parallel to the line of intersection of P1 andP2 is orthogonal to both n1 and n2. Such a vector can by foundvia the cross product:

27

n1 × n2 =

∣∣∣∣∣∣i j k1 1 11 0 2

∣∣∣∣∣∣=

∣∣∣∣ 1 10 2

∣∣∣∣ i− ∣∣∣∣ 1 11 2

∣∣∣∣ j+ ∣∣∣∣ 1 11 0

∣∣∣∣k= [(1)(2)− (1)(0)]i− [(1)(2)− (1)(1)]j+ [(1)(0)− (1)(1)]k

= (2− 0)i− (2− 1)j+ (0− 1)k = 2i− j− k,

So the line of intersection goes through the point (1, 5, 1) and is

parallel to n1 × n2 and thus has parametric equations−→` (t) =

〈1, 5, 1〉+ t〈2, –1, –1〉, that is, x = 1 + 2t, y = 5− t, z = 1− t.(b) This is similar to problem 13 as we are given that the plane

passes through A(1, 2, 3) and contains the line L : x = 1 + 2t,y = 5− t, z = 1− t. Since this plane contains L, it goes throughany two points on it. In particular, if we let t = 0, then we getthe point B(1 + 0, 5− 2(0), 1 + 0) = B(1, 5, 1), and if we let t = 1,then we get the point C(1 + 2(1), 5 − 1, 1 − 1) = C(3, 4, 0). Our

plane contains the vectors−→AB = 〈1− 1, 5− 2, 1− 3〉 = 〈0, 3, –2〉

and−→AC = 〈3 − 1, 4 − 2, 0 − 3〉 = 〈2, 2, –3〉. Therefore, a normal

perpendicular to the plane is given by

−→AB ×

−→AC =

∣∣∣∣∣∣i j k0 3 –22 2 –3

∣∣∣∣∣∣=

∣∣∣∣ 3 –22 –3

∣∣∣∣ i− ∣∣∣∣ 0 –22 –3

∣∣∣∣ j+ ∣∣∣∣ 0 32 2

∣∣∣∣k= [(3)(–3)− (–2)(2)]i− [(0)(–3)− (–2)(2)]j+ [(0)(2)− (3)(2)]k

= [–9− (–14)]i− [0− (–4)]j+ (0− 6)k = 5i− 4j− 6k,

and so the plane’s equation is 5(x− 1)− 4(y − 2)− 6(z − 3) = 0.

28

37. Intuitively, the acute angle of intersection of P1 and P2 matches theacute angle of intersection of their normals, which are n1 = 〈3, –2, 5〉and n2 = 〈–1, –1, 2〉. Note that

cos θn1,n2 =n1 •n2

||n1|| ||n2||

=(3)(–1) + (–2)(–1) + (5)(2)√

32 + (–2)2 + 52√

(–1)2 + (–1)2 + 22

=–3 + 2 + 10√

9 + 4 + 25√1 + 1 + 4

=9√38√6> 0

so that θn1,n2 is acute. Therefore, the angle between P1 and P2 is

cos–1(

9√38√6

).

38. The acute angle of intersection of P1 and P2 matches the acute angleof intersection of their normals, which are n1 = 〈3, –2, –5〉 and n2 =〈–1, –1, 2〉. Note that

cos θn1,n2 =n1 •n2

||n1|| ||n2||

=(3)(–1) + (–2)(–1) + (–5)(2)√

32 + (–2)2 + (–5)2√

(–1)2 + (–1)2 + 22

=–3 + 2− 10

√9 + 4 + 25

√16

=–11√38√6< 0

so that θn1,n2 is obtuse. Therefore, the angle between P1 and P2 is

π − cos–1(

–11√38√6

)= cos–1

(11√38√6

).

39. (a) Note that d = ||projnQP|| = |QP •n|||n||

.

29

(b) Note that Q(0, 1, 1) is a point on the plane, which has a normalvector of n = 〈1, 2, 3〉. HenceQP = 〈2−0, –1−1, 4−1〉 = 〈2, –2, 3〉,and so

d =|QP •n|||n||

=|(2)(1) + (–2)(2) + (3)(3)|√

12 + 22 + 32=|2− 4 + 9|√1 + 4 + 9

=7√14.

40. (a) Normal vectors of P1 and P2 are n1 = 〈2, –4, 5〉 and n2 =⟨1, –2, 5

2

⟩=

12〈2, –4, 5〉, respectively. Since n1 and n2 are scalar multiples of

each other, it follows that n1 and n2, and consequently P1 and P2

are parallel.

(b) Consider Q(–1, 0, 0), a point on P1, and P (0, 0, 2), a point on P2.Then QP = 〈0 − (–1), 0 − 0, 2 − 0〉 = 〈1, 0, 2〉. Intuitively, thedistance between P1 and P2 is the same as the distance betweenP and P1. By our work in the previous problem,

d =|QP •n1|||n1||

=|(1)(2) + (0)(–4) + (2)(5)|√

22 + (–4)2 + 52=|2 + 0 + 10|√4 + 16 + 25

=12√45.

30