homework1–solution …faculty.uaeu.ac.ae/jaelee/class/1120/10fa_hws.pdfcalculusiiforengineering...

United Arab Emirates UniversityCollege of Sciences

Department of Mathematical Sciences

HOMEWORK 1 – SOLUTION

Section 10.1 Vectors in the Plane

Calculus II for EngineeringMATH 1120 SECTION 04 CRN 235102:00 – 4:00 on Monday&WednesdayDue Date: Wednesday, October 6, 2010

Calculus II for Engineering HOMEWORK 1 – SOLUTION Fall, 2010

1. Compute�3a+ 2b for a = h 3; 2 i and b = h 3;�13 i.

Answer.

�3a+ 2b = �3 h 3; 2 i+ 2 h 3;�13 i = h �9;�6 i+ h 6;�26 i

= h �9 + 6;�6� 26 i = h �3;�32 i : �

2. Determine whether the vectors a = h 1;�2 i and b = h 2; 1 i are parallel.

Answer. We recall that a and b are parallel if and only if there is a scalar s such that a = sb. If a = sb,then

h 1;�2 i = s h 2; 1 i = h 2s; s i ; 1 = 2s and � 2 = s:

There is no such a scalar s satisfying both 2s = 1 and s = �2. Therefore, given two vectors cannot beparallel. �

3. Find the vector with initial pointA(2; 3) and terminal pointB(5; 4).

Answer. �!AB = h 5� 2; 4� 3 i = h 3; 1 i : �

4. Find a vector with the magnitude 4 in the same direction as the vector v = 2i� j.

Answer. Let u be such a vector, i.e., the one with the magnitude 4 in the same direction as the vectorv = 2i� j. Since u is parallel to v, there should be a positive scalar s (negative scalar gives the oppositedirection) such that u = sv. Since u has the magnitude 4, we have

4 = kuk = ksvk = jsjkvk = jsjk2i� jk = jsjp5;

i:e:; jsj = 4p5=

4p5

5; i:e:; s =

4p5

5:

Therefore, we deduce such a vector u =4p5

5(2i� j) : �

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5. The thrust of an airplane’s engines produces a speed of 600 mph in still air. The wind velocity is givenby h �30; 60 i. In what direction should the airplane head to fly due east?

Answer. Let v = h x; y i and w = h �30; 60 i be the velocities of the airplane and the wind, respec-tively. Since we want the airplane to move due east, the sum of two vectors v and w should satisfyv +w = h c; 0 i, where the vector h c; 0 i points due east with the positive constant c. The equation im-plies

h c; 0 i = v +w = h x; y i+ h �30; 60 i = h x� 30; y + 60 i ;i:e:; c = x� 30; and y = �60:

The airplane produces a speed of 600 mph and so we get kvk = 600, which implies

600 = kvk = h x; y i =qx2 + y2; 6002 = x2 + y2 = x2 + (�60)2;

x = �p6002 � 602 = �180

p11:

Because of the condition that x� 30 = c > 0, we choose x = +180p11, hence,

v = h x; y i =D180

p11;�60

E= 60

D3p11;�1

E:

This points right and down at an angle of tan�1 �13p11

!� �5:73917� � �0:100167 (radian) south of

east. �

6. Suppose a vector a has magnitude kak = 3 and vector b has magnitude kbk = 4.

(6.1) What is the largest possible magnitude for the vector a+ b?

Answer. The largest magnitude of a+ b is 7 (if the vectors point in the same direction). �

(6.2) What is the smallest possible magnitude for the vector a+ b?

Answer. The smallest magnitude is 1 (if the vectors point in the opposite directions). �

(6.3) What will be the magnitude of a+ b if a and b are perpendicular?

Answer. If the vectors are perpendicular, then a + b can be viewed as the hypotenuse of a righttriangle with sides a and b, so it has length

p32 + 42 = 5. �

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7. A vector 4i� 2j is given. Here i = h 1; 0 i and j = h 0; 1 i.(7.1) Find two unit vectors parallel to the given vector.

Answer. Let v = 4i� 2j = 2 (2i� j). Then it has the magnitude

kvk = k2 (2i� j) k = 2k (2i� j) k = 2q22 + (�1)2 = 2

p5:

The unit vector in the same direction as v can be found by

v

kvk =2 (2i� j)

2p5

=1p5(2i� j) :

The unit vector in the opposite direction as v is

� v

kvk = � 1p5(2i� j) :

Thus, two unit vectors parallel to v are� 1p5(2i� j). �

(7.2) Write the given vector as the product of its magnitude and a unit vector.

Answer. Using the results above, we have

v = kvk v

kvk = 2p5

2p5i� 1p

5j

!: �

8. If v 2 V2 lies in the first quadrant of the xy–plane andmakes the angle � = �=3with the positive x–axisand the magnitude 4, then find v in the component form.

Answer. Theunit vectormaking the angle � = �=3with thepositivex–axis is givenby h cos(�=3); sin(�=3) i.(Why?) Hence, the desired vector v 2 V2 is obtained by

v = 4 h cos(�=3); sin(�=3) i = 4

*1

2;

p3

2

+=D2; 2

p3E: �

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9. (Think!) Draw the vectors a = h 3; 2 i, b = h 2;�1 i and c = h 7; 1 i. By using the sketch involvedwith those vectors, show that there exist scalars s and t such that c = sa + tb. Can you prove theexistence of such s and t algebraically? Justify your answer.

Answer. The equationc = sa+tb implies h 7; 1 i = s h 3; 2 i+ t h 2;�1 i, i.e., 7 = 3s+2t and1 = 2s�t.Solving the equations for s and t, we get s = 9

7and t = 11

7. �

10. Find the correct figure of the suma+bwitha = �2i�j and b = �3i+2j, where i and j are standardbasis vectors of V2.

Answer. A simple computation shows

a+ b = �2i� j � 3i+ 2j = �5i+ j = h �5; 1 i :

That is, a + b should point in the direction h �5; 1 i. It is easy to see that the red–colored vector in (2)is the vector h �5; 1 i. Hence, the answer is (2). �

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

H 1 L

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

H 2 L

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

H 3 L

-5 -4 -3 -2 -1 1 2 3 4 5

-5

-4

-3

-2

-1

1

2

3

4

5

H 4 L

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United Arab Emirates UniversityCollege of Sciences



Section 10.2 Vectors in the Space



1. Find two unit vectors parallel to the vector 4i� 2j + 4k.

Answer. Let v = 4i� 2j + 4k = 2 (2i� j + 2k). Then it has the magnitude

kvk = k2 (2i� j + 2k) k = 2k2i� j + 2kk = 2q22 + (�1)2 + 22 = 2

p9 = 6:

The unit vector in the same direction as v can be found by

v

kvk =2 (2i� j + 2k)

6=

1

3(2i� j + 2k) :

The unit vector in the opposite direction as v is

� v

kvk = �1

3(2i� j + 2k) :

Thus, two unit vectors parallel to v are�1

3(2i� j + 2k). �

2. Identify the plane y = 4 as parallel to the xy–plane, xz–plane or yz–plane and sketch a graph.

Answer. The plane y = 4 is parallel to the xz–plane and passes through (0; 4; 0). �

3. Find the displacement vectors��!PQ and�!QR and determinewhether the pointsP (2; 3; 1),Q(0; 4; 2) andR(4; 1; 4) are colinear (on the same line).

Answer. The displacement vectors are

��!PQ = h 0� 2; 4� 3; 2� 1 i = h �2; 1; 1 i ; �!

QR = h 4� 0; 1� 4; 4� 2 i = h 4;�3; 2 i :

There does not exist any scalar s such that��!PQ = s�!QR, because no scalar s satisfies simultaneously

�2 = 4s; 1 = �3s; 1 = 2s:

It implies that two vectors��!PQ and�!QR are not parallel. That is, the pointsP ,Q andR are not colinear.�

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4. Use vectors to determine whether the points (2; 1; 0), (5;�1; 2), (0; 3; 3) and (3; 1; 5) form a square.

Answer. We recall that if a square has the side length s, then it has the diagonal of length sp2 by the

Pythagorean Theorem.

Let P = (2; 1; 0), Q = (5;�1; 2), R = (0; 3; 3) and S = (3; 1; 5). There are six pairs of vectors (��!PQ,�!PR, �!PS, �!QR, �!QS, �!RS) and four of them will correspond to sides. Further ��!PQ, �!PR and �!PS shouldform two sides and one diagonal of the square, because they have the same initial point. When wecompute the lengths, we get

k��!PQk = k h 3;�2; 2 i k =p17; k�!PRk = k h �2; 2; 3 i k =

p17;

k�!PSk = k h 1; 0; 5 i k =p26:

It implies �!PS should be the diagonal. However, sincep26 6= p

17p2, i.e., k�!PSk 6= k��!PQkp2, so �!PS

cannot be the diagonal of a square. That is, those given points cannot form a square. �

5. In the accompanying figure, two ropes are attached to a 300–pound crate. Rope A exerts a force ofh 10;�130; 200 i pounds on the crate, and rope B exerts a force of h �20; 180; 160 i pounds on thecrate.

(5.1) If no further ropes are added, find the net force on the crate and the direction it will move.

Answer. Let the force due to rope A be a = h 10;�130; 200 i, the force due to rope B be b =

h �20; 180; 160 i, and write the force due to gravity asw = h 0; 0;�300 i. Then the net force is

a+ b+w = h 10;�130; 200 i+ h �20; 180; 160 i+ h 0; 0;�300 i = h �10; 50; 60 i : �

(5.2) If a third rope C is added to balance the crate, what force must this rope exert on the crate?

Answer. In order to compensate, rope C must exert a force of h 10;�50;�60 i or78.74 (= k h 10;�50;�60 i k) pounds in direction h 1;�5;�6 i. �

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(5.3) We want to move the crate up and to the right with a constant force of h 0; 30; 20 i pounds. If athird rope C is added to accomplish this, what force must the rope exert on the crate?

Answer. Let the force due to rope C be c. We want the net force to be

a+ b+ c+w = h 0; 30; 20 i ; i:e:; c+ h �10; 50; 60 i = h 0; 30; 20 i ;

where a, b andw defined in (1) above are used. That is,

c = h 0; 30; 20 i � h �10; 50; 60 i = h 10;�20;�40 i :

So rope C must exert a force of h 10;�20;�40 i or 45.8 (= k h 10;�20;�40 i k) pounds in direc-tion h 1;�2;�4 i. �

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United Arab Emirates UniversityFaculty of Science



Section 10.3 Dot Product and Section 10.4 Cross Product



Section 10.3 Dot Product

1. Compute a � b.(1.1) a = h 3; 2; 0 i and b = h �2; 4; 3 i.

Answer.a � b = h 3; 2; 0 i � h �2; 4; 3 i = 3(�2) + 2(4) + 0(3) = 2: �

(1.2) a = 2i� k and b = 4j � k.

Answer.a � b = (2i� k) � (4j � k) = 2(0) + 0(4) + (�1)(�1) = 1: �

2. Compute the angle between the vectors.

(2.1) a = h 2; 0;�2 i and b = h 0;�2; 4 i.

Answer.

cos � =a � b

kakkbk =h 2; 0;�2 i � h 0;�2; 4 i

k h 2; 0;�2 i kk h 0;�2; 4 i k = � 2p10;

� = cos�1 � 2p

10

!� 2:25552 �

(2.2) a = 3i+ j � 4k and b = �2i+ 2j + k.

Answer.

cos � =a � b

kakkbk =(3i+ j � 4k) � (�2i+ 2j + k)

k3i+ j � 4kk � 2i+ 2j + kk =�8

3p26;

� = cos�1 � 8

3p26

!� 2:12114 �

3. Determine whether the vectors are orthogonal.

(3.1) a = h 4;�1; 1 i and b = h 2; 4; 4 i.

Answer.a � b = h 4;�1; 1 i � h 2; 4; 4 i = 8 6= 0:

So a and b are not orthogonal. �(3.2) a = 6i+ 2j and b = �i+ 3j.

Answer.a � b = (6i+ 2j) � (�i+ 3j) = 0:

So a and b are orthogonal. �4. Find a vector perpendicular to the given vector.

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(4.1) h 4;�1; 1 i.

Answer. Let v = h a; b; c i be a vector perpendicular to u = h 4;�1; 1 i. Then we should havev � u = 0, i.e.,

0 = v � u = h a; b; c i � h 4;�1; 1 i = 4a� b+ c; i:e:; 4a� b+ c = 0:

There are so many numbers, a, b, and c satisfying the equation. We choose just one, a = 1, b = 4

and c = 0. That is, v = h 1; 4; 0 i is one vector perpendicular to u = h 4;�1; 1 i. �(4.2) 6i+ 2j � k.

Answer. Let v = h a; b; c i = ai + bj + ck be a vector perpendicular tow = 6i + 2j � k. Thenby the same argument as above, we deduce

0 = v �w = (ai+ bj + ck) � (6i+ 2j � k) = 6a+ 2b� c; i:e:; 6a+ 2b� c = 0:

There are somany numbers, a, b, and c satisfying the equation. We choose just one, a = 1, b = �1and c = 4. That is, v = h 1;�1; 4 i = i � j + 4k is one vector perpendicular tow = 6i + 2j �k. �

5. Find Compb a and Projb a.

(5.1) a = 3i+ j and b = 4i� 3j.

Answer.

Compb a =a � bkbk =

(3i+ j) � (4i� 3j)

k4i� 3jk =9

5

Projb a = (Compb a)b

kbk =9

5

4i� 3j

k4i� 3jk =9

25(4i� 3j) �

(5.2) a = h 3; 2; 0 i and b = h �2; 2; 1 i.

Answer.

Compb a =a � bkbk =

h 3; 2; 0 i � h �2; 2; 1 ik h �2; 2; 1 i k = �2

3

Projb a = (Compb a)b

kbk = �2

3

h �2; 2; 1 ik h �2; 2; 1 i k = �2

9h �2; 2; 1 i �

6. A constant force of h 60;�30 i pounds moves an object in a straight line from the point (0; 0) to thepoint (10;�10). Compute the work done.

Answer. Thedisplacement vector isd = h 10� 0;�10� 0 i = h 10;�10 i. The force isF = h 60;�30 i.By the formula, the workW done is obtained by

W = F � d = h 10;�10 i � h 60;�30 i = 10(60) + (�10)(�30) = 900: �

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7. Label each statement as true or false. If it is true, briefly explainwhy; if it is false, give a counterexample.

(7.1) If a � b = a � c, then b = c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. a = h 1; 0; 0 i, b = h 0; 1; 0 i and c = h 0; 0; 1 i satisfies a � b = 0 = a � c. But, obviously,b 6= c. �

(7.2) If b = c, then a � b = a � c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TAnswer. If b = c, then b� c = 0 and so

a � (b� c) = a � 0 = 0; i:e:; a � b� a � c = 0; i:e:; a � b = a � c: �

(7.3) a � a = kak2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TAnswer. The formula a � b = kakkbk cos �, where � is the angle between a and b, implies

a � a = kakkak cos 0 = kak2; i:e:; a � a = kak2:

One may compute the dot product with a = h a; b; c i and prove the equality. �(7.4) If kak > kbk, then a � c > b � c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . F

Answer. We observe a = h 2; 0 i and b = h 0; 1 i satisfy the inequality kak = 2 > 1 = kbk.However, with c = h 0; 3 i, we get a � c = 0 < 3 = b � c. �

(7.5) If kak = kbk, then a = b. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. We observe i = h 1; 0; 0 i, j = h 0; 1; 0 i and k = h 0; 0; 1 i have kik = 1 = kjk = kkk.However, obviously, i 6= j 6= k. �

8. By the Cauchy–Schwartz Inequality, ja � bj � kakkbk. What relationship must exist between a and b

to have the equality ja � bj = kakkbk?

Answer. (1) a = 0 or b = 0. (2) The cosine of the angle between the vectors is �1. This happensexactly when the vectors point in the same or opposite directions. In other words, when a = sb forsome scalar s. �

9. By the Triangle Inequality, ka+ bk � kak+ kbk. What relationship must exist between a and b tohave the equality ka+ bk = kak+ kbk?

Answer. (1) a = 0 or b = 0. (2) Vectors a and b must be parallel with the same direction so thata = sb for some positive scalar s. �

10. The orthogonal projection of vector a along vector b is defined as Orthb a = a� Projb a. Sketch apicture showing vectors a, b, Projb a and Orthb a, and explain what is orthogonal about Orthb a.

Page 3 of 9


Answer. Orthb a is the component of a that is orthogonal to b:

b � Orthb a = b � (a� Projb a) = b � a� b � Projb a

= b � a� b � (a � b)bkbk2 = b � a� (a � b)(b � b)kbk2 = b � a� a � b = 0;

where b � b = kbk2 is used above. �11. A car makes a turn on a banked road. If the road is banked at 15�, show that a vector parallel to the

road is h cos 15�; sin 15� i. If the car has weight 2500 pounds, find the component of the weight vectoralong the road vector. This component of weight provides a force that helps the car turn.

Answer. The vector b = h cos 15�; sin 15� i represents the direction of the banked road. The weight ofthe car isw = h 0;�2500 i. The component of the weight in the direction of the bank is

Compbw =w � bkbk = �2500 sin 15� � �647:0 lbs

toward the inside of the curve. �

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Section 10.4 Cross Product

12. Compute the determinant: ��0 2 �11 �1 2

1 1 2

��Answer. ��

0 2 �11 �1 2

1 1 2

��= 0(�2� 2)� 2(2� 2)� (1 + 1) = �2: �

13. Compute the cross product a� b.

(13.1) a = h 2;�2; 0 i and b = h 3; 0; 1 i.

Answer.

a� b = h 2;�2; 0 i � h 3; 0; 1 i =

��i j k

2 �2 0

3 0 1

��= �2i� 2j + 6k = �2 h 1; 1;�3 i �

(13.2) a = �2i+ j � 3k and b = 2j � k.

Answer.

a� b = h 2; 1;�3 i � h 0; 2;�1 i =

��i j k

�2 1 �30 2 �1

��= 5i� 2j � 4k = h 5;�2;�4 i �

14. Find two unit vectors orthogonal to the two given vectors.

(14.1) a = h 0; 2; 1 i and b = h 1; 0;�1 i.

Answer. We recall

Orthogonal Vector

a�b is orthogonal to both a and b. Thus,� a� b

k a� b k are two unit vectors orthogonal to both

a and b.

a� b = h 0; 2; 1 i � h 1; 0;�1 i = h �2; 1;�2 i ; � a� b

k a� b k = �1

3h �2; 1;�2 i �

(14.2) a = �2i+ 3j � 3k and b = 2i� k.

Answer.

a� b = h �2; 3;�3 i � h 2; 0;�1 i = h �3;�8;�6 i ; � a� b

k a� b k = � 1p109

h �3;�8;�6 i�

Page 5 of 9


15. Use the cross product to determine the angle between the vectors, assuming that 0 � � � �=2.

(15.1) a = h 2; 2; 1 i and b = h 0; 0; 2 i.

Answer. We recall

Angle Between Two Vectors

k a� b k = k a k k b k sin �; i:e:; sin � =k a� b kk a k k b k ; i:e:; � = sin�1

k a� b kk a k k b k

!

where 0 � � � � is the angle between a and b.

a� b = 4 h 1;�1; 0 i ; k a k = 3; k b k = 2; k a� b k = 4p2;

� = sin�1 4p2

6

!� 1:23096 rad � 70:5288� �

(15.2) a = i+ 3j + 3k and b = 2i+ j.

Answer.

a� b = h �3; 6;�5 i ; k a k =p19; k b k =

p5; k a� b k =

p70;

� = sin�1 p

70p95

!� 1:03213 rad � 59:1369� �

16. Find the distance from the pointQ(1; 3; 1) to the line through (1; 3;�2) and (1; 0;�2).

Answer. The distance d from the point Q to the line through the points P and R is obtained by thefollowing formula.

Distance from a Point to a Line

d =

��!PQ��!PR

�!PR Letting P = (1; 3;�2) andR = (1; 0;�2), we have

��!PQ = 3 h 0; 0; 1 i ; �!

PR = �3 h 0; 1; 0 i ; ��!PQ��!PR = 9 h 1; 0; 0 i ; d =

9 k h 1; 0; 0 i k3 k h 0; 1; 0 i k = 3: �

17. If you apply a force of magnitude 30 pounds at the end of an 8–inch–long wrench at an angle of �=3 tothe wrench, find the magnitude of the torque applied to the bolt.

Answer. We recall

Page 6 of 9


TorqueTorque τ is defined to be the cross product of the position vector r and force vector F , i.e.,

τ = r � F ; k τ k = k r k k F k sin �

where � is the angle between r and F .

Given information: k F k = 30 and � = �=3 and k r k = 8 inch = 8=12 = 2=3 feet. Thus we have

k τ k = 2

3(30) sin

�

3= 10

p3 ft–lbs �

18. Label each statement as TRUE or FALSE. If it is true, briefly explain why. If it is false, give a counterex-ample.

(18.1) If a� b = a� c, then b = c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. We observe that for a = h 1; 0; 0 i, b = h 1; 1; 1 i and c = h 2; 1; 1 i,

a� b = h 0;�1; 1 i = a� c; but b 6= c: �

(18.2) a� b = �b� a. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TAnswer. We observe that for a = h a1; a2; a3 i and b = h b1; b2; b3 i,

a� b = h a2b3 � a3b2; a1b3 � a3b1; a1b2 � a2b1 i = �b� a: �

(18.3) a� a = kak2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. a� a = h 0; 0; 0 i is a vector, while k a k is a scalar. As a counterexample, for a = i, weget i� i = 0 = h 0; 0; 0 i 6= 1 = k i k2. �

(18.4) a � (b� c) = (a � b)� c. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. (a � b) � c is not possible because a � b is a scalar. A cross product must involve twovectors. �

(18.5) If the force is doubled, the torque doubles. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TAnswer. Torque is the cross product of direction and force.

r � (2F ) = 2(r � F ) = 2τ : �

19. Find the area of the parallelogram with two adjacent sides formed by h �2; 1 i and h 1;�3 i.

Answer. We recall

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AreaThe area of the parallelogram with two adjacent sides formed by a and b is the magnitude of theircross product:

A = k a� b k = k a k k b k sin �

where 0 � � � � is the angle between a and b.

The vectors h �2; 1 i and h 1;�3 i are in the plane (R2). The cross product is defined only for thevectors in the space (R3). But we observe h �2; 1; 0 i and h 1;�3; 0 i in the space can correspond tothose vectors in the plane, respectively. So we compute the area with these vectors:

A = k h �2; 1; 0 i � h 1;�3; 0 i k = k 5 h 0; 0; 1 i k = 5: �

20. Find the area of the triangle with vertices (0; 0; 0), (0;�2; 1) and (1;�3; 0).

Answer. Letting P (0; 0; 0),Q(0;�2; 1) andR(1;�3; 0), by the formula, the area of the parallelogramwith two adjacent sides formed by��!PQ and�!PR is the magnitude of their cross product:

Aparallelogram = ��!PQ��!

PR = k h 0;�2; 1 i � h 1;�3; 0 i k = k h 3; 2; 1 i k =

p14:

Since the area of the triangle is the half of the area of the parallelogram, hence the desired area is

Atriangle =

��!PQ��!PR

2

=

p14

2: �

21. Find the volume of the parallelepipedwith three adjacent edges formed by h 0;�1; 0 i, h 0; 2;�1 i, andh 1; 0; 2 i.

Answer. We recall

VolumeThe volume of the parallelepiped determined by the vectors a, b, and c is the magnitude of theirscalar triple product:

V = jc � (a� b)jLettinga = h 0;�1; 0 i, b = h 0; 2;�1 i and c = h 1; 0; 2 i, the formula implies the volumeV = 1. �

22. Use geometry to identify the cross product. (Do not compute!)

(22.1) j � (j � k).

Answer. Geometry implies

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Cross Product on Standard Basis Vectors

i� j = k j � k = i k � i = j

j � i = �k k � j = �i i� k = �jHence, we deduce

j � (j � k) = j � i = �k �(22.2) (j � i)� k.

Answer. By the same argument as above, we get

(j � i)� k = �k � k = 0 �

23. Use the parallelepiped volume formula to determine whether the vectors h 1; 1; 2 i and h 0;�1; 0 i andh 3; 2; 4 i are coplanar.

Answer. Letting a = h 1; 1; 2 i, b = h 0;�1; 0 i and c = h 3; 2; 4 i, the formula above implies thevolume V = 2. Since the volume of the parallelepiped is not zero, it means those three vectors do notlie on the same plane. That is, they are not coplanar. �

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United Arab Emirates UniversityFaculty of Science



Section 10.5 Lines and Planes in Space



1. Find (a) parametric equations and (b) symmetric equations of the line.

(1.1) The line through (3;�2; 4) and parallel to h 3; 2;�1 i.

Answer.

(a) x = 3 + 3t; y = �2 + 2t; z = 4� t (b)x� 3

3=

y + 2

2=

z � 4

�1 �

(1.2) The line through (�1; 0; 0) and parallel to the line x+ 1

�2 =y

3= z � 2.

Answer. The direction comes from the given line, i.e., h �2; 3; 1 i.

(a) x = �1� 2t; y = 0 + 3t; z = 0 + t (b)x+ 1

�2 =y

3=

z

1�

(1.3) The line through (�3; 1; 0) and perpendicular to both h 0;�3; 1 i and h 4; 2;�1 i.

Answer. h 0;�3; 1 i � h 4; 2;�1 i = h 1; 4; 12 i is in the direction perpendicular to both vectors.

(a) x = �3 + t; y = 1 + 4t; z = 0 + 12t (b)x+ 3

1=

y � 1

4=

z

12�

(1.4) The line through (0;�2; 1) and normal to the plane y + 3z = 4.

Answer. h 0; 1; 3 i is normal to the plane.

(a) x = 0 + 0t = 0; y = �2 + t; z = 1 + 3t (b) x = 0;y + 2

1=

z � 1

3�

2. State whether the lines are parallel or perpendicular and find the angle between the lines:

L : x = 4� 2t; y = 3t; z = �1 + 2t

M : x = 4 + s; y = �2s; z = �1 + 3s:

Answer. The vectors parallel to lines L andM are respectively vL = h �2; 3; 2 i and vM = h 1;�2; 3 i.Since there is no constant c satisfying vL = h �2; 3; 2 i = c h 1;�2; 3 i = cvM , so the lines L andM arenot parallel.

By the formula on the dot product, we have

cos � =vL � vM

k vL k k vM k =�2� 6 + 6p

17p14

= � 2p238

; � = cos�1 � 2p

238

!� 1:7 rad;

which is not �=2. Thus, the lines L andM are not perpendicular and the angle between the lines L andM is about 1.7 rad. �

3. Determine whether the lines are parallel, skew or intersect:

L : x = 3 + t; y = 3 + 3t; z = 4� t

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M : x = 2� s; y = 1� 2s; z = 6 + 2s:

Answer. The vectors parallel to linesL andM are respectivelyvL = h 1; 3;�1 i andvM = h �1;�2; 2 i.Since there is no constant c satisfying vL = h 1; 3;�1 i = c h �1;�2; 2 i = cvM , so the lines L andMare not parallel. To determine whether or not the lines intersect, we set the x–, y– and z–values equal:

3 + t = 2� s; 3 + 3t = 1� 2s; 4� t = 6 + 2s;

simply; (i) t+ s = �1; (ii) 3t+ 2s = �2; (iii) t+ 2s = �2:

From (i) and (ii), we get t = 0 and s = �1. Putting t = 0 and s = �1 into (iii), the equation (iii)

holds. This implies that the lines L andM intersect when t = 0 and s = �1:

x = 3; y = 3; z = 4:

Hence, the lines intersect at the point (3; 3; 4). �4. Find an equation of the given plane.

(4.1) The plane containing the point (�2; 1; 0) with normal vector h �3; 0; 2 i.

Answer.

�3(x+ 2) + 0(y � 1) + 2(z � 0) = 0; simply; 3x� 2z + 6 = 0: �

(4.2) The plane containing the points (�2; 2; 0), (�2; 3; 2) and (1; 2; 2).

Answer. Letting P (�2; 2; 0), Q(�2; 3; 2) and R(1; 2; 2), we deduce ��!PQ = h 0; 1; 2 i and �!PR =

h 3; 0; 2 i and ��!PQ � �!PR = h 2; 6;�3 i is normal to the plane. Hence, using the point P (�2; 2; 0)

(one can useQ orR), the equation of the plane is obtained as

2(x+ 2) + 6(y � 2)� 3(z � 0) = 0; simply; 2x+ 6y � 3z � 8 = 0: �

(4.3) The plane containing the point (3;�2; 1) and parallel to the plane x+ 3y � 4z = 2.

Answer. The vector normal to the plane is h 1; 3;�4 i. Hence, the equation of the plane is

(x� 3) + 3(y + 2)� 4(z � 1) = 0; simply; x+ 3y � 4z + 7 = 0: �

(4.4) The plane containing the point (3; 0;�1) and perpendicular to the planes x + 2y � z = 2 and2x� z = 1.

Answer. Let v1 and v2 be the vectors normal to the planes: v1 = h 1; 2;�1 i and v2 = h 2; 0;�1 i.Normal vector must be perpendicular to the normal vectors of both planes. So, the plane that weare looking for has the normal vector v1�v2 = h �2;�1;�4 i. Hence, the equation of the plane is

�2(x� 3)� (y � 0)� 4(z + 1) = 0; simply; 2x+ y + 4z = 2: �

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5. Sketch the given plane.

(5.1) 2x� y + 4z = 4.

Answer. The equation represents a planewhichhas thenormal vector h 2;�1; 4 i andpasses throughthe pointP (2; 0; 0),Q(0;�4; 0) andR(0; 0; 1). We connect those three points and sketch the planeroughly. �

P

Q

R

Normal Vector

-4

-2

0

2

4X-axis

-4

-2

0

2

4Y-axis

-4

-2

0

2

4

Z-axis

(5.2) x+ y = 1.

Answer. The equation represents a plane such that(i) it is parallel to the z–axis and perpendicular to the xy–plane,(ii) it has the normal vector h 1; 1; 0 i,(iii) it passes through the point P (1; 0; 0),Q(0; 1; 0). �

P

QNormal Vector

-4

-2

0

2

4X-axis

-4-2

02

4Y-axis

-4

-2

0

2

4

Z-axis

6. Find the intersection of the planes 3x+ y � z = 2 and 2x� 3y + z = �1.

Answer. Solve the equations for z and equate:

3x+ y � 2 = z = �2x� 3y � 1; i:e:; 3x+ y � 2 = �2x� 3y � 1;

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i:e:; 5x+ 4y = 1; i:e:; y =�5x+ 1

4:

Putting it into the first equation, we get

2 = 3x+�5x+ 1

4� z; i:e:; z =

7x� 7

4:

Using x = t as a parameter, we deduce the line:

x = t; y =1

4� 5

4t; z = �7

4+

7

4t:

If one use y = s as a parameter, one can get the line:

x =1

5+

2

5s; y = s; z = �7

5+

11

5s:

One can use z = u as a parameter and get the equation of the line. �7. Find the distance between the given objects.

(7.1) The point (1; 3; 0) and the plane 3x+ y � 5z = 2.

Answer. The distance formula implies

d =j3(1) + 1(3)� 5(0)� 2jq

32 + 12 + (�5)2=

4p35

: �

(7.2) The planes x+ 3y � 2z = 3 and x+ 3y � 2z = 1.

Answer. Two planes are parallel. We choose a point P (1; 0; 0) on the plane x + 3y � 2z = 1.Applying the distance formula to the point P and the other plane, we get

d =j1(1) + 3(0)� 2(0)� 3jq

32 + 12 + (�2)2=

2p14

: �

8. Find an equation of the plane containing the lines:

L : x = 4 + t; y = 2; z = 3 + 2t

M : x = 2 + 2s; y = 2s; z = �1 + 4s:

Answer. The vectors parallel to the lines are respectively vL = h 1; 0; 2 i and vM = h 2; 2; 4 i. Thenormal vector to the plane is vL � vM = h �4; 0; 2 i. We choose one point (4; 2; 3) from L. (One canchoose any point from either line.) Hence, the equation of the plane is

�4(x� 4) + 0(y � 2) + 2(z � 3) = 0; simply; 2x� z = 5: �

9. State whether the statement is true or false (not always true).

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(9.1) Two planes either are parallel or intersect. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TAnswer. Two planes can be even both parallel and intersect if the planes coincide. �

(9.2) The intersection of two planes is a line. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. It can be a plane if the planes coincide, or can be empty. �

(9.3) The intersection of three planes is a point. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. It can be a point, a line, or a plane, or can be empty. �

(9.4) Lines that lie in parallel planes are always skew. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . TAnswer. It is true, unless the parallel planes coincide. �

(9.5) The set of all lines perpendicular to a given line forms a plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. It is false. However, it is true if we take all lines perpendicular to a given line through agiven point. �

(9.6) There is one line perpendicular to a given plane. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . FAnswer. It is false. There is one line perpendicular to a given plane through each point of theplane. �

(9.7) The set of all points equidistant from two given points forms a plane. . . . . . . . . . . . . . . . . . . . . . . T10. Determine whether the given lines are the same:

L : x = 1 + 4t; y = 2� 2t; z = 2 + 6t

M : x = 9� 2s; y = �2 + s; z = 8� 3s:

Answer. The vectors parallel to the lines are vL = h 4;�2; 6 i and vM = h �2; 1;�3 i, respectively.Since vL = �2vM , the lines L andM are parallel. The point (9;�2; 8) lies on the lineM when s = 0.We solve for t in the x coordinate of the lineL to see that 1+4t = 9 implies t = 2. It means when t = 2

and s = 0, the lines L andM have the same x coordinate 9. What about the y and z coordinates whent = 2 and s = 0? At this time t = 2 and s = 0, the lineL has the y and z coordinates, y = 2�2(2) = �2and z = 2 + 6(2) = 14, respectively. However, the line M has the y and z coordinates, y = �2 andz = 8. That is,

t = �2 & s = 0 =) L : (x; y; z) = (9;�2; 14); M : (x; y; z) = (9;�2; 8):

It implies that these lines L andM are not the same. �

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