Math 133, Chapter 11 Practice

1. The cranes in the following figure are holding a 10650-pound steel girder in midair. WhereF1 is at a 34◦ angle with the horizontal and F2 is at a 53◦ angle with horizontal (in thediagram below A = 34◦ and B = 53◦). Find the magnitudes of the vectors F1 and F2,round your answers to the nearest pound.

Solution: Let a denote the magnitude of F1 and let b denote the magnitude of F2,and let w be the downward weight vector for the girder. Notice also that F1 forms a(180− 34)◦ = 146◦ angle in standard position with the positive x-axis. Then

F1 = a〈cos(146◦), sin(146◦)〉, F2 = b〈cos(53◦), sin(53◦)〉, w = 〈0,−10650〉

Then F1 +F2 +w = 0 because the girder is held in place the vectors sum to the zerovector. Considering the i-component of the vectors we obtain

(1) a cos(146◦) + b cos(53◦) = 0 and so b =−a cos(146◦)

cos(53◦).

Now, j component of sum of the vectors F1, F2 and w is 0 and so we have

(2) a sin(146◦) + b sin(53◦)− 10650 = 0 and so a sin(146◦) + b sin(53◦) = 10650

Substituting the relation from (1) into (2) we obtain

a sin(146◦)− acos(146◦)

cos(53◦)· sin(53◦) = 10650

and so a =10650

sin(146◦)− cos(146◦) tan(53◦)≈ 6418.1258. Thus

b =−a cos(146◦)

cos(53◦)=−6418.1258 cos(146◦)

cos(53◦)≈ 8841.3669

Thus, to the nearest pound ‖F1‖ ≈ 6418 pounds and ‖F2‖ ≈ 8841 pounds.

2. A plane is flying with a bearing of 302◦ (measured clockwise from north). Its speed withrespect to the air is 900 kilometers per hour. The wind at the plane’s altitude is from

the southwest at 100 kilometers per hour. What is the true direction of the plane, andwhat is its speed with respect to the ground?

Solution: See 11.1, #83 from 10th edition of Larson at Calc Chat.

3. Find a vector v with magnitude 5 in the same direction as 〈−1, 8〉.

Solution: First, ‖〈−1, 8〉‖ =√

(−1)2 + 82 =√

65, and thus the desired vector v is

v =5√65〈−1, 8〉 =

⟨− 5√

65,

40√65

⟩

4. Find a vector v such that ‖v‖ = 2 and v forms a 150◦ angle with the positive x-axis.

Solution: v = 2〈cos(150◦), sin(150◦)〉 = 〈−√

3, 1〉.

5. A gun with a muzzle velocity of 2350 feet per second is fired at an angle 5◦ abovehorizontal. Find the horizontal and vertical components of the velocity.

Solution: The horizontal component of velocity is (2350)(cos 5◦) ≈ 2341.1 feet persecond.

The vertical component of velocity is (2350)(sin 5◦) ≈ 204.8 feet per second.

6. To carry a 100-lb cylindrical weight, two workers lift on the ends of short ropes tied toan eyelet on the top center of the cylinder. One rope makes a 20◦ angle away from thevertical and the other makes a 30◦ angle (see figure below).

(a) Find each rope’s tension.

(b) Find the vertical component of each worker’s force.

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Solution: (a) We can write the vectors as F1 = a〈sin 30◦, cos 30◦〉 and F2 =b〈− sin 20◦, cos 20◦〉, and W = 〈0,−100〉. The sum of the forces is 〈0, 0〉 and so

−a sin 20◦ + b sin 30◦ = 0 and a cos 20◦ + b cos 30◦ = 100

The first equation implies b = 2a sin 20◦. Substituting this into the second implies

a =100

cos 20◦ + 2 cos 30◦ sin 20◦≈ 65.27 and then b ≈ 44.6476.

Thus the tension of the 20◦ rope is 65.27 pounds, and the tension on the 30◦ rope is44.65 pounds.

(b) The vertical component on the 20◦ rope is 65.27 cos 20◦ ≈ 61.33 pounds, and thevertical component on the 30◦ rope is 44.65 cos 30◦ ≈ 38.67 pounds. (These shouldadd to 100 pounds.)

7. (a) Find the midpoint of (−7,−7,−5) and (3,−5, 5).

(b) Find the distance from the midpoint in (a) to (−7,−7,−5).

(c) Write the equation of a sphere whose end points of a diameter are (−7,−7,−5) and(3,−5, 5).

Solution: (a) The midpoint is(−7 + 3

2,−7 +−5

2,−5 + 5

2

)= (−2,−6, 0).

(b) Using the distance formula d =√

(x2 − x1)2 + (y2 − y1)2 + (z2 − z1)2 for thedistance between the two points (x1, y1, z1) and (x2, y2, z2) we find the distance is

d =√

(3−−7)2 + (−5−−7)2 + (5−−5)2 =√

25 + 1 + 25 =√

51

(c) The center of the sphere is the midpoint (−2,−6, 0) of the two points, and theradius is the distance

√51 from (b). Recalling that a sphere with radius r and center

(x0, y0, z0) has equation

(x− x0)2 + (y − y0)2 + (z − z0)2 = r2

we obtain that the equation of the desired sphere is

(x+ 2)2 + (x+ 6)2 + z2 = 51

8. Find the center and radius of the sphere given by

x2 − 8x+ y2 + 10y + z2 = −40

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Solution: We complete the squares to write the equation in standard form

x2 − 8x+ 42 + y2 + 10y + 52 + z2 = −40 + 42 + 52

and so the equation of the sphere in standard form is

(x− 4)2 + (y + 5)2 + z2 = 1

This sphere has center (4,−5, 0) and radius√

1 = 1.

9. A television camera weighing 120 pounds is supported by a tripod. The tops of the legsjoin at the point (0, 0, 4) in the xyz-plane, and the bases of the legs are at the points

(0,−1, 0), (√32, 12, 0) and (−

√32, 12, 0) respectively. Represent the force exerted on each leg

of the tripod as a vector.

Solution: See Example 7, Section 11.2 of, 10th edition of Larson at at Calc Chat.

10. (a) Find a unit vector in the same direction as w = 〈−3, 4,−1〉.(b) Find a vector of length 6 in the same direction as w = 〈−3, 4,−1〉.(c) Find a vector of length 6 in the opposite direction as w = 〈−3, 4,−1〉.

Solution: (a) Given any nonzero vector w, a unit vector in the same direction as wis given by u = 1

‖w‖w. In this case, ‖w‖ =√

(−3)2 + (4)2 + (−1)2 =√

26. So thedesired unit vector is

u =1√26〈−3, 4,−1〉 =

⟨−3√

26,

4√26,−1√

26

⟩(b) The desired vector is 6u where u is as in (a):

6u =

⟨−18√

26,

24√26,−6√

26

⟩(c) The desired vector is −6u where u is as in (a):

−6u =

⟨18√26,−24√

26,

6√26

⟩

11. Use vectors to find the point R that lies 120

of the way from the point P (−3,−1, 2) tothe point Q(−6, 0, 4).

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Solution: We will find R by finding the vector v =−→PQ and then multiplying it by

120

and then adding that vector to the point P . Then

v =−→PQ = 〈−3, 1, 2〉

and so we find

R = (−3,−1, 2)+1

20〈−3, 1, 2〉 =

(−3 +

−3

20,−1 +

1

20, 2 +

2

20

)= (−3.15,−0.95, 2.10)

12. With the help of dot products, show that ‖u+v‖2 +‖u−v‖2 = 2‖u‖2 + 2‖v‖2. Explainwhy this is called the parallelogram identity.

Solution: Use the fact for any vector w properties of dot products ensure ‖w‖2 =w ·w. Thus

‖u + v‖2 + ‖u− v‖2 = (u + v) · (u + v) + (u− v) · (u− v)

= u · u + 2(u · v) + v · v + u · u− 2(u · v) + v · v= 2(u · u) + 2(v · v)

= 2‖u‖2 + 2‖v‖2.

This is the parallelogram identity, because a parallelogram whose sides are parallel tou and v has diagonals u + v and u− v. Thus the sum of the squares of the lengthsof the four sides is equal to the sum of the squares of the lengths of the diagonals.

13. (a) Do the points (2, 9, 1), (3, 11, 4), (0, 10, 2) and (1, 12, 5) form the vertices of a paral-lelogram? If so, is it a rectangle?

(b) Find a unit vector in the same direction as v = 〈−4, 3, 7〉. Then find a vector ofmagnitute 10 in the direction opposite of v.

Solution: (a) Yes, see solution to 11.2#69 of the 10th edition of Larson at Calc Chat,the points form the vertices of a parallelogram. No, it is not a rectangle because thedot product of vectors representing sides sharing a vertex is not 0.

(b) First, ‖v‖ =√

16 + 9 + 49 =√

74, so a unit vector in the same direction as vis u = 1√

74〈−4, 3, 7〉 and a vector of magnitude 10 in the opposite direction of v is

w = 1√74〈4,−3,−7〉.

14. Consider the vectors u = 〈3, 1, 3〉 and v = 〈4, 5, 4〉.(a) Find ‖u‖ and ‖v‖.(b) Find the dot product u · v.

(c) Find the angle between u and v. Round your answer to the nearest 0.01 degree.

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Solution: (a) The magnitudes of the vectors are:

‖u‖ =√

(3)2 + (1)2 + (3)2 =√

19 and ‖v‖ =√

(4)2 + (5)2 + (4)2 =√

57

(b) The dot product is

u · v = (3)(4) + (1)(5) + (3)(4) = 29

(c) The angle θ between the vectors satisfies

cos θ =u · v‖u‖‖v‖

=29√

19√

57=

29√1083

and so θ = arccos

(29√1083

)≈ 28.21◦

15. Find the projection of the vector v = 〈2,−3, 3〉 onto the vector w = 〈4, 2, 3〉, find alsothe component of v that is orthogonal to w.

Solution: For this, we will need the dot product

v ·w = (2)(4) + (−3)(2) + (3)(3) = 11

and ‖w‖2 = (4)2 + (2)2 + (3)2 = 29. Then the projection of v onto w is the vector

v1 = Projwv =

(v ·w‖w‖2

)w =

11

29〈4, 2, 3〉 =

⟨44

29,22

29,33

29

⟩Then the component of v that is orthongonal to w is v2 = v − Projwv and so

v2 = 〈2,−3, 3〉 −⟨

44

29,22

29,33

29

⟩=

⟨14

29,−109

29,54

29

⟩(To check your answer, notice that v1 is clearly parallel to w since v1 is a multipleof w. Check also that v1 +v2 = v and that v2 ·w = 0 and so v2 is orthogonal to w.)

16. A constant force F = 〈−3,−6, 0〉 moves an obect along a straight line from the pointP (2, 7, 1) to the point Q(3, 0,−2). Find the work done if distance is measured in metersand the magnitude of the force is measured in Newtons.

Solution: The vector with intial point P and terminal point Q is given by

v =−→PQ = 〈1,−7,−3〉

and then the work done is given by F · v, that is

work = 〈−3,−6, 0〉 · 〈1,−7,−3〉 = (−3)(1) + (−6)(−7) + (0)(−3) = 39 Nm

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17. Suppose u = 〈4,−1, 4〉 and v = 〈4, 4, 0〉. Find u× v and v × u

Solution: First we compute

u× v =

∣∣∣∣∣∣i j k4 −1 44 4 0

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

= 〈−16, 16, 20〉

From properties of cross products

v × u = −(u× v) = −〈−16, 16, 20〉 = 〈16,−16,−20〉

18. Find the area of a parallelogram with vertices A = (5, 2, 0), B = (3, 6, 4), C = (5,−2, 3)and D = (3, 2, 7) and find the area of the triangle with vertices ABC.

Solution: Let u =−→AB = 〈−2, 4, 4〉 and v =

−→AC = 〈0,−4, 3〉. Then

u× v =

∣∣∣∣∣∣i j k−2 4 4

0 −4 3

∣∣∣∣∣∣ = 〈28, 6, 8〉

and so ‖u × v‖ =√

(28)2 + (6)2 + (8)2. The parallelogram has area√

884; the

triangle area is

√884

2.

19. Find the volume of a parallelepiped with adjacent edges u = 〈1, 3, 1〉, v = 〈0, 6, 6〉 andw = 〈−4, 0,−4〉.

Solution: The volume is the absolute value of the triple scalar product

u · (v ×w) =

∣∣∣∣∣∣1 3 10 6 6−4 0 −4

∣∣∣∣∣∣ = −72.

So the volume is 72 units cubed.

20. A bicycle pedal is pushed straight downward by a foot with a 30 Newton force. Theshaft of the pedal is 22 cm long. If the shaft is 11 degrees above the horizontal, what isthe magnitude of the torque about the point where the shaft is attached to the bicycle?

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Solution: The torque vector can be found by using a cross-product, but the magni-tude follows from trigonometry as follows, where 22 cm is converted to 0.22 m:

magnitude of torque = (30)(0.22) cos(11◦) ≈ 6.47874 Nm

21. Find (a) parametric and (b) symmetric equations for the line passing through the pointsP (−3,−1,−4) and Q(−2,−2,−7).

Solution: A direction vector for the line is the vector v =−→PQ. Thus

v = 〈−2− (−3),−2− (−1),−7− (−4)〉 = 〈1,−1,−3〉

(a) Recalling that parametric equations for a line through a point P (x0, y0, z0) in adirection v = 〈a, b, c〉 can be obtained as

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

we use P (−3,−1,−4) and v = 〈1,−1,−3〉 to obtain parametric equations

x = −3 + 1t, y = −1 − 1t, z = −4 − 3t

(b) Solving the parametric equations for the parameter, we obtain the symmetricequations

x+ 3

1=y + 1

−1=z + 4

−3

22. Find an equation for the plane that passes through the points A(0,−2, 0), B(−3, 2, 2),and C(0,−3, 3).

Solution: Let u =−→AB = 〈−3, 4, 2〉 and v =

−→AC = 〈0,−1, 3〉. We will use the cross

product u× v to find a normal vector n for the plane. Then

n = u× v =

∣∣∣∣∣∣i j k−3 4 2

0 −1 3

∣∣∣∣∣∣ = 〈14, 9, 3〉

Recall that a plane with normal vector n = 〈a, b, c〉 passing through a point (x0, y0, z0)has equation

a(x− x0) + b(y − y0) + c(z − z0) = 0

In our case, we use the point (0,−2, 0) with n = 〈14, 9, 3〉 to obtain the equation

14(x− 0) + 9(y + 2) + 3(z − 0) = 0.

and so 14x+ 9y + 3z = −18.

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23. Given a point Q(x0, y0, z0) and a plane ax+ by+ cz+ d = 0, show that the distance from

Q to the plane is|ax0 + by0 + cz0 + d|√

a2 + b2 + c2.

Solution: See Section 11.5 in your text.

24. Find the distance from the point P (−3, 0, 4) to the line

x = −4 + 3t, y = −2 + 3t, z = 5 + 1t

Solution: Let Q be the point (−4,−2, 5) that is obtained by letting t = 0 on theline. Let

u =−→PQ = 〈−1,−2, 1〉

and let v = 〈3, 3, 1〉 be a direction vector for the line. Then the distance from P tothe line is given by

d =‖u× v‖‖v‖

So we next find u× v:

u× v =

∣∣∣∣∣∣i j k−1 −2 1

3 3 1

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

= 〈−5, 4, 3〉

and thus the distance is

d =‖〈−5, 4, 3〉‖‖〈3, 3, 1‖

=

√50√19

25. Find parametric and equations for the line passing through the point P (2,−2,−5) andthat is perpendicular to the plane 3x + 6y − 13z = −4. The find the point Q at whichthe line intersects the xy-plane.

Solution: The line passes through the point (2,−5,−5) and has direction vectorv = 〈3, 6,−13〉. Thus parametric equations for the line are

x = 2 + 3t, y = −2 + 6t, z = −5− 13t

The line passes through the xy-plane when z = 0. Thus −5 − 13t = 0, and sot = −5/13. Thus

x = 2 +(3)(−5)

13=

11

13and y = −2 +

(6)(−5)

13=−56

13

and so Q is the point(1113, −56

13, 0).

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26. The two planes x − 7y − 1z = 2 and −x + 9y + 5z = 0 are not parallel, so the mustintersect along a line. Find parametric equations for the line of intersection.

Solution: (Solution 1 via Systems of Equations). The equations of the planes form asystem of two linear equations in three variables, so the system has either no solution(the planes are parallel and don’t intersect) or it has infinitely many solutions (theplanes are not parallel and they intersect in a line). The parametric form of thesolutions will be the equation of the line of intersection. The augmented matrix for

the system is

[1 −7 −1 2−1 9 5 0

]which we reduce as follows.

[1 −7 −1 2−1 9 5 0

]−→

R1 +R2

[1 −7 −1 20 2 4 2

]−→ 1

2R2

[1 −7 −1 20 1 2 1

]

From the last row have y+ 2z = 1 and so we let z = t, and then y = 1− 2t, and thenthe first row implies

x = 2 + 7y + 1z = 2 + 7(1− 2t) + 1t = 9− 13t

Thus parametric equations for the line of intersection are

x = 9− 13t, y = 1− 2t, z = t

Solution: (Solution 2 using Cross-Products). The line of intersection will be per-pendicular to the normal vector of both planes. So a direction vector for the line willbe the cross product of the normal vectors of the planes which are n1 = 〈1,−7,−1〉and n2 = 〈−1, 9, 5〉. Then

n1 × n1 =

∣∣∣∣∣∣i j k1 −7 −1−1 9 5

∣∣∣∣∣∣= [(−2)(5)− (9)(−1)]i− [(1)(5)− (−1)(−1)]j + [(1)(5)− (9)(−1)]k

= 〈−26,−4, 2〉

any nonzero multiple of the above vector will work, and so we will use

v =1

2〈−26,−4, 2〉 = 〈−13,−2, 1〉.

as a direction vector for the line of intersection. Now we need a point on the line, sowe need one solution to the equations

x− 7y − 1z = 2 and − x+ 9y + 5z = 0

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which we put in an augmented matrix.[1 −7 −1 2−1 9 5 0

]−→

R1 +R2

[1 −7 −1 20 2 4 2

]−→ 1

2R2

[1 −7 −1 20 1 2 1

]

The second row implies y + 2z = 1, so we set z = 0, and then y = 1, then the firstrow implies x = 2 + (7)(1) = 9. We now have a point and a direction vector, so theline of intersection has parametric equations

x = 9− 13t, y = 1− 2t, z = t

27. Find the distance from the point Q(−4,−3,−3) to the plane 3x− 5y − 7z = 9.

Solution: First, find a point P on the plane, for example, (3, 0, 0). Then−→PQ =

〈7,−3,−3〉 and let n = 〈3,−5,−7〉 be the natural normal vector for the plane. Thenthe distance is

d =|−→PQ · n|‖n‖

=|(3)(7) + (−5)(−3) + (−7)(−3)|√

32 + 52 + 72=

57√83

28. Find the distance between the two parallel planes 4x−2y+5z = 4 and 4x−2y+5z = 12.

Solution: First, find a point P on the first plane, for example, (1, 0, 0). Find thedistance from that point to the other plane. For that we observe Q(3, 0, 0) is on the

second plane. Then−→PQ = 〈−2, 0, 0〉 and let n = 〈4,−2,−5〉 be the natural normal

vector for the plane(s). Then the distance is

d =|−→PQ · n|‖n‖

=|(4)(−2)|√42 + 22 + 52

=8√45

29. Sketch the surfaces given by:

(i)x2

9+y2

16+z2

9= 1.

(ii) y2 = 4x2 + 9z2.

(iii) 4x2 − 4y + z2 = 0.

(iv) 4x2 − y2 + 4z2 = 4.

(v) 4x2 − y2 + 4z = 0.

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Solution: See graphs given for Exercises 1 - 6 in Section 11.6. (i) is (c); (ii) is (b);

(iii) is (d); (iv) is (f); (v) is (a) the saddle function z =y2

4− x2.

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