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OHSx XM521 Multivariable Differential Calculus:Homework Solutions §13.1

(37) If a bug walks on the sphere

x2 + y2 + z2 + 2x− 2y − 4z − 3 = 0

how close and how far can it get from the origin ?

Solution: Complete the square of the sphere equation and group constants on one side,to get:

(x+ 1)2 + (y − 1)2 + (z − 2)2 = 3 + 1 + 1 + 4 = 32.

This is the equation of a sphere of radius 3 centered at (−1, 1, 2). The distance, d, fromthe origin to a point P (x, y, z) is given by:

dOP =√x2 + y2 + z2.

It is possible to try to minimize and maximize this equation subject to the constraint that(x, y, z) must lie on the above sphere. But the techniques to do min-max problems withmultiple variables have not been developed at this point in the course. Geometrically, thedistance from the origin to the center of the sphere, C, is:

dOC =√

(−1)2 + (1)2 + (2)2 =√

6.

Consider the triangle formed by the origin, the center of the sphere, and a point P (x, y, z)on the sphere. By the Law of Cosines:

d2OP = d2OC + d2CP − 2dOCdCP cos θ,

where dCP = radius of sphere = 3 and θ = the angle formed by OCP . To maximize dOP ,θ = π so that:

dOPmax =√d2OC + d2CP + 2dOCdCP =

√(dOC + dCP )2 = dOC + dCP =

√6 + 3.

This corresponds to moving from O to the center of sphere C and on the extension of lineOC to P . The minimum corresponds to θ = 0, so that:

dOPmin =√d2OC + d2CP − 2dOCdCP =

√(dOC − dCP )2 = |dOC − dCP | = |

√6− 3|.

(41) As shown in the accompanying figure, a bowling ball of radius R is placed inside a boxjust large enough to hold it, and it is secured for shipping by a Styrofoam sphere intoeach corner of the box. Find the radius of the largest Styrofoam sphere that can be used.[HINT: Take the origin of a Cartesian coordinate system at a corner of the box with thecoordinate axes along the edges.]

Solution: A box just large enough to hold a sphere of radius R is a cube with edges oflength 2R where the center of each face of the cube is tangent to the sphere. Place this

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cube so that one corner of it is at the origin of a Cartesian coordinate system. The distancefrom the origin to the center of the sphere, CR = (R,R,R), is:

dOCR=

√x2 + y2 + z2 =

√R2 +R2 +R2 =

√3R.

Let the radius of the Styrofoam sphere be r. Since this sphere will need to be placed ineach corner of the cube, consider just the sphere in the corner of the origin. This littlesphere must be tangent to the big sphere to hold it in place. It must also be tangent tothe three faces of the cube in the origin corner. By the above analysis, the distance fromthe origin to the center of the Styrofoam sphere, Cr = (r, r, r), is:

dOCr =√x2 + y2 + z2 =

√r2 + r2 + r2 =

√3r.

Let a be the distance from the origin to the little sphere along the line OCR so that fromgeometry:

dOCR= R+ 2r + a =

√3R.

Also:

dOCr= r + a =

√3r,

so that a = (√

3− 1)r. Substituting into the equation for dOCR:

dOCR= R+ 2r + (

√3− 1)r =

√3R,

so that r(√

3 + 1) = (√

3− 1)R. Solving for r and rationalizing:

r =(√

3− 1)2R

(√

3 + 1)(√

3− 1)= (2−

√3)R.

(43) Show that for all values of θ and φ, the point

(a sinφ cos θ, a sinφ sin θ, a cosφ)

lies on the sphere x2 + y2 + z2 = a2.

Solution: Let θ and φ be arbitrary. If x = a sinφ cos θ, y = a sinφ sin θ, and z = a cosφ,then using the trig identity sin2 + cos2 = 1, we have

x2 + y2 + z2 = (a sinφ cos θ)2 + (a sinφ sin θ)2 + (a cosφ)2

= (a sinφ)2(sin2 θ + cos2 θ) + (a cosφ)2

= (a sinφ)2 + (a cosφ)2

= a2.

Therefore, for all values of θ and φ, the point (x, y, z) of the given form satisfies x2+y2+z2 =a2. So all such points lie on this sphere.

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(38) Let r = 〈x, y〉 and r0 = 〈x0, y0〉. In each part, describe the set of all points (x, y) in 2-spacethat satisfy the stated condition (a) ||r− r0|| = 1.

Solution: In terms of components, r−r0 = 〈x−x0, y−y0〉, and ||r−r0|| =√

(x− x0)2 + (y − y0)2.

So ||r − r0|| = 1 if and only if√

(x− x0)2 + (y − y0)2 = 1. A point (x, y) satisfies thisequation if and only if its distance to (x0, y0) is 1. Therefore, the set of points (x, y) suchthat r = 〈x, y〉 satisfies ||r− r0|| = 1 is a circle of radius 1 centered at (x0, y0).

(b) ||r− r0|| ≤ 1.

Solution: By the same reasoning as in part (a), ||r− r0|| ≤ 1 if and only if (x− x0)2 +(y − y0)2 ≤ 1. A point (x, y) satisfies this equation if and only its distance to (x0, y0) isless than or equal to 1. The set of such points is the disk (the circle and its interior) ofradius 1 centered at (x0, y0).

(c) ||r− r0|| > 1.

Solution: By the same reasoning as in part (a) and (b), ||r − r0|| > 1 if and only if(x− x0)2 + (y − y0)2 > 1. A point (x, y) satisfies this equation if and only its distance to(x0, y0) is greater than 1. The set of such points consistns of those points lying strictlyoutside of the disk (the circle and its interior) of radius 1 centered at (x0, y0).

(49) The accompanying figure shows a 250-lb traffic light supported by two flexible cables. Themagnitude of the forces that the cables apply to the eye ring are called the cable tensions.Find the tensions in the cables if the traffic light is in static equillibrium (defined aboveexercise 47).

Solution: Static equillibrium for a given point implies∑

i Fi = 0 where each Fi is a forceapplied to the point and the sum is carried over all such forces Fi. Use this equation torelate the forces on the eyering:

F1 + F2 + F250 = 0

where F1 is the tension in the line making a 30o angle with the eyering, F2 is the tension inthe line making a 45o angle, and F250 is the downward pointing force of the 250 lb trafficlight. Locating the origin of an xy coordinate system at the eyering, look at the 〈x, y〉component form of this vector equation:

||F1||〈cos 150o, sin 150o〉+ ||F2||〈cos 45o, sin 45o〉+ ||F250||〈cos(−90o), sin(−90o)〉 = 〈0, 0〉,

where the identity v = ||v||〈cosφ, sinφ〉 has been used to find the various 〈x, y〉 componentsof the vectors. Simplifying the above expression gives:

〈||F1|| cos 150o + ||F2|| cos 45o + ||F250|| cos(−90o), ||F1|| sin 150o + ||F2|| sin 45o + ||F250|| sin(−90o)〉 = 〈0, 0〉.

Equating components gives the following two equations:

||F1|| cos 150o + ||F2|| cos 45o + ||F250|| cos(−90o) = 0;

||F1|| sin 150o + ||F2|| sin 45o + ||F250|| sin(−90o) = 0.

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Substituting for various trignometric values, using ||F250|| = 250 lbs, and solving for||F1||, ||F2||, gives:

−√

3||F1||+√

2||F2|| = 0;

||F1||+√

2||F2|| = 2(250).

Solving gives ||F1|| = 250(√

3− 1) ≈ 183 lbs, and ||F2|| ≈ 225 lbs.

(52) A vector w is said to be a linear combination of the vectors v1, v2, and v3 if w can beexpressed as w = c1v1 + c2v2 + c3v3, where c1, c2, and c3 are scalars.

(a) Find scalars c1, c2, and c3 to express 〈−1, 1, 5〉 as a linear combination of the vectorsv1 = 〈1, 0, 1〉, v2 = 〈3, 2, 0〉, and v3 = 〈0, 1, 1〉.

Solution: Substitute the various vectors into the vector equation w = c1v1 + c2v2 + c3v3

to get:〈−1, 1, 5〉 = c1〈1, 0, 1〉+ c2〈3, 2, 0〉+ c3〈0, 1, 1〉.

Multiply constants into vectors and add components on the right side to get:

〈−1, 1, 5〉 = 〈c1 + 3c2, 2c2 + c3, c1 + c3〉.

Equate components on both sides to get following 3 equations in 3 unknowns:

−1 = c1 + 3c2;

1 = 2c2 + c3;

5 = c1 + c3.

Subtract the third equation from first to get:

−6 = 3c2 − c3.

Add this equation to the second equation above to get:

−5 = 5c2.

So c2 = −1, implying −6 = −3− c3, or c3 = 3. Finally 5 = c1 + 3, so c1 = 2.

(b) Show that the vector 2i + j − k cannot be expressed as a linear combination of thevectors v1 = i− j , v2 = 3i + k , and v3 = 4i− j + k.

Solution: Suppose there are scalars c1, c2, c3 such that

〈2, 1,−1〉 = c1〈1,−1, 0〉+ c2〈3, 0, 1〉+ c3〈4,−1, 1〉.

Scalar multiply componentwise and add components on the right side to get:

〈2, 1,−1〉 = 〈c1 + 3c2 + 4c3,−c1 − c3, c2 + c3〉.

Equate components on both sides to obtain the following 3 equations in 3 unknowns:

2 = c1 + 3c2 + 4c3;

1 = −c1 − c3;

−1 = c2 + c3.

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Add the first and second equations together to get:

3 = 3c2 + 3c3.

So 1 = c2 + c3. But from the third equation above −1 = c2 + c3. These two equationsare inconsistent and will not have any solutions for c2 and c3. (The graphs of these twoequations in the c2c3-plane are two parallel lines with no intersection point.) Hence 2i+j−kis not a linear combination of v1,v2,v3.

(53) Use a theorem from plane geometry to show that if u and v are vectors in 2-space or3-space, then

||u + v|| ≤ ||u||+ ||v||,

which is called the triangle inequality for vectors. Give some examples to illustratethis inequality.

Solution: Consider the triangle formed by the vectors u, v, and u+v, where u and u+vhave a common initial point, and the initial point of v is the terminal point of u. Let θ bethe angle between the sides defined by u and v. (Note: As vectors, the angle φ betweenu and v is the angle formed between u and v when drawn with common initial point. Sothe angle θ defined above is the supplement of φ, although this won’t matter.)

Apply the Law of Cosines to this triangle to get:

||u + v||2 = ||u||2 + ||v||2 − 2||u|| · ||v|| cos θ. (1)

We now obtain a more useful bound on the right side of the above equation. Since | cos θ| ≤1,

||u||2+||v||2−2||u||·||v|| cos θ ≤ ||u||2+||v||2+2||u||·||v||| cos θ| ≤ ||u||2+||v||2+2||u||·||v||.

So, combining with (1) yields

||u + v||2 ≤ ||u||2 + ||v||2 + 2||u|| · ||v|| = (||u||+ ||v||)2.

Since the terms being squared on both sides are nonnegative, we can take the square rootof each side to obtain the desired inequality:

||u + v|| ≤ ||u||+ ||v||.

For an example of this relation, consider the vectors u = 〈1, 0〉, v = 〈0, 1〉, u + v = 〈1, 1〉 ,so that ||u|| = 1, ||v|| = 1, and ||u+v|| =

√2. The inequality holds since

√2 ≤ (1+1) = 2.

(57) Use vectors to prove that the line segment joining the midpoint of two sides of a triangleis parallel to the third side and half as long.

Solution: Let u, v, w be three vectors that overlay the sides u, v, and w of an arbitrarytriangle, and assume that the terminal point of u is the initial point of v, the terminalpoint of v is the initial point of w, and the terminal point of w is the initial point of u.The midpoints of sides u and v can be represented as:

mu = 0.5u;

mv = u + 0.5v.

The directed line segment from the midpoint of u to the midpoint of v is

mv −mu = 0.5(u + v).

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Now u+v+w = 0 since tracing the three sides of a triangle makes a loop around a closedpath resulting in 0 net displacement. Hence w = −(u + v) and so mv −mu = −0.5w.Since mv −mu is a scalar multiple of w and ||mv −mu|| = 0.5||w||, the line segmentjoining the midpoint of two sides of a triangle is parallel to the third side and half as long.

(58) Use vectors to prove that the midpoints of the sides of a quadrilateral are the vertices ofa parallelogram.

Solution: Let u, v, w , x be four vectors that overlay the sides u, v, w, and x of anarbitrary quadrilateral, and assume that each of the vectors u, v, w , x has terminal pointequal to the initial point of the following vector. The midpoints of the various sides canbe represented as:

mu = 0.5u;

mv = u + 0.5v;

mw = u + v + 0.5w;

mx = u + v + w + 0.5x.

The line segments connecting the midpoints of sides u and v as well as w and x can berepresented as the vector difference of the respective midpoint vectors:

mv −mu = 0.5(u + v);

mx −mw = 0.5(w + x).

Now u + v + w + x = 0 since tracing the four sides of the quadrilateral makes a looparound a closed path resulting in 0 net displacement. Using this to simplify the above twoequations results in mv−mu = −(mx−mw). Therefore these two line segments formed byjoining the midpoints of the quadrilateral are parallel as well as equal in length. A similarargument applies to the line segments formed by joining the midpoints of u and x as wellas v and w respectively, to show that these segments are also parallel and equal in length.Since the opposite sides are parallel and equal in length, the figure is a parallelogram.

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(14) True or False? If a•b = a•c and if a 6= 0, then b = c. Justify your conclusion.

Solution: False. Let a = 〈1, 1, 1〉, b = 〈1, 0, 0〉, and c = 〈0, 1, 0〉. a•b = a•c = 1 anda 6= 0, but b 6= c.

(20) Show that two nonzero vectors v1 and v2 are orthogonal if and only if their directioncosines satisfy:

cosα1 cosα2 + cosβ1 cosβ2 + cos γ1 cos γ2 = 0.

Solution: Consider v1•v2 with v1 and v2 expressed in terms of their direction cosines:

v1•v2 = (||v1||(cosα1i+ cosβ1j+ cos γ1k))•(||v2||(cosα2i+ cosβ2j+ cos γ2k))

= ||v1||||v2||(cosα1 cosα2 + cosβ1 cosβ2 + cos γ1 cos γ2).

If v1 and v2 are orthogonal and nonzero, then v1•v2 = 0, ||v1|| 6= 0, and ||v2|| 6= 0; socosα1 cosα2 + cosβ1 cosβ2 + cos γ1 cos γ2 = 0 from the above equation for v1•v2. On theother hand, if cosα1 cosα2 + cosβ1 cosβ2 + cos γ1 cos γ2 = 0, and v1 and v2 are nonzero,then by the above equation v1•v2 = 0, implying that v1 and v2 must be orthogonal.

(37) Let u and v be adjacent sides of a parallelogram. Use vectors to prove that the diagonalsof the parallelogram are perpendicular if the sides are equal in length.

Solution: Given u and v as the adjacent sides of a parallelogram (drawn with terminalpoint of u equal to the initial point of v), the diagonals of the parallelogram can berepresented as u+ v and u− v. Since a•b = b•a and a•a = ||a||2 for all vectors a and b,the dot product of the diagonals can be written as

(u+ v)•(u− v) = u•u− u•v + v•u− v•v = u•u− v•v = ||u||2 − ||v||2.

So, if ||u|| = ||v||, then (u+v)•(u−v) = 0, implying that these diagonals are perpendicular.

(40) Prove u•v = 14 ||u+ v||2 − 1

4 ||u− v||2.

Solution: since ||a||2 = a•a and b•c = c•b for all a, b, and c,

||u+ v||2 = (u+ v)•(u+ v) = u•u+ u•v + v•u+ v•v = ||u||2 + ||v||2 + 2u•v.

Similarly,||u− v||2 = ||u||2 + ||v||2 − 2u•v.

So

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4||u+ v||2 − 1

4||u− v||2 =

||u||2 + ||v||2 + 2u•v − (||u||2 + ||v||2 − 2u•v)4

=4u•v4

= u•v,

as desired.

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(41) Show that if v1, v2, and v3 are mutually perpendicular nonzero vectors in 3-space, and ifa vector v in 3-space is expressed as:

v = c1v1 + c2v2 + c3v3,

then the scalars c1, c2, and c3 are given by the formulas:

ci = (v•vi)/||vi||2, i = 1, 2, 3.

Solution: For v1, consider:

v•v1 = (c1v1 + c2v2 + c3v3)•v1 = c1v1•v1 + c2v2•v1 + c3v3•v1.

We know that v1•v1 = ||v1||2, and v1•v2 = v1•v3 = 0 because these vectors are mutuallyperpendicular. Substituting these relations into the above equation and solving for c1yields

c1 = (v•v1)/||v1||2.

Similar results hold for v2 and v3.

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(26) Show that in 3-space the distance d from a point P to the line L through points A and Bcan be expressed as

d =|| ~AP × ~AB|||| ~AB||

.

Solution: The distance d is equal to the length of the component of the vector ~APorthogonal to the vector ~AB. With θ as the angle between these two vectors, d can beexpressed as:

d = || ~AP || sin θ =|| ~AB|| · || ~AP || sin θ

|| ~AB||=|| ~AP × ~AB|||| ~AB||

.

(31) What can you say about the angle between nonzero vectors u and v if u•v = ||u× v||?

Solution: Let θ be the angle between the nonzero vectors u and v. Using u•v =||u||||v|| cos θ and ||u× v|| = ||u||||v|| sin θ, the condition u•v = ||u× v|| becomes cos θ =sin θ, or tan θ = 1. This is satisfied if and only if θ = π/4. (Recall that the angle betweentwo vectors must be no greater than π.)

(32) Show that if u and v are vectors in 3-space, then

||u× v||2 = ||u||2||v||2 − (u•v)2.

[Note: This result is sometimes called Lagrange’s identity.]

Solution: If either u or v is zero, then clearly each side of the above equation is zero. Sowe may assume u and v are nonzero. Using ||u × v|| = ||u||||v|| sin θ, sin2 θ = 1 − cos2 θ,and cos θ = u•v/||u||||v||, we can write:

||u× v||2 = ||u||2||v||2 sin2 θ = ||u||2||v||2(1− cos2 θ) = ||u||2||v||2(

1−( u•v||u||||v||

)2).

Multiplying out the right-most portion of the above gives:

||u× v||2 = ||u||2||v||2 − (u•v)2,

as desired.

(38) Prove part (b) of Theorem 13.4.1 for 3× 3 determinants. [Just give the proof for the firsttwo rows.] Then use (b) to prove (a).

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Solution: Part (b) of Theorem 13.4.1 states that “Interchanging two rows of a deter-minant multiplies its value by −1.” We first show that interchanging the first two rowsmultiplies the determinant by −1.∣∣∣∣∣∣

a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ = a1

∣∣∣∣ b2 b3c2 c3

∣∣∣∣− a2 ∣∣∣∣ b1 b3c1 c3

∣∣∣∣ + a3

∣∣∣∣ b1 b2c1 c2

∣∣∣∣= a1(b2c3 − b3c2)− a2(b1c3 − b3c1) + a3(b1c2 − b2c1)

= a1b2c3 − a1b3c2 − a2b1c3 + a2b3c1 + a3b1c2 − a3b2c1

= −(b1(a2c3 − a3c2)− b2(a1c3 − a3c1) + b3(a1c2 − a2c1)

)= −

(b1

∣∣∣∣ a2 a3c2 c3

∣∣∣∣− b2 ∣∣∣∣ a1 a3c1 c3

∣∣∣∣ + b3

∣∣∣∣ a1 a2c1 c2

∣∣∣∣)

= −

∣∣∣∣∣∣b1 b2 b3a1 a2 a3c1 c2 c3

∣∣∣∣∣∣ .Now we consider the case of interchanging the second and third rows. We use the fact thatinterchanging the two rows of a 2× 2 matrix multiplies its determinant by −1, a fact thatis trivial to check. Expanding in terms of its 2× 2 determinants,∣∣∣∣∣∣

a1 a2 a3b1 b2 b3c1 c2 c3

∣∣∣∣∣∣ = a1

∣∣∣∣ b2 b3c2 c3

∣∣∣∣− a2 ∣∣∣∣ b1 b3c1 c3

∣∣∣∣ + a3

∣∣∣∣ b1 b2c1 c2

∣∣∣∣= −a1

∣∣∣∣ c2 c3b2 b3

∣∣∣∣ + a2

∣∣∣∣ c1 c3b1 b3

∣∣∣∣− a3 ∣∣∣∣ c1 c2b1 b2

∣∣∣∣= −

(a1

∣∣∣∣ c2 c3b2 b3

∣∣∣∣− a2 ∣∣∣∣ c1 c3b1 b3

∣∣∣∣ + a3

∣∣∣∣ c1 c2b1 b2

∣∣∣∣)

= −

∣∣∣∣∣∣a1 a2 a3c1 c2 c3b1 b2 b3

∣∣∣∣∣∣ .Finally, notice that interchanging the first and third rows of a matrix can be accomplishedby interchanging rows 1 and 2, then interchanging rows 2 and 3, then interchanging rows 1and 2. Each interchange multiplies the determinant of the original matrix by −1. So threeinterchanges multiplies the determinant by (−1)3 = −1.

Part (a) of Theorem 13.4.1 states “If two rows of a determinant are the same, then thevalue of the determinant is 0.”.

To prove this, let M be a matrix having two rows that are the same. Let M ′ be the matrixobtained by interchanging these two rows. Clearly, M = M ′, so

det(M) = det(M ′).

However, by part (b),det(M ′) = −det(M).

These two equations imply that det(M) = −det(M), which can happen only if det(M) = 0.

(41) Prove: If a, b, c, and d lie in the same plane when positioned with a common initial point,then

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(a× b)× (c× d) = 0.

Solution: Let P be the plane containing a, b, c, and d when positioned with a commoninitial point. Let x = a×b and y = c×d. x must be orthogonal to the plane P containinga and b. Similarly, y is orthogonal to the plane P containing c and d. In order for x and yto both be orthogonal to this same plane P , they must be scalar multiples of one another,so that x = ky for some scalar k. Therefore, x× y = x× (kx) = k(x× x) = 0, since thecross product of any vector with itself is 0. Hence (a× b)× (c× d) = 0.

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(23) Where does the line x = 1 + t, y = 3− t, z = 2t intersect the cylinder x2 + y2 = 16?

Solution: Substitute the parametric equations for x and y into the equation for thecylinder to find the values for t at the intersection point:

x2 + y2 = (1 + t)2 + (3− t)2 = 2t2 − 4t+ 10 = 16.

This implies that t must satisfy t2 − 2t− 3 = (t− 3)(t+ 1) = 0, so t = 3 or t = −1 at theintersection points. Substituting these t values into the equation for the line, we see thatthe points of intersection are (4, 0, 6) and (0, 4,−2). (Note: We can check our result bydirectly verifying that these two points do satisfy the equation for the cylinder.)

(31) Determine whether the points P1(6, 9, 7), P2(9, 2, 0), and P3(0,−5,−3) lie on the same line.

Solution: Points P1, P2, and P3 lie on the same line if and only if the directed linesegments ~P1P2 and ~P1P3 are parallel vectors, which is true if and only if ~P1P2 = c ~P1P3

for some scalar c. Since ~P1P2 = 〈3,−7,−7〉 and ~P1P3 = 〈−6,−14,−10〉, there is no c that

satisfies ~P1P2 = c ~P1P3, so the points P1, P2, and P3 do not all lie on the same line.

(33) Show that the linesL1 : x = 3− t, y = 1 + 2tL2 : x = −1 + 3t, y = 9− 6t.

are the same.

Solution: Represent the lines L1 and L2 via the following vector equations:

L1 : r1 = ro1 + v1t = 〈3, 1〉 + 〈−1, 2〉tL2 : r2 = ro2 + v2t = 〈−1, 9〉+ 〈3,−6〉t.

Since v2 = −3v1, the lines L1 and L2 are parallel. L1 and L2 also share the common point〈−1, 9〉 (when t = 4 in L1 and t = 0 in L2). Because parallel lines can share a commonpoint only if they are the same line, L1 and L2 must be the same line.

(39) Show that the lines

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

are parallel, and find the distance between them.

Solution: Represent the lines L1 and L2 via the following vector equations:

L1 : r1 = ro1 + v1t = 〈2, 0, 1〉 + 〈−1, 2, 1〉tL2 : r2 = ro2 + v2t = 〈1, 3, 5〉+ 〈2,−4,−2〉t.

Since v2 = −2v1, the lines L1 and L2 are parallel.

To find the distance between the lines, use the result from Exercise 26 in §13.4 whichstates that in 3-space the distance d from a point P to the line L through points A and Bcan be expressed as

1

d =|| ~AP × ~AB|||| ~AB||

.

Let A = (2, 0, 1), B = (1, 2, 2) (this corresponds to t = 1 in L1), and P = (1, 3, 5), so

that ~AB = 〈−1, 2, 1〉 and ~AP = 〈−1, 3, 4〉. This gives ~AP × ~AB = 〈−5,−3, 1〉, so that

|| ~AP × ~AB|| =√

52 + 32 + 12 =√

35. || ~AB|| =√

6; so d =√

35/6.

(42) Let L be the line that passes through the point (x0, y0, z0) and is parallel to the vectorv = 〈a, b, c〉 where a, b, and c are nonzero. Show that a point (x, y, z) lies on the line L ifand only if

x− x0a

=y − y0b

=z − z0c

.

These equations, which are called the symmetric equations of L, provide a nonparamet-ric representation of L.

Solution: The line L can be represented parametrically as L : x = x0 + at, y = y0 + bt,z = z0 + ct. Suppose (x, y, z) lies on this line. Then there exists some t1 such thatx = x0 + at1, y = y0 + bt1, and z = z0 + ct1. Solving for t1 in each of these equations gives

t1 =x− x0a

; t1 =y − y0b

; t1 =z − z0c

.

In short,x− x0a

=y − y0b

=z − z0c

,

(and all are equal to t1). This completes the first part of the proof.

To prove the other direction of the if and only if, suppose that (x, y, z) satisfies

x− x0a

=y − y0b

=z − z0c

.

Denote this common value by t0. Then

t0 =x− x0a

; t0 =y − y0b

; t0 =z − z0c

.

Solving for (x, y, z) gives x = x0 + at0, y = y0 + bt0, z = z0 + ct0. Therefore, by lettingt = t0, we see that (x, y, z) lies on the line L : x = x0 + at, y = y0 + bt, z = z0 + ct.

(45) Let L1 and L2 be the lines whose parametric equations are

L1 : x = 1 + 2t, y = 2− t, z = 4− 2tL2 : x = 9 + t, y = 5 + 3t, z = −4− t.

(a) Show that L1 and L2 intersect at the point (7,−1,−2).

Solution: To determine if L1 passes through (7,−1,−2), we look for a t value thatsatisfies the equations for L1 with (x, y, z) replaced by (7,−1,−2). It’s easy to see thatt = 3 does satisfy these equations, so L1 passes through (7,−1,−2) at t = 3. Similarly, wefind that L2 passes through (7,−1,−2) at t = −2.

2

(b) Find, to the nearest degree, the acute angle between L1 and L2 at their intersection.

Solution: L1 is parallel to the vector v1 = 〈2,−1,−2〉 and L2 is parallel to the vectorv2 = 〈1, 3,−1〉. (Note that if the angle θ between v1 and v2 is acute, then θ is the acuteangle between L1 and L2. If θ is obtuse, then the acute angle between L1 and L2 is 180−θ.)Using the dot product we find

v1•v2 = 〈2,−1,−2〉•〈1, 3,−1〉 = 1.

Therefore,1 = ||v1|| · ||v2|| cos θ = 3

√11 cos θ.

Solving for θ givesθ = cos−1(1/3

√11) ≈ 84o.

(c) Find parametric equations for the line that is perpendicular to L1 and L2 and passesthrough their point of intersection.

Solution: The desired line must be perpendicular to v1 and v2 as given above. So thisline is parallel to v1 × v2 = 〈7, 0, 7〉, which is parallel to 〈1, 0, 1〉. Since the desired linepasses through the point (7,−1,−2) (which is the point of intersection of L1 and L2), ithas parametric equations x = 7 + 1 · t, y = −1 + 0 · t, z = −2 + 1 · t.

3


(16) Determine whether the line and plane are parallel, perpendicular, or neither.

(a)x = 3− t, y = 2 + t, z = 1− 3t;

2x + 2y − 5 = 0.

Solution: v = 〈−1, 1,−3〉 is a vector parallel to the given line, and n = 〈2, 2, 0〉 is anormal vector to the given plane. Since v•n = 0, the line is parallel to the given plane.The line is not contained in the plane because there is no value of t that results in valuesfor x, y, z that also satisfy the plane equation.

(b)x = 1− 2t, y = t, z = −t;

6x− 3y + 3z = 1.

Solution: v = 〈−2, 1,−1〉 is a vector parallel to the given line, and n = 〈6,−3, 3〉 is anormal vector to the given plane. Since n = −3v, the line is parallel to the normal of thegiven plane, and so the line is perpendicular to the plane.

(c)x = t, y = 1− t, z = 2 + t;

x + y + z = 1.

Solution: v = 〈1,−1, 1〉 is a vector parallel to the given line, and n = 〈1, 1, 1〉 is anormal vector to the given plane. Since v•n 6= 0 and n is not some multiple of v, the lineis not parallel to the plane and it is not perpendicular to the plane.

(25) Find an equation of a plane that satisfies the stated conditions.

The plane through (1, 2,−1) that is perpendicular to the line of intersection of the planes2x + y + z = 2 and x + 2y + z = 3.

Solution: Let P1 : 2x+y+z = 2 and P2 : x+2y+z = 3 represent the two given planes.The line of intersection of the planes must be perpendicular to their normals n1 = 〈2, 1, 1〉and n2 = 〈1, 2, 1〉, and therefore parallel to n = n1 × n2 = 〈−1,−1, 3〉, which must benormal to the desired plane. (Alternatively, the line of intersection can be found by solvingthe simultaneous equations given by the two planes.) In general, a plane passing through(x0, y0, z0) with normal vector 〈a, b, c〉 has equation a(x− x0) + b(y − y0) + c(z − z0) = 0.In this case the desired plane passes through (1, 2,−1) and has normal vector 〈−1,−1, 3〉.Therefore, −1(x − 1) − 1(y − 2) + 3(z + 1) = 0, or x + y − 3z = 6, is an equation for thedesired plane.

1


The plane through the points P1(−2, 1, 4), P2(1, 0, 3) that is perpendicular to the plane4x− y + 3z = 2.

Solution: Because P1 and P2 lie in the desired plane, ~P1P2 = 〈3,−1,−1〉 is parallel tothat plane. Let n1 = 〈4,−1, 3〉 be a normal vector to the given plane. Then a normal to

the desired plane is any (nonzero) vector perpendicular to both ~P1P2 and n1. Therefore,

n = ~P1P2 × n1 = 〈3,−1,−1〉 × 〈4,−1, 3〉 = 〈−4,−13, 1〉

is a normal to the desired plane. Since this plane passes through the point (1, 0, 3), it hasequation −4(x− 1)− 13y + (z − 3) = 0, or 4x + 13y − z = 1.


The plane whose points are equidistant from (2,−1, 1) and (3, 1, 5).

Solution: The midpoint m between the two given points, P1(2,−1, 1) and P2(3, 1, 5),must lie in the desired plane and is given by m(x, y, z) = 1

2 (x1 + x2, y1 + y2, z1 + z2) =12 (2 + 3,−1 + 1, 1 + 5) = ( 5

2 , 0, 3). The directed line segment ~P1P2 = 〈1, 2, 4〉 must beperpendicular to the desired plane and can be used as a normal vector n for the desiredplane. Therefore, an equation for the desired plane is (x − 5

2 ) + 2y + 4(z − 3) = 0, orx + 2y + 4z = 29/2.


The plane that contains the line x = 3t, y = 1 + t, z = 2t and is parallel to the intersectionof the planes 2x− y + z = 0 and y + z + 1 = 0.

Solution: Let L be the given line, which is parallel to v = 〈3, 1, 2〉. Let P1 : 2x−y+z = 0and P2 : y + z + 1 = 0 represent the two given planes with normals n1 = 〈2,−1, 1〉 andn2 = 〈0, 1, 1〉. The line of intersection of P1 and P2 is parallel to n1 × n2 = 〈−2,−2, 2〉,which is parallel to w = 〈1, 1,−1〉. So a normal n to the desired plane is any (nonzero)vector perpendicular to both v and w. One such vector is

n = v ×w = 〈3, 1, 2〉 × 〈1, 1,−1〉 = 〈−3, 5, 2〉.

Since the desired plane contains the line x = 3t, y = 1+ t, z = 2t, which contains the point(0, 1, 0), the desired plane passes through this point. Therefore, −3(x − 0) + 5(y − 1) +2(z − 0) = 0, or 3x− 5y − 2z = −5, is an equation for the desired plane.

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(18) Identify and sketch the quadric surface

y2 − x2

4− z2

9= 1.

Solution: This is a hyperboloid of two sheets. It has no trace in the xz-plane since

setting y = 0 results in −x2

4 −z2

9 = 1, which has no real solutions. The traces in planesparallel to the xz-plane with |y| ≥ 1 are ellipses. For example, when |y| = 2 the traces aregiven by the following equation:

x2

12+

z2

27= 1.

The two sheets of the hyperboloid open in the positve and negative y-directions withy-intercepts at (0, 1, 0) and (0,−1, 0).

(25) The given equation represents a quadric surface whose orientation is different from thosein Table 13.7.1. Identify and sketch the surface

2y2 − x2 + 2z2 = 8.

Solution: This is a hyperboloid of one sheet. Dividing the above equation by 8 resultsin

y2

4− x2

8+

z2

4= 1.

This hyperboloid opens in the positive and negative x-directions. The trace in the yz-plane (x = 0) is the circle y2 + z2 = 4. The traces in the planes |x| = 2

√2 (parallel to the

yz-plane) are circles given by y2 + z2 = 8.

(35) Identify the surface 9x2 + y2 + 4z2 − 18x + 2y + 16z = 10, and make a rough sketch thatshows its position and orientation.

1

Solution: Completing the squares, we obtain

9(x2 − 2x + 1) + (y2 + 2y + 1) + 4(z2 + 4z + 4) = 10 + 9 + 1 + 16,

which simplifies to9(x− 1)2 + (y + 1)2 + 4(z + 2)2 = 36,

or(x− 1)2

4+

(y + 1)2

36+

(z + 2)2

9= 1.

This is the equation of an ellipsoid centered at (1,−1,−2) and with axes of length 4 in thex-direction, length 12 in the y-direction, and length 6 in the z-direction.

(37) Exercises 37 and 38 are concerned with the ellipsoid 4x2 + 9y2 + 18z2 = 72.

(a) Find an equation of the elliptical trace in the plane z =√

2.

Solution: Substituting z =√

2 into the above equation, subtracting 36 from both sides,

and then dividing both sides by 36 gives x2

9 + y2

4 = 1 as the equation of the elliptical trace.

(b) Find the length of the major and minor axes of the ellipse in part (a).

Solution: The major axis is along the x-direction and of length 2 · 3 = 6. The minoraxis is along the y-direction and of length 2 · 2 = 4.

2

(c) Find the coordinates of the foci of the ellipse in part (a).

Solution: The ellipse x2/a2 +y2/b2 = 1 with a > b has foci at coordinates (±c, 0) wherec =√a2 − b2. In this case, c =

√9− 4 =

√5. So the coordinates (in 3-space) of the foci

are (√

5, 0,√

2) and (−√

5, 0,√

2).

(d) Describe the orientation of the focal axis of the ellipse in part (a) relative to thecoordinate axes.

Solution: The focal axis is parallel to the x-axis and in the plane z =√

2.

(39) Exercises 39-42 refer to the hyperbolic parabloid z = y2 − x2.

(a) Find an equation of the hyperbolic trace in the plane z = 4.

Solution: Substituting z = 4 into the above equation and dividing both sides of equation

by 4 results in y2

4 −x2

4 = 1 for the hyperbolic trace.

(b) Find the vertices of the hyperbola in part (a).

Solution : The vertices are at y = 2 and y = −2, so (in 3-space) the coordinates are(0, 2, 4) and (0,−2, 4).

(c) Find the foci of the hyperbola in part (a).

Solution : The hyperbola y2/b2 − x2/a2 = 1 has foci at coordinates (0,±c) wherec =√a2 + b2. So in this case the coordinates (in 3-space) of the foci are (0,±c, 4) where

c =√a2 + b2 =

√4 + 4 = 2

√2.

(d) Describe the orientation of the focal axis of the hyperbola in part (a) relative to thecoordinate axes.

Solution: The focal axis is parallel to the y-axis and in the plane z = 4.

(45) In Exercises 45 and 46, find an equation for the surface generated by revolving the curveabout the axis.

y = 4x2 (z = 0) about the y-axis.

Solution: Revolving any point (x0, y0) on the given curve about the y-axis results ina circle in the plane y = y0 with radius equal to x0 =

√y0/4, and centered at (0, y0, 0).

This circle consists of the set of points {(x, y, z)∣∣∣ y = y0, x

2 + z2 = (√y0/4)2 = y0/4}.

The generated surface consists of all such circles where y0 ≥ 0. So the generated surface

consists of the set of points {(x, y, z)∣∣∣ x2 + z2 = y/4, y ≥ 0}, which (because squares are

3

always nonnegative) can be more concisely written {(x, y, z)∣∣∣ 4x2 + 4z2 = y}. In short,

the surface has equation 4x2 + 4z2 = y.

(49) If a spherex2

a2+

y2

a2+

z2

a2= 1

of radius a is compressed in the z-direction, then the resulting surface, called an oblatespheroid, has an equation of the form

x2

a2+

y2

a2+

z2

c2= 1

where c < a. Show that the oblate spheroid has a circular radius a in the xy-plane andan elliptical trace in the xz-plane with major axis of length 2a along the x-axis and minoraxis of length 2c along the z-axis.

Solution: In the xy-plane z = 0, so the spheroid equation becomes:

x2

a2+

y2

a2= 1,

or simplyx2 + y2 = a2.

4

This is the standard equation of a circle with radius a in the xy-plane.

In the xz-plane y = 0, so the spheroid equation becomes:

x2

a2+

z2

c2= 1.

This is the standard equation for an ellipse in the xz-plane with major axis of length 2aalong the x-axis and minor axis of length 2c along the z-axis.

5


(18) In Exercises 15-22, an equation is given in cylindrical coordinates. Express the equationin rectangular coordinates and sketch the graph.

z = r cos θ.

Solution: Since x = r cos θ, the equation becomes z = x. In the xz-plane, this is the linez = x. In 3-space, this is a plane defined by x− z = 0 with normal vector n = 〈1, 0,−1〉.

(20) In Exercises 15-22, an equation is given in cylindrical coordinates. Express the equationin rectangular coordinates and sketch the graph.

r = 2 sec θ.

Solution: r = 2 sec θ = 2/ cos θ, which can be rewritten as r cos θ = 2. Since x = r cos θ,the equation in rectangular coordinates becomes x = 2, which is obviously a plane.

(28) In Exercises 23-30, an equation is given in spherical coordinates. Express the equation inrectangular coordinates and sketch the graph.

ρ sinφ = 1.

1

Solution: ρ sinφ = r =√x2 + y2 so the equation becomes

√x2 + y2 = 1, which is

equivalent to x2 + y2 = 1. In the xy-plane, this is an equation of the unit circle, which in3-space extrudes along the z-direction to a cylinder of radius 1.

(30) In Exercises 23-30, an equation is given in spherical coordinates. Express the equation inrectangular coordinates and sketch the graph.

ρ− 2 sinφ cos θ = 0.

Solution: Using the fact that sinφ cos θ = x/ρ, this equation becomes ρ − 2xρ = 0.

Multiplying by ρ and using ρ2 = x2 + y2 + z2, this becomes x2 + y2 + z2 − 2x = 0.Completing the square for x, this becomes (x− 1)2 + y2 + z2 = 1. This is the equation ofa sphere with radius 1 centered at (1, 0, 0).

(40) In Exercises 31-42, an equation of a surface is given in rectangular coordinates. Find anequation of the surface in (a) cylindrical coordinates and (b) spherical coordinates.

x2 + y2 − z2 = 1.

Solution (a): Since r2 = x2 + y2, the equation becomes r2 − z2 = 1 in cylindricalcoordinates. This is an equation for a hyperboloid of one sheet.

2

Solution (b): Since r2 = ρ2 sin2 φ and z = ρ cosφ, the equation becomes ρ2 sin2 φ −ρ2 cos2 φ = 1. Using cos 2φ = cos2 φ− sin2 φ, this simplifies to ρ2 cos 2φ = −1.

(42) In Exercises 31-42, an equation of a surface is given in rectangular coordinates. Find anequation of the surface in (a) cylindrical coordinates and (b) spherical coordinates.

x2 + y2 + z2 = 2z.

Solution (a): Since r2 = x2 + y2, the equation becomes r2 + z2 = 2z. Subtracting 2zfrom both sides and completing the square for z gives r2 + (z − 1)2 = 1. This is a sphereof radius 1 centered at (0, 0, 1) (this coordinate is valid either in cylindrical or rectangularcoordinates).

Solution (b): Since ρ2 = x2+y2+z2 and z = ρ cosφ, this equation becomes ρ2 = 2ρ cosφ,which simplifies to ρ = 2 cosφ.

(43) In Exercises 43-46, describe the region in 3-space that satisfies the given inequalities.

r2 ≤ z ≤ 4.

Solution: The constraint r2 ≤ z implies that the points must be on or above (within)the circular parabloid z = r2 = x2 + y2. Therefore the region in 3-space consists of thosepoints that lie on or directly above (within) the circular parabloid z = r2 = x2 + y2 andon or below the plane z = 4. The projection of this region onto the xy-plane is the diskgiven by the inequality x2 + y2 ≤ 4.

(44) In Exercises 43-46, describe the region in 3-space that satisfies the given inequalities.

0 ≤ r ≤ 2 sin θ, 0 ≤ z ≤ 3.

Solution: Using sin θ = y/r, the equation r = 2 sin θ becomes r = 2y/r. Multiplyingboth sides by r, using r2 = x2 + y2, subtracting 2y from both sides, and completing thesquare for the y term, this becomes x2 + (y− 1)2 = 1, which is a circle of radius 1 centeredat (0, 1) in the xy-plane. Since r ≤ 2 sin θ, the region in 3-space must be those points(x, y, z) such that (x, y) lies on or within the circle x2 + (y − 1)2 = 1 and 0 ≤ z ≤ 3. This

3

is a cylinder of radius 1 and height 3, parallel to the z-axis and with base centered at (0, 1)in the xy-plane.

(49) The accompanying figure shows a right circular cylinder of radius 10 cm spinning at 3revolutions per minute about the z-axis. At time t = 0, a bug at the point (0, 10, 0) beginswalking straight up the face of the cylinder at the rate of 0.5 cm/min.

(a) Find the cylindrical coordinates of the bug after 2 min.

Solution: Since the bug will stay on the surface of the cylinder, r = 10 cm for the bugeven when the cylinder rotates. After 2 min., the cylinder will undergo (3rev/min)(2min)=6revolutions, so the θ coordinate will be the same as its initial value. Since the bug startson the y-axis, θ = π

2 radians. Since the bug moves in the vertical direction at a rate of0.5 cm/min., z =(0.5cm/min)(2 min)=1 cm. Therefore the cylindrical coordinates are (10cm,π2 rad,1 cm).

(b) Find the rectangular coordinates of the bug after 2 min.

Solution: Since the 6 revolutions will bring the bug back to the same x and y coordinatesas it initially starts from, x = 0 cm and y = 10 cm. The z coordinate will be the same asin (a): z = 1 cm. Therefore the rectangular coordinates are (0, 10, 1) cm.

(c) Find the spherical coordinates of the bug after 2 min.

Solution: Since ρ =√r2 + z2, the value of ρ can be calculated by converting the

cylindrical coordinates from (a): ρ =√

(10)2 + 12 =√

101 cm. The θ value will be the samein spherical and cylindrical coordinates: θ = π

2 radians. The value of φ can be calculatedby converting from cylindrical coordinates using the formula tanφ = r/z = 10/1 = 10; soφ = tan−1(10) ≈ 1.47 radians. Therefore, the spherical coordinates are (

√101 cm, π

2 rad,1.47 rad).

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