LB 220 Homework 2 (due Tuesday, 01/22/13)
Directions. Please solve the problems below. Your solutions must begin
with a clear statement (or re-statement in your own words) of the problem.
You solutions should be clear, legible, and demonstrate at minimum partial
progress towards a complete solution to the problem. Please refer to the
syllabus for the policy on grading (communication, completeness, and
correctness) and late homework (homework is due at the start of class, late
homework is assessed a 20% penalty if submitted within the next 48 hours)
Collaboration. I encourage you to discuss the homework problems with
your classmates. However, each student must write and submit his or her
own homework solutions.
Calculators. You can use calculators to determine a numerical
approximation to an answer to an application question, but you should use
exact values until the very last step in the problem. Calculators are not,
however, permitted on any quizzes or exams.
1. A boat sails south with the help of a wind blowing in the direction
S36◦ E with a magnitude of 400 lb. Find the work done by the wind
as the boat moves 120 ft.
−−→
Solution: The equation for work W is W = F · P Q, where F is the
force applied to an object which moves from P to Q. The boat is
moving moving from an initial point, which may be chosen to be the
origin O(0, 0) to a terminal point Q(0, −120) which is 120 ft due
south. The magnitude of the force applied is 400 lb (the “pound”
here refers to units of force– the pound-force lbf , to be precise; for
more on this, please read the solution to exercise #5 on the first
homework.) Using the physicists’ definition of the scalar product, we
have that
−−→
Work = kFkkP Qk cos 36◦ = (400)(120) cos 36◦ ≈ 38, 833 ft · lb.
The answer in the metric system, using 400 lb ≈ 4.45 N and
1 ft ≈ 0.3048 m, is approximately equal to 5, 2671 N · m or,
equivalently, 5, 2671 J.
2. Which of the following expressions is meaningful? If it is, state
whether the result is a vector or a scalar; if it is not, explain why.
(a)
(b)
(c)
(d)
(e)
(f)
a · (b × c)
a × (b · c)
a × (b × c)
a · (b · c)
(a · b) × (c · d)
(a × b) · (c × d)
Solution:
(a) is meaningful, the result is a scalar
(b) is meaningless since the computation involved the vector product
of a vector and a scalar
(c) is meaningful, the result is a scalar, though be careful with the
vector product– it is not associative
(d) is meaningless since the computation involves the scalar product
of two scalars; however, if all three vectors are 1-dimensional vectors
(i.e. vectors in R1 , then the expression is meaningful and simply is
the product of these three numbers.
(e) is meaningless since the computation involves the vector product
of two scalars
(f) is meaningful, the result is a scalar
3. Determine the two unit vectors which are orthogonal to the plane
through the points P (1, 0, 1), Q(−2, 1, 3), and R(4, 2, 5). Then
compute the area of triangle 4P QR.
−−→ −→
Solution: The vector P Q × P R is orthogonal to the plane containing
4P QR:
i j k
−−→ −→ P Q × P R = −3 1 2 = h0, 18, −9i.
3 2 4
The magnitude of this vector is the area of the parallelogram
spanned by the two vectors; half this value is the area of the triangle.
So, the area of the triange is equal to
√
1p 2
1p
9 5
2
2
2
2
0 + 18 + 9 =
0+4·9 +9 =
.
2
2
2
The two unit vectors orthogonal to the plane are
1
2
1
√ h0, 18, −9i = h0, √ , − √ i and its negative.
9 5
5
5
4. Prove the following statement: if P is a point not on a line L that
passes through the points Q and R, then the distance from P to L is
−−→
−−→
equal to ka × bk/kak, where a is represented by QR and b by QP .
Solution: The distance d from P to L is equal to the length of a line
segment joining P to L so that the segment is orthogonal to L.
Suppose this line segment meets L in a point called N . Let θ be the
−−→
measure of the angle ∠P QN . Then d = kQP k sin θ. (There are two
cases: either the angle is acute or the angle is obtuse, but since
sin (π − θ) = sin θ both cases yield the same equation.) Next cleverly
multiply by 1:
−−→
−−→ −−→
−−→
kQRk
kQP × QRk
d = kQP k sin θ −−→ =
,
−−→
kQRk
kQRk
the last inequality following from the physicists’ definition of the
vector product.
There is one small tecnical point: what do we do if N = P ? In this
case, we simply use R in place of P and everything works out in an
analogous way.
5. Find a vector equation and parametric equations for the line of
intersection of the planes x + 2y + 3z = 1 and x − y + z = 4.
Solution: There are several approaches to this problem. Here is one
approach. If P (0, y, z) is a point common to both planes, then
2y + 3z = 1 and −y + z = 4. Multiplying the second equation by 2,
adding the result to the first equation, and solving for z yields
z = 9/5. Substituting back in to either equation yields y = −11/5.
Therefore, P (0, −11/5, 9/5) is a point in the intersection of the two
planes. We could do this again starting with some point Q(1, y, z) so
that 1 + 2y + 3z = 1 and 1 − y + z = 4. Solving this system of two
equations reveals that Q(1, −9/5, 6/5). Therefore, a vector equation
−−→
−−→
of the line of intersection is r = OP + tP Q:
r = h0, −11/5, 9/5i + th1, 2/5, −3/5i,
which is equivalent to x = t, y = −11/5 + 2t/5, z = 9/5 − 3t/5.
Another approach is to stop after finding the point P and then use
the vector product of normal vectors to the planes to determine a
vector parallel to the line of intersection:
i j k
v = 1 2 3 = h5, 2, −3i
1 −1 1 −−→
So, the line has equation r = OP + tv:
r = h0, −11/5, 9/5i + th5, 2, −3i,
which is equivalent to x = 5t, y = −11/5 + 2t, z = 9/5 − 3t.
A third approach is to set x = t and to solve the pair of equations
t + 2y + 3z = 1 and t − y + z = 4 for y and z in terms of t. As before,
mulitply the second equation by 2 and add it to the first. This yields
3t + 5z = 9. Therefore, z = −3t/5 + 9/5. Substiting back into either
equation yields y = 2t/5 − 11/5.
It may seem odd that there can be two such very different looking
equations for the same line. Parametrizations are not unique. We
can check that these are indeed the same line by checking whether
−−→
−−→
P Q and v are parallel. Indeed they are: 5P Q = v.
6. Determine an equation of the plane which contains the line x = 1 + t,
y = −2 + t, z = −t and which is parallel to the plane x + y + 2z = 5.
Solution: The approach is to use the normal vector h1, 1, 2i of the
plane and any point on the line. For instance, if t = 0, then
P (1, −2, 0) lies on the line. Therefore, the following is an equation of
a plane parallel to the given plane and containing P :
(x − 1) + (y + 2) + 2z = 0.
Finally, we need to verify that the line is indeed contained in this
plane. You can either substitute in the equations:
(1 + t − 1) + (−2 + t + 2) + 2(−t) = 0
and check it really equals zero. Or you might verify that the vector
h1, 1, −1i which is parallel to the line is orthogonal to the normal
vector h1, 1, 2i:
h1, 1, −1i · h1, 1, 2i = 0.
7. Determine the cosine of the angle between the planes
5x + 2y + 3z = 0 and x − y + z = 2.
Solution: The cosine of the angle between the planes is the same as
the cosine of the angle between a pair of normal vectors, each
corresponding to one of the planes.
h5, 2, 3ih1, −1, 1i = 5 − 2 + 3 = 6
On the other hand, using the physicists’ definition of the dot
product, the above is equal to
√
√
25 + 4 + 9 1 + 1 + 1 cos θ.
Therefore cos θ =
√6
114
√
=
114
19 .
8. Let L1 be the line through the origin and the point (1, 1, 1). Let L2
be the line through the points (2, 3, 4) and (−1, −1, 0). Determine
the distance between L1 and L2 .
Solution: The strategy is as follows. First, observe that the shortest
line segment joining a point of L1 to a point of L2 must meet both
lines at right angles. So, we first find a vector orthogonal to both
lines by using the vector product:
i j k
n = 1 1 1 = h0, −1, 1i.
3 4 4 Next, write an equation for the plane containing L1 and normal to n.
For instance, using the point O(0, 0, 0) of L1 , the following is such an
equation:
−y + z = 0.
Finally, choose any point of L2 , for instance P (2, 3, 4) and compute
the distance from P to the plane above. This distance is the same as
the distance between the two lines. (This may not be clear, but if
you try visualizing this with a notebook as a prop for a plane and a
pair of pens as props for the lines, you will soon be convinced this is
correct.) Using the formula from from the book, this distance d
satisfies
−−→
1
|OP · n|
d=
=√ .
knk
2
Remark: Here is a nice formula that I learned from a student: if a is
a vector parallel to a line L and b a vector parallel to a line M and if
P is a point on L and Q is a point on M , then if L and M are skew
lines then the distance between L and M is given by
−−→
|P Q · (a × b)|
.
ka × bk
The proof follows the same outline as in the problem above. You
need to be careful with such a formula, however: it does not give the
correct answer if the line are not skew. For example, if the lines are
parallel, then the above quantity is undefined (since the cross
product in the denominator is zero). Similarly, if the lines intersect
then their distance is zero, but the formula above is non-zero.
9. Sketch the traces of the surface x2 + y 2 − z 2 = 1 and explain why the
surface looks like a hyperboloid of one sheet. If the equation is
changed to x2 − y 2 + z 2 = 1, how is the surface changed? What if it
is changed to x2 + y 2 + 2y − z 2 = 0?
Solution: The trace of the surface in the plane
√ z = c has equation
x2 + y 2 = 1 + c2 , which is a circle of radius 1 + c2 .
The curves x2 + y 2 = 1 + c2 for c = 0, 1, 2, 3, 4.
The trace of the surface in yz-plane has equation y 2 − z 2 = 1, which
is a hyperbola.
The curve y 2 − z 2 = 1.
The trace of the surface in the xz-plane has equation x2 − z 2 = 1,
which is a hyperbola. From this we conclude that the surface is a
hyperboloid. It is a hyperboloid of one sheet because for every plane
z = c, the trace is a circle. If it were a two-sheeted hyperboloid, then
one of these planes would not intersect the surface at all.
If the equation is changed to x2 − y 2 + z 2 = 1, then the roles of y and
z have been interchanged. This is realized by a reflection in the plane
y = z in R3 . So, the resulting surface has the same shape (it is still a
one-sheeted hyperboloid), but the y-axis is now the radial axis.
If the equation is changed to x2 + y 2 + 2y − z 2 = 0, then the
hyperboloid is translated from being centered at the origin to being
centered at the point (0, −1, 0). This can be seen by completing the
square: x2 + (y + 1)2 − z 2 = 1.
10. Sketch the traces of the surface −x2 − y 2 + z 2 = 1 and explain why
the surface looks like a hyperboloid of two sheets. If the equation is
changed to x2 − y 2 − z 2 = 1, how is the surface changed? What if it
is changed to x2 + 2x − y 2 − z 2 = 0?
Solution: The trace of the surface in the plane
√ z = c has equation
x2 + y 2 = c2 − 1, which is a circle of radius c2 − 1 if c2 > 1. If
c2 = 1, then the trace is a single point. if c2 < 1, then the trace is
empty. For this reason this will be a hyperboloid of two-sheets. As in
the problem above, the traces in the xz- and yz-planes are
hyperbolas.
If the equation is changed to x2 − y 2 − z 2 = 1, then the roles of x and
z have been interchanged. This is realized by a reflection in the plane
x = z in R3 . So, the resulting surface has the same shape (it is still a
two-sheeted hyperboloid), but the x-axis is now the radial axis.
If the equation is changed to x2 + 2x − y 2 − z 2 = 0, then the
hyperboloid in the previous paragraph (having the x-axis as its radial
axis) is translated from being centered at the origin to being centered
at the point (−1, 0, 0). This can be seen by completing the square:
(x + 1)2 − y 2 − z 2 = 1. So, for instance, the two points at the base of
the sheets are (−1, 0, 0) and (1, 0, 0).