Calculus III
Practice Problems 1: Answers
2I J.
Answer. Let α be the angle W makes with the horizontal. We have tan α 1 2, so α 26 6 . Thus the
vector V makes an angle of 18 4 with the horizontal, and we can take
V cos 18 4 I sin18 4 J 949I 316J
Here’s another solution. Since W has the same length as W and is orthogonal to W, we can take V as the
diagonal of the rectangle spanned by W and W (see the figure).
Thus, we take V W W 3I J. Note that the first solution is the unit vector in the direction of
W W .
1. Find the components of a vector V making an angle of 45 with W
V
V
PSfrag replacements
45
V
V
2. Show that the diagonals of a rhombus intersect at right angles. (A rhombus is a parallelogram with all
sides of the same length).
W V W. To show the diagonals
V W V W V V V W W V W W V W 0
Answer. Let V and W be the sides of the rhombus. The diagonals are V
are orthogonal, we calculate
2
2
since V and W have the same length.
3. A river ferry runs at a speed of 6 knots, across a river with a current of 2 knots. Assume that the river has
shores which are parallel straight lines. In what direction should the barge head in order to cross the river
perpendicular to the shores?
Answer. The ferry should head upstream at some angle α to the vertical, so that the resultant of the ferry
velocity V and the stream velocity C is directly vertical. (See the figure). Since C is horizontal of magnitude
2, C 2I, and since V
6, V
6 sin α I 6 cos α J. For V C to be a multiple of J, we must have
6 sin α 2, so sin α 1 3, and α 19 47 .
V
C
PSfrag replacements
V
C
4. Find the area of the parallelogram determined by the vectors V 6I
coordinates of the vertex of this parallelogram farthest from the origin?
Answer. The area is det V W
64
7 3
45. The vector V
to the furthest vertex, so that vertex has coordinates (9,-3).
7J W W
9I 3I
4J. What are the
3J goes from the origin
5. A plane flies at a ground speed of 480 mph. The jet stream comes from 30 north of west at 60 mph. In
what direction should the pilot direct the plane so as to be heading directly east? How many miles will the
plane cover in 2 hours?
Answer. Let W be the velocity vector of the wind. Then, by the figure, we have
W
60
3
I
2
1
J
2
Let V aI bJ be the velocity vector of the plane (relative to ground). The information we have is that
V 480, and the sum V W point in the direction of I, so its J-component is zero. This gives the equation
b 30 0, so b 30. Then a 2 302 4802 , so a 479 1. Then
V
479 1I
30J
and V
W
479 1 30 3 I 531J
In two hours the plane travels 1062 miles.
PSfrag replacements
V
6. Let V
with V.
2I
C
C
V
V
W
3J. Find the vector W to the left of V, of the same length as V, and making an angle of 60
Answer. Recall that, for any vector V, V is perpendicular to, and left of V, and of the same length. Thus,
any vector of the form W cos α V sin α V has the same length as V and V and makes an angle α with
V.
PSfrag replacements
V
C
C
V
V
W
V
W
α
V
We can see this from the figure above, but it also can be computed:
2 cos α sin α V V sin α V V since V V 0, and
W V V cos α W V cos α so α is the angle between W and V. In our case, V 3I 2J and α π 3, so the answer is
π
π
2
3
3
3 2 3
I
J
W cos V sin V
3
3
2
2
W 2
cos2 α V 2
2
2
2
2
7. Find a base L M (in a base, the vectors are of length 1 and orthogonal) so that L is on the line y 5x and
M is left of L. Any vector X xI yJ can be written in the form X uL vM. Find u v as functions of
x y.
Answer. A unit vector on the line y
For X
xI yJ
uL
I 5J 26. Thus
I 5J
L M L 5I26 J 26
5x is L
vM, we have
u
X L x 5y
26
v X M 5x26 y 7 L : x 2y 11?
Answer. Let V be the vector on the y-axis joining L to L . Then V 2J, and the distance we seek is the
length of the projection of V in the direction orthogonal to the lines. The vector N I 2J is orthogonal to
the lines, thus
d V N 4 N
5
8. What is the distance between the two parallel lines L 1 : x
1
2
2y
2
y 10.
"!
Answer. Since Q(5,0) is on the line L, the vector PQ 2I J joins P to a point in Q. Thus, the distance
#
!
sought is the length of the projection of PQ in the direction normal to L. The vector N 2I 2J points in
this direction. Thus
"! PQ N 2I J 2I 2J 1
d
N
2 2
2
9. Find the distance from the point P(3,-2) to the line L : 2x
$ 8 % 6(
PSfrag replacements
V
C
C
V
V
W
V
V
W
α
$ 5 % 0(
$ 2 %'& 6(
L
10. Find a point x y on the line 2x y 10 such that the triangle with vertices (0,0), (5,0), x y has area
equal to 15. How many such points are there?
Answer. Let P be the point; since it is on the given line, y 2x 10. The hypothesis tells us that the vectors
xI 2x 10 J and 5I must span a parallelogram of area 30. Thus
det xI 2x 10 J 5I ) 30 which gives us 5 2x 10 30 or, the two equations 2x 10 6 2x 10 6. Thus the possible solutions for P are (8,6), (2,-6).