Quiz 1 Practice Problems
Practice problems are similar, both in difficulty and in scope, to the type of
problems you will see on the quiz. Problems marked with a ? are ‘‘for your
entertainment’’ and are not essential.
SUGGESTED REFERENCE MATERIAL:
As you work through the problems listed below, you should reference sections
11.1 and 11.2 of the recommended textbook (or the equivalent section in your
alternative textbook/online resource) and your lecture notes.
EXPECTED SKILLS:
• Be able to determine the location of a point in space using rectangular
coordinates.
• Be able to find the distance between and the midpoint of the line
segment connecting two points in space.
• Know the standard equation of a sphere and be able to find the center
and radius of a sphere.
• Be able to sketch cylindrical surfaces.
• Be able to perform arithmetic operations on vectors and understand
the geometric consequences of the operations.
• Know how to compute the magnitude of a vector and normalize a
vector.
• Be able to use vectors in the context of geometry and force problems.
1
PRACTICE PROBLEMS:
Problems 1-3 refer to the rectangular box shown below. The base
of the rectangular box is the xy-plane.
z
1
1
1
2
2
3
B
4
3
5
x
y
A
C
1. Find the coordinates of the eight corners of the box.
2. Find the midpoint of the diagonal AB.
3. Consider the triangle with vertices A, B, and C.
(a) Compute the length of each of the three sides.
(b) Verify that 4ABC is a right triangle.
(c) Compute the angle between AB and AC.
4. Consider the triangle with vertices A(5, –2, –1), B(7, 0, 3), and C(9, –4, 1).
(a) Show that the triangle is an equilateral triangle.
(b) Compute the area of the triangle.
√ [Hint: The area of an equilateral
3 2
triangle with side length s is
s .]
4
2
5. Find an equation of the sphere whose center is (3, 0, 2) and which has a
diameter of 6.
6. Find an equation of the sphere whose center is (4, 2, –1) and which
passes through the origin.
7. Find an equation of the sphere which contains points A(1, 3, 2) and
B(4, 3, 7) and such that the distance between A and B is equal to the
diameter of the sphere.
8. Does the origin lie inside the sphere (x − 1)2 + (y + 2)2 + (z + 3)2 = 13?
Justify your answer.
9. Consider the cube with center at the origin which has edges of length 2
that are parallel to the coordinate planes.
(a) Find the equation of the sphere which is inscribed in this cube.
(b) Find the equation of the sphere which is circumscribed around the
cube.
10. Find equations of the tangent spheres of equal radii whose centers are
(2, 3, 1) and (5, –3, 2), respectively.
11. Sketch the following surfaces in space
(a) 3x + 4y = 12
x2 y 2
+
=1
4
9
(c) z = x2
(b)
(d) z = ey
3
12. Describe all points in space whose coordinates satisfy the inequality
x2 + z 2 − 4x − 8z + 13 > 0.
13. Consider the surface x2 + y 2 + z 2 − 4x − 12y − 8z = k, where k is a real
number. For which values of k will the surface be a sphere?
14. Consider the points P1 (2, 3) and P2 (5, –1). Find the components of the
−−→
−−→
vector P1 P2 . Sketch P1 , P2 , P1 P2 , and an equivalent vector with its
initial point at the origin.
15. Find the components of vector v and sketch an equivalent vector with
its initial point at the origin.
y
4
3
v
2
1
1
2
3
4
4
5
6
7
8
9
x
16. Sketch the vector u + v + w and express it in component form.
y
5
v
4
w
3
2
u
1
1
2
3
4
5
6
7
8
9
x
−
−
−
17. The figure below is a parallelogram. Express →
w in terms of →
u and →
v.
18. Consider the points P1 (1, 2, 3) and P2 (5, 4, 6). Find the components of
−−→
the vector P1 P2 .
19. Let u = 3i + 2j − k, v = –2i + 4j + 3k, and w = 7i + 4j + k. Compute
each of the following:
(a) 2u − 3w
(b) ||u + w||
5
(c) ||u|| + ||v||
(d) ||2u||
1 v (e) ||v|| 20. For each of the following, find a vector which satisfies the given conditions.
(a) A unit vector which is in the opposite direction of v = 3i + 4j.
(b) A unit vector which is in the same direction as the vector from
P1 (1, 0, 5) to P2 (3, –1, 2).
(c) A vector which is in the opposite direction of v = h1, 2, 3i and
whose magnitude is half that of v.
(d) A vector which is in the same direction of w = i − 2j + 3k and
√
which has a length of 5.
(e) A vector in 2-space which makes an angle of θ = π/6 with the
positive x-axis and which has a magnitude of 4.
(f) A vector in 2-space which makes an angle of θ = 210◦ with the
positive x-axis and which has a length of 2.
21. Find the value(s) of a so that the vectors v = ha2 , 6i and w = h4a, 2i
are parallel.
6
−
−
22. Vectors →
v and →
w , shown below, are unit vectors. Find the components
→
−
→
−
of v + w .
23. For each of the following, find the magnitude of the resultant force and
the angle that it makes with the positive x-axis
7
? Each edge of a cubical box has length 1 m. The box contains nine spherical
balls with the same radius r. The center of one ball is at the center of the cube
and it touches the other eight balls. Each of the other eight balls touches
three sides of the box. Thus, the balls are tightly packed in the box. Find r.
? A plane is capable of flying at a speed of 180 km/h in still air. The pilot takes
off from an airfield and heads due north according to the plane’s compass.
After 30 minutes of flight time, the pilot notices that, due to the wind, the
plane has actually traveled 80 km at an angle 5◦ east of north.
(a) What is the wind velocity?
(b) In what direction should the pilot have headed to reach the intended
destination?
8
SOLUTIONS
1. Notice that the box is the set of all points (x, y, z) such that 1 ≤ x ≤ 3,
2 ≤ y ≤ 5, and 0 ≤ z ≤ 1. This yields the eight following corners:
(1, 2, 0), (1, 2, 1), (1, 5, 0), (1, 5, 1), (3, 2, 0), (3, 2, 1), (3, 5, 0), (3, 5, 1).
(Observe that these are all the possible ways to make an ordered triple
by taking an endpoint from the x-interval, then the y-interval, and
then the z-interval. As there are 2 choices for each of the 3 coordinates,
it makes since that there are 23 = 8 corners.)
2. A is the point (3, 2, 0) and B is the point (1, 5, 1), so the midpoint of
the diagonal AB is
3+1 2+5 0+1
,
,
2
2
2
=
7 1
2, ,
2 2
.
3. We already identified A and B. C is the point (1, 5, 0).
(a) Use the distance formula:
AB =
p
p
√
√
(1 − 3)2 + (5 − 2)2 + (1 − 0)2 = (–2)2 + 32 + 12 = 4 + 9 + 1 = 14
AC =
BC =
p
√
√
(1 − 3)2 + (5 − 2)2 + (0 − 0)2 = 4 + 9 = 13
p
p
(1 − 1)2 + (5 − 5)2 + (0 − 1)2 = 02 + 02 + (–1)2 = 1
9
(b) Since AC 2 + BC 2 = 13 + 1 = 14 = AB 2 , it follows by the
Pythagorean theorem that 4ABC is a right triangle.
(c) Since 4ABC is a right triangle, we know that
√
13
AC
adjacent
=
=√
cos(∠BAC) =
hypotenuse
AB
14
p
and so ∠BAC = cos–1 ( 13/14).
4. (a) In order for 4ABC to be equilateral, we need AB = AC = BC.
Indeed:
AB =
p
√
√
√
(7 − 5)2 + (0 − (–2))2 + (3 − (–1))2 = 22 + 22 + 42 = 4 + 4 + 16 = 24
AC =
p
p
√
(9 − 5)2 + (–4 − (–2))2 + (1 − (–1))2 = 42 + (–2)2 + 22 = 24
BC =
p
(9 − 7)2 + (–4 − 0)2 + (1 − 3)2 =
p
√
22 + (–4)2 + (–2)2 = 24.
(b) Using the formula in the hint, the area of 4ABC is
√
√
√
3 √ 2
3
( 24) =
· 24 = 6 3.
4
4
5. Since the diameter of a sphere is twice its radius, the sphere has a radius
of 3, and so its equation is (x − 3)2 + y 2 + (z − 2)2 = 9.
10
6. The equation of the sphere is (x − 4)2 + (y − 2)2 + (z + 1)2 = r2 , with
r yet to be determined. But since the origin is a point on the sphere,
plugging zero into x, y, and z will satisfy the above equation. Thus
(0 − 4)2 + (0 − 2)2 + (0 + 1)2 = r2 ⇒ 16 + 4 + 1 = r2
and so the sphere has equation (x − 4)2 + (y − 2)2 + (z + 1)2 = 21.
7. Since AB is a diameter of the sphere, the center of the sphere is the
midpoint of AB, which is
1+4 3+3 2+7
,
,
2
2
2
=
5
9
, 3,
2
2
.
The radius is half the length of the diameter, which is
1p
1√ 2
1√
1√
1
AB =
(4 − 1)2 + (3 − 3)2 + (7 − 2)2 =
3 + 02 + 52 =
9 + 25 =
34.
2
2
2
2
2
Hence the equation of the sphere is (x − 5/2)2 + (y − 3)2 + (z − 9/2)2 =
√
( 34/2)2 = 34/4 = 17/2.
8. No. The distance between the origin (0, 0, 0) and the center of the
sphere (1, –2, –3) is
p
√
√
12 + (–2)2 + (–3)2 = 1 + 4 + 9 = 14,
which is larger than
√
13, the radius of the sphere.
11
9. Note that because the cube is centered at the origin and that it has
edge length 2, the cube is the set of all points (x, y, z) such that –1 ≤
x, y, z ≤ 1, the graph of which is provided below.
z
1
–1
–1
1
1
y
x
–1
Further note that any inscribed/circumscribed spheres must be centered
at the origin as well.
(a) The sphere inscribed in the cube is, by definition, tangent to each
of the cube’s faces. The distance from the origin to any one of the
faces is 1. Therefore the inscribed sphere also has a radius of 1,
and thus its equation is x2 + y 2 + z 2 = 1.
(b) Each of the cube’s vertices must be a point on the sphere circumscribed around the cube, again by definition. Hence the distance
between any one of the cube’s vertices and the origin is also the
radius of the sphere. Taking the point (1, 1, 1) for instance, its dis√
√
tance from the origin is 11 + 12 + 12 = 3, and so the equation
of the circumscribed sphere is x2 + y 2 + z 2 = 3.
10. If the two spheres are tangent, then their only point of intersection is
the midpoint of the line segment that has the centers of the spheres as
its endpoints (why?), which is
12
2 + 5 3 + 3 1 + (–2)
,
,
2
2
2
=
7
3
, 3,
2
2
.
Furthermore, since the two spheres have the same radius and they
intersect at the midpoint given above, then their radius is half the
distance between their centers, which is
1p
1p 2
1√
1√
(5 − 2)2 + (–3 − 3)2 + (2 − 1)2 =
3 + (–6)2 + 12 =
9 + 36 + 1 =
46.
2
2
2
2
The square of this is 46/4 = 23/2, and so the equations of the two spheres
are (x−2)2 +(y −3)2 +(z −1)2 = 23/2 and (x−5)2 +(y +3)2 +(z −2)2 =
23/2.
11. (a) This is the cylindrical surface whose graph in the xy-plane is the
line 3x + 4y = 12. (If you rewrite this as y = – 34 x + 4, then you
see more specifically the line with slope –3/4 and y-intercept 4.)
Extend its graph indefinitely in the z-direction to get the whole
picture.
13
(b) This is the cylindrical surface whose graph in the xy-plane is the
ellipse x2 /4 + y 2 /9 = 1. (More specifically, it is an ellipse in the
xy-plane centered at the origin with minor axis along the x-axis
of length 2 and major axis along the y-axis of length 3.) Extend
its graph indefinitely in the z-direction to get the whole picture.
(c) This is the cylindrical surface whose graph in the xz-plane is the
parabola z = x2 . Extend its graph indefinitely in the y-direction
to get the whole picture.
(d) This is the cylindrical surface whose graph in the yz-plane is the
natural exponential function. Extend its graph indefinitely in the
x-direction to get the whole picture.
14
12. Complete the square to get from x2 + z 2 − 4x − 8z + 13 > 0 to
(x2 − 4x + 4) + (z 2 − 8z + 16) + 13 > 4 + 16 ⇒ (x − 2)2 + (z − 4)2 > 7.
Now (x − 2)2 + (z − 4)2 = 7 is a cylindrical surface. (More specifically,
√
it is a cylinder of radius 7 extending infinitely in the y-direction and
such that its graph in the xz-plane is the circle centered at (2, 4) of
√
radius 7.) So all points (x, y, z) that satisfy (x − 2)2 + (x − 4)2 > 7
lie outside of this cylinder.
13. Complete the square to get from x2 + y 2 + z 2 − 4x − 12y − 8z = k to
(x2 − 4x + 4) + (y 2 − 12y + 36) + (z 2 − 8z + 16) = k + 4 + 36 + 16
⇒ (x − 2)2 + (y − 6)2 + (z − 4)2 = k + 56
In order for the above to be the equation of a sphere, we need k +56 > 0.
So k > –56.
15
14. The components of a vector can be attained by subtracting the initial
point from the terminal point:
−−→ −−→ −−→
P1 P2 = OP2 − OP1 = h5, –1i − h2, 3i = h3, –4i.
y
P1
3
2
1
1
2
3
4
x
5
–1
P2
–2
–3
–4
15. The vector v has initial point (4, 1) and terminal point (9, 4). Hence
v = h9 − 4, 4 − 1i = h5, 3i.
y
4
3
v
2
1
1
2
3
4
5
6
7
8
16. We can attain u + v + w graphically as follows:
16
9
x
y
5
v
4
u+v
3
w
2
u
1
(u + v) + w
1
2
3
4
5
6
7
8
9
x
As (u + v) + w = u + v + w starts at the origin and ends at (6, 1), it
follows that u + v + w = h6, 1i.
−
−
17. Note that the initial point of →
w is also the initial point of both →
u and
→
−
−
v and that the terminal point of →
w is the point of intersection of the
−
−
u and →
v.
two diagonals of the parallelogram determined by vectors →
−
−
−
−
−
w is parallel to the diagonal vector →
u +→
v , and →
w −→
u is parallel
So →
−
−
to the other diagonal vector →
v −→
u . There are thus scalars a and b
−
−
−
−
−
−
−
such that →
w = a(→
u +→
v ) and →
w −→
u = b(→
v −→
u ). Hence
−
−
−
−
−
−
−
−
a→
u + a→
v =→
w = b(→
v −→
u)+→
u = (–b + 1)→
u + b→
v.
By comparing like terms on both ends of the above equation, we find
that a = –b + 1 and a = b. Hence a = –a + 1 ⇒ 2a = 1 ⇒ a = 1/2.
−
−
−
Hence →
w = 1 (→
u +→
v ). (This intuitively makes sense since the diagonals
2
of a parallelogram bisect each other.)
18. The components of a vector can be attained by subtracting the initial
point from the terminal point:
17
−−→ −−→ −−→
P1 P2 = OP2 − OP1 = h5, 4, 6i − h1, 2, 3i = h4, 2, 3i.
19. (a)
2u − 3w = 2(3i + 2j − k) − 3(7i + 4j + k)
= 6i + 4j − 2k − 21i − 12j − 3k = –15i − 8j − 5k
(b)
||u + w|| = ||(3i + 2j − k) + (–2i + 4j + 3k)||
= ||i + 6j + 2k||
√
= 12 + 62 + 22
√
√
= 1 + 36 + 4 = 41
(c)
||u|| + ||w|| = ||3i + 2j − k|| + ||–2i + 4j + 3k||
p
p
= 32 + 22 + (–1)2 + (–2)2 + 42 + 32
√
√
√
√
= 9 + 4 + 1 + 4 + 16 + 9 = 14 + 29
√
(d) ||2u|| = |2| ||u|| = 2 14.
1 1 ||v|| = ||v|| = 1.
(e) v = ||v||
||v|| ||v||
20. (a) The desired vector is –
1
3
4
1
v = –√
(3i + 4j) = – i − j.
2
2
||v||
5
5
3 +4
−−→
(b) Note that P1 P2 = h3 − 1, –1 − 0, 2 − 5i = h1, –2, –3i, which has
p
√
√
2 + (–1)2 + (–3)2 =
magnitude 2
4 + 1 + 9 = 14. The desired
2
1
3
.
vector is thus √ , – √ , – √
14
14
14
(c) This is simply – 21 v = h– 12 , –1, – 23 i.
18
(d) Find the unit vector in the same direction as w and multiply it
√
√
by 5 to get a vector in that direction of length 5 :
√
5
w
||w||
√
= 5
i − 2j + 3k
p
12 + (–2)2 + 32
!
√
√
√
5
2 5
3 5
= √ i − √ j + √ k.
14
14
14
√ (e) The desired vector is 4 cos(π/6), 4 sin(π/6)i = h2 3, 2 .
√
(f) The desired vector is h2 cos(210◦ ), 2 sin(210◦ )i = h– 3, –1i.
21. Our goal is the find all a such that v = kw for some scalar k. So we
want ha2 , 6i = h4ak, 2ki. Equating components, we see that we need
6 = 2k ⇒ k = 3. Hence ha2 , 6i = h12a, 6i. This gives a2 = 12a ⇒
a2 − 12a = 0 ⇒ a(a − 12) = 0, and so a = 0 or a = 12.
22.
→
−
−
v +→
w = hcos(45◦ ), sin(45◦ )i + hcos(–30◦ ), sin(–30◦ )i
*√ √ + *√
+ *√
+
√ √
2 2
3 1
2+ 3 2−1
,
+
,–
=
,
.
=
2 2
2
2
2
2
23. (a) The resultant force is h–40 lb, 20 lbi, which has magnitude
√
402 + 202 lb =
√
p
√
√
(2 · 20)2 + 202 lb = 4 · 202 + 202 lb = 5 · 202 lb = 20 5 lb.
Also tan θ = 20/(–40) = –1/2. Since the resultant is in the second quadrant while the range of the inverse tangent function is
(–π/2, π/2), which puts a point in the first and fourth quadrants,
it follows that the angle that the resultant makes with the positive
x-axis is π + tan–1 (–1/2) = π − tan–1 (1/2).
19
(b) The resultant force is
√
h0, 50 2 Ni+h400 cos(45◦ ) N, 400 sin(45◦ ) Ni+h200 cos(–45◦ ) N, 200 sin(–45◦ ) Ni =
√
√
√
√
√
√
√
= h0, 50 2 Ni+h200 2 N, 200 2 Ni+h100 2 N, –100 2 Ni = h300 2 N, 150 2 Ni,
which has magnitude
q
p
√
√
(300 2)2 + (150 2)2 N = 2(2 · 150)2 + 2(150)2 N
√
= 8 · 1502 + 2 · 1502 N
√
√
= 10 · 1502 N = 150 10 N.
√
√
Also tan θ = (150 2)/(300 2) = 1/2. Since the resultant is in the
first quadrant, we can simply say that the angle that it makes
with the positive x-axis is tan–1 (1/2).
20