Homework 2
problems: 3.2, 3.23, 3.53, 3.63
Problem 3.2
Two points in the xy plane have the following polar coordinates (2.50 m, 30.0°) and
(3.80 m, 120.0°). Determine (a) the Cartesian coordinates of these points and (b) the
distance between them.
y
By
B-A
B
A
Ay
120°
30°
x
Ax
Bx
(a)
From geometrical consideration (or from the relationship between the polar
and the Cartesian coordinates)
A = A x , A y = [2.5m cos 30°,2.5m sin 30°] = [2.17,1.25]m
B = Bx , B y = [3.8m cos120°,3.5m sin 120°] = [− 1.90,3.29]m
[
[
]
]
(b)
solution 1
The simplest way is to find the magnitude of vector B-A from the scalar
components of this vector
B − A = [− 1.9,3.29]m − [2.17,1.25]m = [4.07,2.04]m
B−A =
(4.07m )2 + (2.04m )2
= 4.55m
solution 2
One can also use the law of cosines
B−A =
(2.5m )2 + (3.8m )2 − 2 ⋅ 2.5m ⋅ 3.8m ⋅ cos(120° − 30°) = 4.55m
Problem 3.23
Consider two vectors A = 3i – 2j and B = – i – 4j . Calculate (a) A + B, (b) A – B,
(c) ⏐A + B ⏐, (d) ⏐A – B ⏐, (e) the direction of A + B and A – B.
A+B
j
i
A
B
A+B
(a),(b)
To find the sum and the difference of the two vectors we should use the
properties of the vectors (associative and commutative properties of addition
and the distributive property of the product over addition of vectors).
A + B = (3i − 2 j) + (− i − 4 j) = 2i − 6 j
A − B = (3i − 2 j) − (− i − 4 j) = 4i + 2 j
(c),(d)
The magnitude of a vector can be easily found from its scalar components
A + B = 22 + (− 6 )2 = 6.32
A − B = 42 + 22 = 4.47
(e)
Knowing the scalar components also allows one to find the direction of the
vectors
−6
θ A + B = arctan
= −71.6°
3
2
θ A − B = arctan = 26.6°
4
Problem 3.53
A vector is given by R = 2i + j + 3k. Find (a) the magnitudes of the x, y, and z
components, (b) the magnitude of R, and (c) the angles between R and the x, y, and z axes.
R
k
j
i
(a) First one has to identify the vector components of vector R:
Ry = 1.00 j
Rz = 3.00 k
Rx = 2.00 i
and to use the scalar components of these three vectors to find their
magnitudes
R x = 22 = 2
R y = 12 = 1
R z = 32 = 3
(b) Using the scalar components
R = 22 + 12 + 32 = 3.74
(c) From the definition of the cosine function
Rx
2
= arccos
= 57.7°
R
3.74
Ry
1
θ y = arccos
= arccos
= 74.5°
R
3.74
R
3
θz = arccos z = arccos
= 36.7°
R
3.74
θ x = arccos
Problem 3.63
A rectangular parallelepiped has the dimensions a, b, and c as in Figure 3.63. (a) Obtain a
vector expression for the face diagonal vector R1. What is the magnitude of this vector?
(b) Obtain a vector expression for the body diagonal vector R2. Note that make R1, ck and
R2 make a right triangle, and prove that the magnitude of R2 is
z
a
a2 + b2 + c2 .
b
Rz
c
R2
Ry
Rx
y
R1
x
(a) Imagine vectors Rx, Ry, and Rz along the edges of the parallelepiped. In the
Cartesian coordinate system, marked in the figure,
Rx = a i , Ry = b j , Rz = c k .
Therefore
R1 = Rx + Ry = a i + b j
and
R1 = R 2x + R 2y = a 2 + b 2
Similarly,
R2 = Rx + Ry + Rz = a i + b j + c k
and
R 2 = R 2x + R 2y + R 2z = a 2 + b 2 + c 2