INITIALS: ___________________
SURNAME: _______MEMO_________
STUDENT CODE: __________________
PHYSICS
ZUG1021
[20 MARKS]
12/03/2019
DUE: START OF LECTURE
TUTORIAL ASSIGNMENT 1
In each of the following questions, show all your steps and indicate all formulas used. Be sure to
write your final answer in each block provided. Not every question will necessarily be marked.
QUESTION 1 [3]
Two vectors are given by
a   4.0 m  i   3.0 m  j  1.0 m  k
and b   1.0 m  i  1.0 m  j   4.0 m  k.
In unit-vector notation, find (a) a  b , (b) a  b , and (c) a third vector c such that a  b  c  0 .
MEMO
All distances in this solution are understood to be in meters.
ˆ m.
(a) a  b  [4.0  (1.0)] ˆi  [(3.0)  1.0] ˆj  (1.0  4.0)kˆ  (3.0iˆ  2.0ˆj  5.0 k)
ˆ m.
(b) a  b  [4.0  (1.0)]iˆ  [( 3.0) 1.0]jˆ  (1.0  4.0)kˆ  (5.0 ˆi  4.0 ˆj  3.0 k)

 



(c) The requirement a  b  c  0 leads to c  b  a , which we note is the opposite of what we found
ˆ m.
in part (b). Thus, c  (5.0 ˆi  4.0 ˆj  3.0 k)
QUESTION 2 [7]
An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be
distinguished from the sky) while returning to base camp. He was supposed to travel due north for
5.6 km, but when the snow clears, he discovers that he actually traveled 7.8 km at 50° north of due
east. (a) How far and (b) in what direction must he now travel to reach base camp?
MEMO

The desired result is the displacement vector, in units of km, A
= (5.6 km), 90º (measured
counterclockwise from the +x axis), or A  (5.6 km)jˆ , where ˆj is the unit vector along the positive y
axis (north). This consists of the sum of two displacements: during the whiteout, B  (7.8 km), 50 , or
B  (7.8 km)(cos 50 ˆi  sin50 ˆj)  (5.01 km)iˆ  (5.98 km)ˆj
and the unknown C . Thus, A  B  C .
(a) The desired displacement is given by C  A  B  ( 5.01 km) ˆi  (0.38 km) ˆj . The magnitude is
( 5.01 km) 2  ( 0.38 km) 2  5.0 km.
(b) The angle is tan  1[( 0.38 km) /( 5.01 km)]  4.3 , south of due west.
QUESTION 3 [3]
In Fig. 3-31, a cube of edge length a sits with one corner at the origin of an xyz coordinate system.
A body diagonal is a line that extends from one corner to another through the center. In unit-vector
notation, what is the body diagonal that extends from the corner at (a) coordinates (0, 0, 0), (b)
coordinates (a, 0, 0), (c) coordinates (0, a, 0), and (d) coordinates (a, a, 0)? (e) Determine the
angles that the body diagonals make with the adjacent edges. (f) Determine the length of the body
diagonals in terms of a.
Figure 3
MEMO
(a) As can be seen from Figure 3-30, the point diametrically opposite the origin (0,0,0) has position vector
a i  a j  a k and this is the vector along the “body diagonal.”
(b) From the point (a, 0, 0), which corresponds to the position vector a î, the diametrically opposite point
is (0, a, a) with the position vector a j  a k . Thus, the vector along the line is the difference
 a ˆi  aˆj  a kˆ .
(c) If the starting point is (0, a, 0) with the corresponding position vector a ˆj , the diametrically opposite
point is (a, 0, a) with the position vector a ˆi  a kˆ . Thus, the vector along the line is the difference
a ˆi  a ˆj  a kˆ .
QUESTION 4 [7]


For the following three vectors, what is 3C  2 A  B ?
A  2.00iˆ + 3.00jˆ  4.00kˆ
B  3.00ˆi + 4.00jˆ  2.00kˆ C  7.00iˆ  8.00ˆj
MEMO
Using the fact that
ˆi  ˆj  k,
ˆ ˆj  kˆ  ˆi, kˆ  ˆi  ˆj
we obtain


2 A  B  2 2.00iˆ  3.00ˆj  4.00kˆ  3.00iˆ  4.00ˆj  2.00kˆ

ˆ
 44.0iˆ  16.0jˆ  34.0k.
Next, making use of
ˆi  ˆi = ˆj  ˆj = kˆ  kˆ = 1
ˆi  ˆj = ˆj  kˆ = kˆ  ˆi = 0
we have

 

3C  2 A  B  3 7.00 ˆi  8.00ˆj  44.0 ˆi  16.0 ˆj  34.0 kˆ

 3[(7.00) (44.0)+(  8.00) (16.0)  (0) (34.0)]  540.