1. Theory
The purpose of this tutorial is to practice working out the vector product of two vectors.
It is called the 'vector product' because the result is a 'vector', i.e. a quantity with both magnitude and direction.
NOTE: Throughout this tutorial we use the notation a to denote a vector quantity. However, there
are alternative notations, such as a, which are used in the PPLATO Interactive Mathematics modules (see Notation Section). 
The magnitude of the vector product of a and b
a×b = a bsin θ
where θ is the angle
between a and b



The direction of the vector product of a and b is perpendicular to both a and b:
such that if we look along a×b
then a rotates towards b
in a clockwise manner


It can be shown that the above definitions of magnitude and direction of a vector product allow us to calculate the x, y and z components of a × b from the individual components of the vectors a and b
The components of the vector a × b are given by the 'determinant' of a matrix with 3 rows, where the components of a = a_{x}i + a_{y}j + a_{z}k and b = b_{x}i + b_{y}j + b_{z}k appear in the 2nd and 3rd rows.
This 3row determinant is evaluated by expansion into 2row determinants, which are themselves then expanded.
Matrix theory is itself very useful, and the scheme is shown below:
a×b 
= 

i 
j 
k 
a_{x} 
a_{y} 
a_{z} 
b_{x} 
b_{y} 
b_{x} 




= 
i 

− j 

+ k 


= 
i(a_{y}b_{z} − a_{z}b_{y})
−j(a_{x}b_{z} − a_{z}b_{x})
+ k(a_{x}b_{y} − a_{y}b_{x})^{ } 
2. Exercises
Click on questions to reveal their solutions
Exercise 1:
Calculate a×b when a = 2, b = 4 and the between a and b is θ = 45°.
Solution:
a×b 
= 
a b sin θ 

= 
(2) (4) sin 45° 

= 
(2) (4) 1/ √2^{ } 

= 
8/ √2^{ } 

= 
8^{ }/ √2^{ } √2^{ } / √2^{ } 

= 
8 √2^{ } /2 

= 
4√2 
Exercise 2:
Calculate a×b when a = 3, b = 5 and the between a and b is θ = 60°.
Solution:
a×b 
= 
a b sin θ 

= 
(3) (5) sin 60° 

= 
(3) (5) √3 /2 

= 
15/2 √3 
Exercise 3:
Calculate a×b when a = 1, b = 3 and the between a and b is θ = 30°.
Solution:
a×b 
= 
a b sin θ 

= 
(1) (3) sin 30° 

= 
(1) (3) 1/2 

= 
3/2 
Exercise 4:
Calculate a×b when a = 2, b = 5 and the between a and b is θ = 35°.
Solution:
a×b 
= 
a b sin θ 

= 
(2) (5) sin 35° 

≈ 
(2) (5) (0.5736) 

≈ 
5.736 
Exercise 5:
Calculate the magnitude of the torque τ = s×F when s = 2m, F = 4N and θ = 30°, where θ is the angle between the position
vector s and the force F.
Solution:
τ 
= 
s×F 

= 
s F sin θ 

= 
(2m) (4N) sin 30° 

= 
(2m) (4N) 1/2 

= 
(8) (N m) 1/2 

= 
4J 
Exercise 6:
Calculate the magnitude of the velocity v = ω×s when ω = 3m s^{−1}, s = 2m and θ = 30°, where θ is the angle between the angular
velocity ω and the position vector s.
Solution:
τ 
= 
ω×s 

= 
ω s sin θ 

= 
(3s^{−1}) (2m) sin 45° 

= 
(3s^{−1}) (2m) 1/√2^{ } 

= 
(6) (m s^{−1}) 1/√2^{ } √2^{ } / √2^{ } 

= 
(6) (m s^{−1}) √2^{ } /2 

= 
3√2^{ }m s^{−1} 
Exercise 7:
If a = 4i + 2j − k and b = 2i − 6j − 3k calculate a vector that is perpendicular to both a and b
Solution:
a×b 
= 


= 
i 

− j 

+ k 


= 
i[(2)(−3) − (−1)(−6)]
−j[(4)(−3) − (−1)(2)]
+ k[(4)(−6) − (2)(2)] 

= 
i[−6 − 6]
−j[−12 + 2]
+ k[−24 − 4] 

= 
−12i + 10j − 28k 
Exercise 8:
Calculate the vector a×b when a = 2i + j − k and b = 3i − 6j + 2k
Solution:
a×b 
= 


= 
i 

− j 

+ k 


= 
i[(1)(2) − (−1)(−6)]
−j[(2)(2) − (−1)(3)]
+ k[(2)(−6) − (1)(3)] 

= 
i[2 − 6]
−j[4 + 3]
+ k[−12 − 3] 

= 
−4i − 7j − 15k 
Exercise 9:
Calculate the vector a×b when a = 3i + 4j − 3k and b = i + 3j + 2k
Solution:
a×b 
= 


= 
i 

− j 

+ k 


= 
i[(4)(2) − (−3)(3)]
−j[(3)(2) − (−3)(1)]
+ k[(3)(3) − (4)(1)] 

= 
i[8 + 9]
−j[6 + 3]
+ k[9 − 4] 

= 
17i − 9j + 5k 
Exercise 10:
Calculate the vector a×b when a = i + 2j − k and b = 3i + 3j + k
Solution:
a×b 
= 


= 
i 

− j 

+ k 


= 
i[(2)(1) − (−1)(3)]
−j[(1)(1) − (−1)(3)]
+ k[(1)(3) − (2)(3)] 

= 
i[2 + 3]
−j[1 + 3]
+ k[3 − 6] 

= 
5i − 4j −3k 
Exercise 11:
Calculate the vector a×b when a = 2i + 4j + 2k and b = i + 5j − 2k
Solution:
a×b 
= 


= 
i 

− j 

+ k 


= 
i[(4)(−2) − (2)(5)]
−j[(2)(−2) − (2)(1)]
+ k[(2)(5) − (4)(1)] 

= 
i[−8 −10]
−j[−4 + 2]
+ k[10 − 4] 

= 
−18i+ 6j+ 6k 
Exercise 12:
Calculate the vector a×b when a = 3i −4 j + k and b = 2i + 5j − k
Solution:
a×b 
= 


= 
i 

− j 

+ k 


= 
i[(−4)(−1) − (1)(5)]
−j[(3)(−1) − (1)(2)]
+ k[(3)(5) − (−4)(2)] 

= 
i[4 − 5]
−j[−3 − 2]
+ k[15 + 8] 

= 
−i+ 5j+ 23k 
Exercise 13:
Calculate the vector a×b when a = 2i − 3j + k and b = 2i + 6j + 4k
Solution:
a×b 
= 


= 
i 

− j 

+ k 


= 
i[(−3)(4) − (1)(1)]
−j[(2)(4) − (1)(2)]
+ k[(2)(1) − (−3)(2)] 

= 
i[−12 −1]
−j[8 − 2]
+ k[2 + 6] 

= 
−13i−6j+ 8k 
Exercise 14:
Calculate the vector a×b when a = 2i + 3k and b = 2i + j + 4k
Solution:
a×b 
= 


= 
i 

− j 

+ k 


= 
i[(0)(4) − (3)(2)]
−j[(2)(4) − (3)(1)]
+ k[(2)(2) − (0)(1)] 

= 
i[0 −6]
−j[8 − 3]
+ k[4 − 0] 

= 
−6i−5j+ 4k 
3. Alternative Notation
In this Tutorial we use symbols like a to denote a vector. In some texts, symbols for vectors are in bold e.g. a instead of a.
In this Tutorial, vectors are given in terms of the unit Cartesian vectors i, j and k.
A common alternative notation involves quoting the Cartesian components within brackets. For example, the vector a = 2i + j + 5k can be written as a = (2, 1, 5).
The scalar product a⋅b is also called a 'dot product' (reflecting the symbol used to denote this type of multiplication). Likewise, the vector product a×b is also called a 'cross product'.
An alternative notation for the vector product is a∧b.
PPLATO material © copyright 2004, University of Salford