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 = axi + ayj + azk and b = bxi + byj + bzk appear in the 2nd and 3rd rows.

This 3-row determinant is evaluated by expansion into 2-row determinants, which are themselves then expanded. Matrix theory is itself very useful, and the scheme is shown below:

a×b =
 
i j k
ax ay az
bx by bx
 
  = i  
 
ay az
by bx
 
− j
 
ax az
bx bx
 
+ k
 
ax ay
bx by
 
  = i(aybz − azby) −j(axbz − azbx) + k(axby − aybx) 

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
4 2 −1
2 −6 −3
 
  = i  
 
2 −1
−6 −3
 
− j
 
4 −1
2 −3
 
+ k
 
4 2
2 −6
 
  = 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
2 1 −1
3 −6 2
 
  = i  
 
1 −1
−6 2
 
− j
 
2 −1
3 2
 
+ k
 
2 1
3 −6
 
  = 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
3 4 −3
1 3 2
 
  = i  
 
4 −3
3 2
 
− j
 
3 −3
1 2
 
+ k
 
3 4
1 3
 
  = 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
1 2 −1
3 3 1
 
  = i  
 
2 −1
3 1
 
− j
 
1 −1
3 1
 
+ k
 
1 2
3 3
 
  = 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
2 4 2
1 5 −2
 
  = i  
 
4 2
5 −2
 
− j
 
2 2
1 −2
 
+ k
 
2 4
1 5
 
  = 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
3 −4 1
2 5 −1
 
  = i  
 
−4 1
5 −1
 
− j
 
3 1
2 −1
 
+ k
 
3 −4
2 5
 
  = 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
2 −3 1
2 1 4
 
  = i  
 
−3 1
1 4
 
− j
 
2 1
2 4
 
+ k
 
2 −3
2 1
 
  = 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
2 0 3
1 2 4
 
  = i  
 
0 3
2 4
 
− j
 
2 3
1 4
 
+ k
 
2 0
1 2
 
  = 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