# 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(aybzazby) −j(axbzazbx) + k(axbyaybx)

# 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 + 2jk 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 + jk 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 + 2jk 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 + 5jk

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 ab 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 ab.