# 1. Theory

The purpose of this tutorial is to practice using the scalar product of two vectors. It is called the 'scalar product' because the result is a 'scalar', i.e. a quantity with magnitude but no associated 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 scalar (or 'dot') product of two vectors a and b is:

 a⋅b = a b cos θ = axbx + ayby + azbz

- where θ is the angle between a and b, and:

 a = axi + ayj + azk b = bxi + byj + bzk

- where i, j and k are the unit vectors in the x, y, and z directions, respectively. The magnitudes of a and b are given by:

a = ax2 + ay2 + az2    and    b = bx2 + by2 + bz2.

# 2. Exercises

Click on questions to reveal their solutions

Exercise 1:

Calculate ab when a = 2i − 3j + 5k , b = i + 2j + 8k

Solution:

 a⋅b = (2)(1) + (−3)(2) + (5)(8) = 2 − 6 + 40 = 36

Exercise 2:

Calculate ab when a = 4i − 7j + 2k , b = 5ij − 4k

Solution:

 a⋅b = (4)(5) + (−7)(−1) + (2)(−4) = 20 + 7 − 8 = 19

Exercise 3:

Calculate ab when a = 2i − 3j + 3k , b = 3i − 2j + 5k

Solution:

 a⋅b = (2)(3) + (3)(−2) + (3)(5) = 6 − 6 + 15 = 15

Exercise 4:

Calculate ab when a = 3i + 6jk , b = 8i − 3jk

Solution:

 a⋅b = (3)(8) + (6)(−3) + (−1)(−1) = 24 − 18 + 1 = 7

Exercise 5:

Show that a is perpendicular to b when a = i + j + 3k , b = i − 7j + 2k

Solution: For the vectors to be perpendicular requires θ = 90°. Therefore, since ab = a b cos θ, ab = 0. To show that this is true, we use ab = axbx + ayby + azbz :

 a⋅b = (1)(1) + (1)(−7) + (3)(2) = 1 − 7 + 6 = 0

Exercise 6:

Show that a is perpendicular to b when a = i + 23j + 7k , b = 26i + j − 7k

Solution: For the vectors to be perpendicular requires θ = 90°. Therefore, since ab = a b cos θ, ab = 0. To show that this is true, we use ab = axbx + ayby + azbz :

 a⋅b = (1)(26) + (23)(1) + (7)(−7) = 26 + 23 − 49 = 0

Exercise 7:

Show that a is perpendicular to b when a = i + j + 3k , b = 2i + 7j − 3k

Solution: For the vectors to be perpendicular requires θ = 90°. Therefore, since ab = a b cos θ, ab = 0. To show that this is true, we use ab = axbx + ayby + azbz :

 a⋅b = (1)(2) + (1)(7) + (3)(−3) = 2 + 7 − 9 = 0

Exercise 8:

Show that a is perpendicular to b when a = 39i + 2j + k , b = i − 23j + 7k

Solution: For the vectors to be perpendicular requires θ = 90°. Therefore, since ab = a b cos θ, ab = 0. To show that this is true, we use ab = axbx + ayby + azbz :

 a⋅b = (39)(1) + (2)(−23) + (1)(7) = 39 − 49 + 7 = 0

Exercise 9:

Calculate the work done Fs given F, s and θ (the angle between the force F and the displacement s) when F = 7N, s = 3m and θ = 0°

Solution: Fs = F s cos θ = (7N)(3m)(1) = 21J.

Note: When the angle θ is zero then Fs = F s and we can simply multiply the magnitudes of F and s.

Exercise 10:

Calculate the work done Fs given F, s and θ (the angle between the force F and the displacement s) when F = 4N, s = 2m and θ = 27°

Solution: Fs = F s cos θ = (4N)(2m) cos 27° ≈ 7.128J.

 Note: F⋅s = F s cos θ can be rewritten as (F cos θ) s. This can be interpreted as being the product of s and the projected component F cos θ along the direction of s. Exercise 11:

Calculate the work done Fs given F, s and θ (the angle between the force F and the displacement s) when F = 5N, s = 4m and θ = 48°

Solution: Fs = F s cos θ = (5N)(4m) cos 48° ≈ 13.38J.

Exercise 12:

Calculate the work done Fs given F, s and θ (the angle between the force F and the displacement s) when F = 2N, s = 3m and θ = 56°

Solution: Fs = F s cos θ = (2N)(3m) cos 56° ≈ 3.355J.

Exercise 13:

Calculate the angle θ between vectors a and b when a = 2i − 3j + 2k , b = i + j + k

Solution: From ab = a b cos θ, we have:

cos θ = ab / a b

If we use the Cartesian representation of the two vectors then we have:

 a⋅b = axbx + ayby + azbz a = √ ax2 + ay2 + az2 b = √ bx2 + by2 + bz2

For the given vectors these are:

 a⋅b = (2)(1) + (−1)(1) + (2)(1) = 2 − 1 + 2 = 3 a = √22 + (−1)2 + 22  =  √4 + 1 + 4  =  √9  =  3 b = √ 12 + 12 + 12  =  √3
∴ cos θ = ab / a b = 3 / 3 3 ≈ 0.5774

so:

θ ≈ cos−1(0.5774) ≈ 54.7°

Exercise 14:

Calculate the angle θ between vectors a and b when a = i + j + k , b = 2i − 3j + k

Solution: From ab = a b cos θ, we have:

cos θ = ab / a b

If we use the Cartesian representation of the two vectors then we have:

 a⋅b = axbx + ayby + azbz a = √ ax2 + ay2 + az2 b = √ bx2 + by2 + bz2

For the given vectors these are:

 a⋅b = (1)(2) + (1)(−3) + (1)(1) = 2 − 3 + 1 = 0 a = √12 + 12 + 12  =  √3 b = √22 + (−3)2 + 12  =  √4 + 9 + 1  =  √14
∴ cos θ = ab / a b = 0 / 3 14  = 0

so:

θ = cos−1(0) = 90°

Exercise 15:

Calculate the angle θ between vectors a and b when a = i − 2j + 2k , b = 2i + 3j + k

Solution: From ab = a b cos θ, we have:

cos θ = ab / a b

If we use the Cartesian representation of the two vectors then we have:

 a⋅b = axbx + ayby + azbz a = √ ax2 + ay2 + az2 b = √ bx2 + by2 + bz2

For the given vectors these are:

 a⋅b = (1)(2) + (−2)(3) + (2)(1) = 2 − 6 + 2 = −2 a = √12 + (−2)2 + 22  =  √1 + 4 + 4  =  √9  =  3 b = √22 + 32 + 12 √4 + 9 + 1  =  √14
∴ cos θ = ab / a b = −2 / 3 14 ≈ −0.1782

so:

θ ≈ cos−1(−0.1782) ≈ 100.3°

NOTE:

 When a⋅b is positive cos θ is positive and θ is an acute angle When a⋅b is negative cos θ is negative and θ is an obtuse angle Exercise 16:

Calculate the angle θ between vectors a and b when a = 5i + 4j + 3k , b = 4i − 5j + 3k

Solution: From ab = a b cos θ, we have:

cos θ = ab / a b

If we use the Cartesian representation of the two vectors then we have:

 a⋅b = axbx + ayby + azbz a = √ ax2 + ay2 + az2 b = √ bx2 + by2 + bz2

For the given vectors these are:

 a⋅b = (5)(4) + (4)(−5) + (3)(3) = 20 − 20 + 9 = 9 a = √52 + 42 + 32  =  √25 + 16 + 9  =  √50  =  3 b = √42 + (−5)2 + 32  =  √16 + 25 + 9  =  √50
∴ cos θ = ab / a b = 9 / 50 50 = 9 / 50  = 0.18

so:

θ = cos−1(0.18) ≈ 79.6°

# 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. For example, a = i + 2j + 3k implies that a can be decomposed into the sum of the following three vectors as shown below:

 a = i + 2j + 3k one step along the x-axis two steps along the y-axis three steps along the z-axis a is the (vector) sum of i and 2j and 3k A common alternative notation for expressing a in terms of these Cartesian components is given by a = (1, 2, 3).