## Scalar Product

Aims: By the end of this chapter, you will be able to
(i) calculate scalar product given two vectors, and
(ii) employ scalar product to solve for the angle between two vectors.

1. A scalar product is also called dot product because the symbol that is used to represent this product of two vectors is simply a dot •.
2. It is call "scalar product" because the result of this operation is a scalar (a number) and NOT a vector.
3. The scalar product of two vectors a and b is defined by
 a • b = | a | | b | cos θ -----(1)
where θ is the angle between the wo vectors a and b.
4. Note that the angle θ is measured between the direction of these two vectors as shown in diagrams below.
5. Looking at the formula above (this is also in the formulae booklet page 5), we see that the right hand side is basically the product of three items: the magnitude of vector a (scalar), magnitude of vector b (scalar) and cos θ (scalar). Thus, the product must be a scalar.
6. The above formula can be expanded to vectors in three dimensions ( i, j, k ) and beyond.
7. In mechanics, work (energy) is the scalar product of Force and Displacement. The scalar product is often used to find the angle between two vectors.
 cos θ = a • b | a | | b |
or simply θ = arccos [ a b / ( | a | | b | ) ]
8. Example 1: Let there be two vectors b1i + b2j + b3k and c1i + c2j + c3k. Then the scalar product of these two vectors is simply:
 (b1i + b2j + b3k) • (c1i + c2j+ c3k) = (b1c1) + (b2c2) + (b3c3) ----(2)

This method works for three and more dimensions as well. Note that this is basically the sum of products between values belonging to the same basis (i or j or k).

9.  Box 1. How to make sense of expression (2) above. From the diagram on the left, we can let vector b = b1i + b2j + b3k and vector c = c1i + c2j + c3k. Applying Cosine Rule will allow us to write 2| b | | c | cos θ = | b |2 + | c |2 - | a |2 2| b | | c | cos θ = (b12+b22+b32)+(c12+c22+c32)-[(b1-c1)2 +(b2-c2)2 + (b3-c3)2 ] 2| b | | c | cos θ = 2b1c1 + 2b2c2 + 2b3c3 2| b | | c | cos θ = 2(b1c1 + b2c2 + b3c3)  | b | | c | cos θ =  (b1c1 + b2c2 + b3c3) Thus, we obtained the required expression (2) above.

10. Example 2: Find the scalar product of 3i + 5j and 7i - 3j.
Solution: (3i + 5j) • (7i - 3j) = (3x7) + (5x(-3)) = 21 -15 = 6.
11. Example 3: Find the scalar product of
 ( -4 ) ( -1/2 ) 8 and -5 10 12

Solution: (-4i + 8j + 10k) • (-1/2i - 5j + 12k) = (2) + (-40) + (120) = 82
12. Example 4: Find the angle that is between the two vectors in example 3.
Solution: θ = arccos [ a b / ( | a | | b | ) ]
From example 3 above, we know the scalar product to be 82. Now, we need to find the magnitude of these two vectors. The magnitude of the first vector is √[ (-4)2 + (8)2 + (10)2 ] = √(180) and
The magnitude of the other vector is √ [ (-1/2)2 + (-5)2 + (12)2]= √(169.25).
θ = arccos [ 82 / ( √(180) √(169.25) ) ] ≈ 62.0o ( 3 s.f) or 1.08 radian ( 3.s.f). [you answer should be either in degree or radian]
13. Example 5: Let u = u1i + u2j + u3k , v = v1i + v2j + v3k and w = w1i + w2j + w3k.
Show that u • (v
+ w) = u v + u w ; that is scalar product is distributive over addition.
Solution:
LHS = (u1i + u2j + u3k) · [(v1+ w1)i + (v2 + w1)j + (v3 + w1) k]
LHS = u1(v1+ w1) + u2(v2+ w2) + u3(v3+ w3)
LHS = u1v1+ u1w1 + u2v2+ u21w2) + u3v3+ u3w3
LHS = u1v1+ u2v2+ u3v3+ u1w1 +u2w2+ u3w3
LHS = u · v
+ u · w
LHS = RHS
Note that this property also holds for vector in other dimensions and over substraction.

### Worksheet.

1. Let vector a and vector b be i and j respectively, i.e a = i + 0j and b = 0i + j.
1. What is the relationship between vector a and vector b? [hint: draw these vectors]
2. What is the scalar product between these two vectors?
3. What is the angle between these two vectors?

2. Let vector a be i + j and vector b be 5i + 5j.
1. What is the relationship between vector a and vector b? [hint: draw these vectors]
2. What is the scalar product between these two vectors?
3. What is the angle between these two vectors?
3. Let vector v be i + j.
1. What is the magnitude of vector v?
2. Find the value of | v |2 ?
3. Calculate v v.
4. What is the relationship between v v and | v |2 ?
4. Calculate the angle between 3i + 5j and 7i - 3j. [Ans: 82.2o ]
5. Show that if a and b are non-zero vector in three dimensions and are perpendicular to each other then ab = 0.
6. (i) Show that if a and b are non-zero vector in three dimensions and parallel to each other then
ab = | a | | b |
(ii) Show that if a and b are non-zero vector in three dimensions and antiparallel to each other then
a
b = -| a | |b |
7. Let u and v be defined as in example 5 above.
(i) Show that u (u - v) - v (u - v) = (u - v) (u - v)

8. (ii) With the help of the diagram above and the property in (i), derive the cosine rule.

Properties in (5) and (6) are true for vectors of other dimensions. The results of the above questions above are important properties of scalar product. Jot down these results in the box given below.

### Important Properties of Scalar Product.

 v • w = w • v   --- commutative u • (v+ w) = (u• v) + (u• w) --- distributive (k v) • w = k (v • w) (write results from 1) (write results from 2 here) (write results from 3iv above here)
• Scalar product is not closed because the scalar product of two vectors is NOT a vector but a scalar number.
• Scalar product has no identity or inverse.