### The Dot Product - Scalar Product

The **dot product** of two vectors is defined as follows:

**A.B** = |A| |B| cos(θ), where |A|, |B| = magnitude of A and B,and θ = angle between the two vectors;

In order to see the impact of the change in angle on the **dot product** of the two vectors below, please move the slider slowly and see the value.

**|A| =4; |B| = 6**

**Angle: 0**

^{0}**180**

^{0}

Since the right hand side of the formula is a number, a scalar, this is also called the **scalar product.**

It is important to note that the angle between the two vectors must be considered in the following way:

#### The Dot Product of Unit Vectors

i.j = |i| |j| cos(90)

= 1 X 1 X 0

= 0

i.i = |i| |i| cos(0)

= 1 X 1 X 1

= 1

j.j = |j| |j| cos(0)

= 1 X 1 X 1

= 1

k.k = |k| |k| cos(0)

= 1 X 1 X 1

= 1

i.k = |i| |k| cos(90)

= 1 X 1 X 0

= 0

j.k = |j| |k| cos(90)

= 1 X 1 X 0

= 0

If two vectors are parallel, the dot product is maximum.

If two vectors are perpendicular, the dot product is zero.

**E.g.1**

If a = 2i + 3j -5k and b = 3i - 5j + 4k, find the a.b.

a.b = (2i + 3j -5k).(3i - 5j + 4k)

= 6i.i - 15j.j -20k.k (i.j = j.k = i.k = 0)

= 6 - 15 - 20

= -29.

**E.g.2**

If a = 3i + 2j -5k and b = 4i - 5j + 3k, find the angle between the two vectors.

a.b = (3i + 2j -5k).(4i - 5j + 3k)

= 12i.i - 10j.j -15k.k (i.j = j.k = i.k = 0)

= 12 - 10 - 15

= -23

|a| = √(9 + 4 + 25) = √38; |b| = √(16 + 25 + 9) = √50;

-13 = √38 √50 cos(θ)

cos(θ) = -0.2982

θ = 107.34

**E.g.3**

Using the dot product, verify that **Pythagoras Theorem** is true.

AC.AC= (AB + BC).(AB + BC)

AC.AC= AB.AB + AB.BC + BC.AB + BC.BC

z z cos(0) = x x.cos(0) + x y cos(90) + x y cos(90) + y y cos(0)

z^{2} = x^{2} + 0 + 0 + y^{2}

z^{2} = x^{2} + y^{2}

This is Pythagoras Theorem.

**E.g.4**

Using the dot product, verify the **cosine rule**.

AC.AC= (AB + BC).(AB + BC)

AC.AC= AB.AB + AB.BC + BC.AB + BC.BC

z z cos(θ) = x x.cos(θ) + x y cos(180-θ) + x y cos(180-θ) + y y cos(θ)

z^{2} = x^{2} - xy cos(θ) - xy cos(θ) + y^{2}

z^{2} = x^{2} + y^{2} - 2xy cos(θ)

This is the cosine rule.

**E.g.5**

Using the dot product, prove that the angle of a semi-circle is a right angle.

AB.BC= (AO + OB).(BO + OC)

= AO.BO + AO.OC + OB.BO + OB.OC

= r r.cos(180-θ) + r r cos(0) + r r cos(180) + r r cos(θ)

= -r^{2} cos(θ) + r^{2} - r^{2} + r^{2} cos(θ)

= 0

Since the dot product of AB and BC is zero, AB is perpendicular to BC.

**Challenge:**