### 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: 00 1800

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)
z2 = x2 + 0 + 0 + y2
z2 = x2 + y2
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)
z2 = x2 - xy cos(θ) - xy cos(θ) + y2
z2 = x2 + y2 - 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)
= -r2 cos(θ) + r2 - r2 + r2 cos(θ)
= 0
Since the dot product of AB and BC is zero, AB is perpendicular to BC.

Challenge:

Using the dot product, show that the diagonals of a square intersect at right angles.

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