# Dot Product of Two Vectors and Applications

Tutorial on the calculation and applications of the dot product of two vectors.

## Dot Product of Two Vectors

Knowing the coordinates of two vectors v = < v1 , v2 > and u = <u1 , u2> , the dot product of these two vectors, denoted v . u, is given by:

**v · u = < v1 , v2 > . <u1 , u2> = v1 × u1 + v2 × u2**

NOTE that the result of the dot product is a __scalar__.

Example 1: Vectors v and u are given by their components as follows

v = < -2 , 3> and u = < 4 , 6>

Find the dot product v · u of the two vectors.

__Solution to example 1:__

v · u = < -2 , 3> . <4 , 6>

= (-2) × (4) + (3) × (6) = -8 + 18 = 10

## Properties of the Dot Product

1. v · u = u · v

2. v · (u + w) = v · u + v · w

3. v · v = || v ||^{ 2}

4. c(v · u) = cv · u = v · cu

Example 2: Find || v || first using the definition || v || = √( v1^{ 2} + v1^{ 2} ) and then using property 3 above where v = <3 , - 4>

__Solution to example 2:__

1. definition: || v || = √( v1^{ 2} + v1^{ 2} ) = √( (3)^{ 2} + (- 4)^{ 2} )

= √(9 + 16) = 5

2. Property 3: || v ||^{ 2} = v · v

= <3 , - 4> . <3 , - 4> = (3) × (3) + (- 4) × (- 4) = 25

Take the square root to find || v || = 5

## Alternative Form of the Dot Product of Two Vectors

In the figure below, vectors v and u have same initial point the origin O(0,0). Points A and B are the terminal points. t is the angle made by the two vectors. Applying the cosine law to triangle OAB, we obtain:

d(A,B)^{ 2} = || v ||^{ 2} + || u ||^{ 2} -2|| v || || u || cos (t)

Use the definition of the distance to find d(A,B) and the definition of the magnitude to find || v || and || u || and substitute in the above

(v1 - u1)^{ 2} + (v2 - u2)^{ 2} = (v1^{ 2} + v2^{ 2}) + (u1^{ 2} + u2^{ 2}) -2 || v |||| u || cos (t)

Expand the squares in the left side and simplify to obtain

v1 u1 + v2 u2 = || v || || u || cos (t)

The left side is the dot product of vectors v and u, hence

v · u = || v || || u || cos (t)

We may use the above property of the dot product to find angle t between two vectors

cos t = v · u / (|| v || || u ||)

NOTE that if cos t = 0 ( t = Pi / 2) the dot product v . u = 0. This leads to:

vectors v and u are orthogonal if and only if v · u = 0.
Example 3: Show that vectors v = <3 , - 4> and u = <4 , 3> are orthogonal

__Solution to example 3:__

Find dot product v · u

v · u = <3 , - 4> . <4 , 3> = (3)*(4) + (-4)*(3) = 0

according to above

cos t = v · u / (|| v || || u ||) = 0

cos t = 0 means that t = Pi / 2 and the two vectors are orthogonal.
Example 4: Find the angle between vectors v = <1 , 1> and u = < - 4 , 3>.

__Solution to example 4:__

Find dot product v · u

v · u = <1 , 1> . < - 4 , 3> = (1)*(- 4) + (1)*(3) = - 1

Find || v || and || u ||

|| v || = √(1^{ 2} + 1^{ 2}) = √ (2)

|| u || = √((-4)^{ 2} + 3^{ 2}) = √ (25) = 5

We now use the formula cos t = v . u / (|| v || || u ||) to find cos t

cos t = v · u / (|| v || || u ||) = - 1 / [ √(2)*5 ]

t = arccos (- 1 / [ √(2)*5 ]) (approximately) = 98.1 ^{ o}

Exercises

1. Given vectors

v = <10 , - 5> and u = <2 , u2>,

Find u2 so that vectors v and u are orthogonal.

2. Find the angle between vectors v and u given below

v = <1 , 1> and u = <-2 , -2>,

__Answers to above exercises__

1. u2 = 4

2. 180^{ o}

## More References and Links

Vector Calculators.

vectors.