Dot Product of Two Vectors and Applications








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

Dot Product of Two Vectors

The dot product of two vectors v = < v1 , v2 > and u = <u1 , u2> denoted v . u, is

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 || = SQRT( v1 2 + v1 2 ) and then using property 3 above where v = <3 , - 4>

Solution to example 2:

1. definition: || v || = SQRT( v1 2 + v1 2 ) = SQRT( (3) 2 + (- 4) 2 )

= SQRT(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)

dot product and cosine rule


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 || = SQRT(1 2 + 1 2) = SQRT (2)

|| u || = SQRT((-4) 2 + 3 2) = SQRT (25) = 5

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

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

t = arccos (- 1 / [ SQRT(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 pages and references related to vectors.

Vector Calculators.

More on vectors.


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Updated: 27 November 2007 (A Dendane)