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)
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.
