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, isNOTE 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 . v2. 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