# Dot Product of Two Vectors and Applications

Tutorial on the calculation and applications of the dot product of two vectors.
## Dot Product of Two VectorsThe dot product of two vectors v = < v1 , v2 > and u = <u1 , u2> denoted v . u, isv . u = < v1 , v2 > . <u1 , u2> = v1 u1 + v2 u2NOTE that the result of the dot product is a scalar.
Example 1: Vectors v and u are given by their components as follows
## Properties of the Dot Product1. 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 VectorsIn 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>.
Exercises
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