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2.3 The dot product  (Page 2/16)

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Proof

Let u = ⟨ u 1 , u 2 , u 3 ⟩ and v = ⟨ v 1 , v 2 , v 3 ⟩ . Then

u · v = ⟨ u 1 , u 2 , u 3 ⟩ · ⟨ v 1 , v 2 , v 3 ⟩ = u 1 v 1 + u 2 v 2 + u 3 v 3 = v 1 u 1 + v 2 u 2 + v 3 u 3 = ⟨ v 1 , v 2 , v 3 ⟩ · ⟨ u 1 , u 2 , u 3 ⟩ = v · u .

The associative property looks like the associative property for real-number multiplication, but pay close attention to the difference between scalar and vector objects:

c ( u · v ) = c ( u 1 v 1 + u 2 v 2 + u 3 v 3 ) = c ( u 1 v 1 ) + c ( u 2 v 2 ) + c ( u 3 v 3 ) = ( c u 1 ) v 1 + ( c u 2 ) v 2 + ( c u 3 ) v 3 = ⟨ c u 1 , c u 2 , c u 3 ⟩ · ⟨ v 1 , v 2 , v 3 ⟩ = c ⟨ u 1 , u 2 , u 3 ⟩ · ⟨ v 1 , v 2 , v 3 ⟩ = ( c u ) · v .

The proof that c ( u · v ) = u · ( c v ) is similar.

The fourth property shows the relationship between the magnitude of a vector and its dot product with itself:

v · v = ⟨ v 1 , v 2 , v 3 ⟩ · ⟨ v 1 , v 2 , v 3 ⟩ = ( v 1 ) 2 + ( v 2 ) 2 + ( v 3 ) 2 = [ ( v 1 ) 2 + ( v 2 ) 2 + ( v 3 ) 2 ] 2 = ‖ v ‖ 2 .

□

Note that by property iv. we have 0 · v = 0 . Also by property iv. if v · v = 0 , then v = 0 .

Using properties of the dot product

Let a = ⟨ 1 , 2 , −3 ⟩ , b = ⟨ 0 , 2 , 4 ⟩ , and c = ⟨ 5 , −1 , 3 ⟩ . Find each of the following products.

  1. ( a · b ) c
  2. a · ( 2 c )
  3. ‖ b ‖ 2
  1. Note that this expression asks for the scalar multiple of c by a · b :
    ( a · b ) c = ( ⟨ 1 , 2 , −3 ⟩ · ⟨ 0 , 2 , 4 ⟩ ) ⟨ 5 , −1 , 3 ⟩ = ( 1 ( 0 ) + 2 ( 2 ) + ( −3 ) ( 4 ) ) ⟨ 5 , −1 , 3 ⟩ = −8 ⟨ 5 , −1 , 3 ⟩ = ⟨ −40 , 8 , −24 ⟩ .
  2. This expression is a dot product of vector a and scalar multiple 2 c :
    a · ( 2 c ) = 2 ( a · c ) = 2 ( ⟨ 1 , 2 , −3 ⟩ · ⟨ 5 , −1 , 3 ⟩ ) = 2 ( 1 ( 5 ) + 2 ( −1 ) + ( −3 ) ( 3 ) ) = 2 ( −6 ) = −12.
  3. Simplifying this expression is a straightforward application of the dot product:
    ‖ b ‖ 2 = b · b = ⟨ 0 , 2 , 4 ⟩ · ⟨ 0 , 2 , 4 ⟩ = 0 2 + 2 2 + 4 2 = 0 + 4 + 16 = 20 .
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Find the following products for p = ⟨ 7 , 0 , 2 ⟩ , q = ⟨ −2 , 2 , −2 ⟩ , and r = ⟨ 0 , 2 , −3 ⟩ .

  1. ( r · p ) q
  2. ‖ p ‖ 2

a. ( r · p ) q = ⟨ 12 , −12 , 12 ⟩ ; b. ‖ p ‖ 2 = 53

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Using the dot product to find the angle between two vectors

When two nonzero vectors are placed in standard position, whether in two dimensions or three dimensions, they form an angle between them ( [link] ). The dot product provides a way to find the measure of this angle. This property is a result of the fact that we can express the dot product in terms of the cosine of the angle formed by two vectors.

This figure is two vectors with the same initial point. The first vector is labeled “u,” and the second vector is labeled “v.” The angle between the two vectors is labeled “theta.”
Let θ be the angle between two nonzero vectors u and v such that 0 ≤ θ ≤ π .

Evaluating a dot product

The dot product of two vectors is the product of the magnitude of each vector and the cosine of the angle between them:

u · v = ‖ u ‖ ‖ v ‖ cos θ .

Proof

Place vectors u and v in standard position and consider the vector v − u ( [link] ). These three vectors form a triangle with side lengths ‖ u ‖ , ‖ v ‖ , and ‖ v − u ‖ .

This figure is two vectors with the same initial point. The first vector is labeled “u,” and the second vector is labeled “v.” The angle between the two vectors is labeled “theta.” There is also a third vector from the terminal point of vector u to the terminal point of vector v. It is labeled “v – u.”
The lengths of the sides of the triangle are given by the magnitudes of the vectors that form the triangle.

Recall from trigonometry that the law of cosines describes the relationship among the side lengths of the triangle and the angle θ . Applying the law of cosines here gives

‖ v − u ‖ 2 = ‖ u ‖ 2 + ‖ v ‖ 2 − 2 ‖ u ‖ ‖ v ‖ cos θ .

The dot product provides a way to rewrite the left side of this equation:

‖ v − u ‖ 2 = ( v − u ) · ( v − u ) = ( v − u ) · v − ( v − u ) · u = v · v − u · v − v · u + u · u = v · v − u · v − u · v + u · u = ‖ v ‖ 2 − 2 u · v + ‖ u ‖ 2 .

Substituting into the law of cosines yields

‖ v − u ‖ 2 = ‖ u ‖ 2 + ‖ v ‖ 2 − 2 ‖ u ‖ ‖ v ‖ cos θ ‖ v ‖ 2 − 2 u · v + ‖ u ‖ 2 = ‖ u ‖ 2 + ‖ v ‖ 2 − 2 ‖ u ‖ ‖ v ‖ cos θ − 2 u · v = −2 ‖ u ‖ ‖ v ‖ cos θ u · v = ‖ u ‖ ‖ v ‖ cos θ .

□

We can use this form of the dot product to find the measure of the angle between two nonzero vectors. The following equation rearranges [link] to solve for the cosine of the angle:

cos θ = u · v ‖ u ‖ ‖ v ‖ .

Using this equation, we can find the cosine of the angle between two nonzero vectors. Since we are considering the smallest angle between the vectors, we assume 0 ° ≤ θ ≤ 180 ° (or 0 ≤ θ ≤ π if we are working in radians). The inverse cosine is unique over this range, so we are then able to determine the measure of the angle θ .

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Source:  OpenStax, Calculus volume 3. OpenStax CNX. Feb 05, 2016 Download for free at http://legacy.cnx.org/content/col11966/1.2
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