## u v = u 1 v 2 + u 2 v 2 + u 3 v 3

Section 12.3 The Dot Product It is often necessary to calculate the angle between two vectors with the same initial point. For instance, in the follo...
Author: Miranda Barnett
Section 12.3

The Dot Product It is often necessary to calculate the angle between two vectors with the same initial point. For instance, in the following diagram, we might need to ﬁnd the value for θ:

In order to ﬁnd the required angle, we will need to learn a new operation on vectors; we can think of this operation as a kind of vector multiplication: Definition 0.0.1. The dot product of vectors ⃗u = ⟨u1 , u2 , u3 ⟩ and ⃗v = ⟨v1 , v2 , v3 ⟩ is given by ⃗u · ⃗v = u1 v2 + u2 v2 + u3 v3 . Notice that the result of the dot product operation on two vectors is a number (scalar), not a vector. The dot product has several important properties. If ⃗u, ⃗v , and w ⃗ are vectors and c is a scalar, then: 1. ⃗u · ⃗v = ⃗v · ⃗u 2. (c⃗u) · ⃗v = c(⃗u · ⃗v ) = ⃗u · (c⃗v ) 3. ⃗u · (⃗v + w) ⃗ = ⃗u · ⃗v + ⃗u · w ⃗ 4. ⃗u · ⃗u = |⃗u|2 5. ⃗0 · ⃗u = 0. The ﬁrst property says that the order of the vectors does not matter when calculating their dot product; it is easy to see that this is true: ⃗u · ⃗v = u1 v2 + u2 v2 + u3 v3 = v1 u1 + v2 u2 + v3 u3 = ⃗v · ⃗u. Property (2) says that the dot product of a vector with a scalar multiple of another vector may be calculated by taking the dot product of the two original vectors, then multiplying by the scalar. Property (3) says that the dot product ”distributes” over addition. Notice that, on the left hand side of the equality, the addition being performed is vector addition, whereas on the right hand side of the equality, the addition is real number addition. In property (5), notice that there are two diﬀerent meanings for 0 on either side of the equality: on the left hand side, ⃗0 is the vector ⃗0 = ⟨0, 0, 0⟩; however, on the right hand side, 0 means the number 0. We need the dot product to be able to determine the angle between two vectors. The theorem below can be proved using the Law of Cosines: 1

Section 12.3

Theorem 3. The angle θ between vectors ⃗a = ⟨a1 , a2 , a3 ⟩ and ⃗b = ⟨b1 , b2 , b3 ⟩ is given by θ = cos−1

(

) ( ) ⃗a · ⃗b a1 b1 + a2 b2 + a3 b3 = cos−1 . |⃗a||⃗b| |⃗a||⃗b|

Alternatively, ⃗a · ⃗b = |⃗a||⃗b| cos θ. In the fraction above, note that the denominator is a product of two numbers, |⃗a| and |⃗b|, so this is just regular multiplication with no vectors involved. Examples: √ √ Given the vectors ⃗u = ⟨1, 2, 1⟩ and ⃗v = ⟨−2, 2 2, 2⟩, ﬁnd the dot product ⃗u · ⃗v and the angle between ⃗u and ⃗v . The dot product is given by ⃗u · ⃗v = 1 · (−2) +

√ √ 2·2 2+1·2

= −2 + 4 + 2 = 4. Again, notice that the ”product” of these two vectors is a scalar. In order to determine the angle between the two vectors, we will need to ﬁnd the length of each vector. √ |⃗u| = = =

√ √

√ (1)2 + ( 2)2 + (1)2

1+2+1 4

=2 and √ |⃗v | =

√ (−2)2 + (2 2)2 + (2)2

√ 4+8+4 √ = 16

=

= 4.

2

Section 12.3

So the angle between the two vectors is ) ⃗u · ⃗v θ = cos |⃗u||⃗v | ( ) 4 −1 = cos 2·4 ( ) 1 = cos−1 2 π = . 3 −1

(

Question: Suppose that the (nonzero) vectors ⃗u and ⃗v are perpendicular. What is the value for the dot product ⃗u · ⃗v ? Since ⃗u and ⃗v are perpendicular, the angle θ between them is precisely ( ) π u · ⃗v −1 ⃗ = θ = cos , 2 |⃗u||⃗v |

π 2.

Since

and we know that cos−1 0 = π2 , we have ⃗u · ⃗v = 0. |⃗u||⃗v | Since the fraction’s value is 0, the numerator must be 0; so we have ⃗u · ⃗v = 0. Note, on the other hand, that if we pick any pair of non-zero vectors ⃗a and ⃗b so that ⃗a · ⃗b = 0, then the angle between them is π2 , i.e. the vectors are perpendicular. In other words, there is a really nice physical interpretation for a dot product of 0: two vectors have dot product 0 if and only if they are perpendicular.

Theorem 7. The nonzero vectors ⃗u and ⃗v are perpendicular or orthogonal if and only if ⃗u·⃗v = 0.

Vector Projections If the vectors ⃗a and ⃗b have the same initial point, we can compare the two vectors by ”projecting” one onto another. In the graph below, imagine shining a ﬂashlight positioned directly above the terminal point of ⃗b down onto the pair of vectors; the projection of ⃗b = ⟨1, 2⟩ onto the vector ⃗a = ⟨3, 3⟩ is the vector determined by the shadow that ⃗b casts on ⃗a:

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Section 12.3

⃗ onto ⃗a = AC ⃗ is the vector AD ⃗ whose Definition 0.0.2. The vector projection of ⃗b = AB terminal point D is determined by dropping a line perpendicular to the ray AC from the point B. We denote the projection by ⃗ = proj⃗a⃗b AD ⃗ is the vector projection of ⃗b onto ⃗a. and we say that AD The vector projection of ⃗b onto ⃗a is calculated as follows: ( ⃗) ( ⃗) ⃗a · b ⃗a · b ⃗a proj⃗a⃗b = . ⃗ a = |⃗a|2 |⃗a| |⃗a| (

) ⃗a · ⃗b Notice that is a scalar, not a vector, since the numerator is a dot product (a real |⃗a|2 number), and the denominator is the square of the length of ⃗a (also a number). Thus the quantity ( ⃗) ⃗a · b ⃗ ⃗a is a scalar multiple of the vector ⃗a. proj⃗a b = |⃗a|2 Sidenote: The order of the vectors in the deﬁnition is very important, as proj⃗b⃗a is in general not the same vector as proj⃗a⃗b. Using the same vectors ⃗a and ⃗b above, we see that proj⃗b⃗a and proj⃗a⃗b are deﬁnitely diﬀerent vectors:

4

Section 12.3

The projection proj⃗a⃗b is always a vector lying on the same line as ⃗a. Its length is ( ) ⃗a · ⃗b |proj⃗a⃗b| = ⃗a |⃗a|2 ( ) ⃗a · ⃗b |⃗a| = |⃗a|2 ( ) |⃗a · ⃗b| = |⃗a| |⃗a|2 (we can pull out the factor ⃗a · ⃗b). But

⃗a · ⃗b since it is a constant; in particular, |⃗a · ⃗b| is the absolute value of |⃗a|2 (

) |⃗a · ⃗b| |⃗a · ⃗b| . |⃗ a | = |⃗a|2 |⃗a|

To sum up the above calculations, the length of the vector projection proj⃗a⃗b of ⃗b onto ⃗a is given by |⃗a · ⃗b| . |proj⃗a⃗b| = |⃗a| The quantity ⃗a · ⃗b |⃗a| is important enough that we have a term to describe it: Definition 0.0.3. The scalar component of ⃗b in the direction ⃗a is given by ⃗a · ⃗b , comp⃗a⃗b = |⃗a| and the length of the vector projection of ⃗b onto ⃗a is the absolute value of this quantity, |comp⃗a⃗b|. 5

Section 12.3

Again using the example above, the scalar component of ⃗b in the direction ⃗a is the length of the black vector below:

Example Find the vector projection of ⃗u = ⟨1, 2, 0⟩ onto ⃗v = ⟨3, −3, 0⟩ and the scalar component of ⃗u in the direction ⃗v . Since

(

) ⃗u · ⃗v proj⃗v ⃗u = ⃗v , |⃗v |2

let’s begin by calculating ⃗u · ⃗v : ⃗u · ⃗v = 1 · (3) + 2 · (−3) + 0 · 0 =3−6+0 = −3. In addition, |⃗v |2 is √ |⃗v |2 = ( 32 + (−3)2 + 02 )2 =9+9 = 18. So the vector projection of ⃗u onto ⃗v is ( ) ⟨ ⟩ −3 1 1 proj⃗v ⃗u = ⟨3, −3, 0⟩ = − , , 0 . 18 2 2 6

Section 12.3

The vectors ⃗u, ⃗v , and proj⃗v ⃗u are graphed below:

The scalar component of ⃗u in the direction ⃗v is given by −3 ⃗u · ⃗v −1 =√ =√ . |⃗v | 18 2 The length of proj⃗v ⃗u is the absolute value of the scalar component of ⃗u in the direction ⃗v , i.e. −1 1 |proj⃗v ⃗u| = √ = √ . 2 2

7

Section 12.3

An ”application” of the dot product is found in physics: ⃗ = P⃗Q Definition 0.0.4. The work W done by a constant force F⃗ acting through a displacement D is given by ⃗ = |F⃗ ||D| ⃗ cos θ, W = F⃗ · D ⃗ where θ is the angle between F⃗ and D.

Example A crate is hauled 8 m up a ramp under a constant force of 200 N applied at an angle of 25◦ to the ramp. Find the work done. Since work is given by ⃗ cos θ, W = |F⃗ ||D| ⃗ = 8, the work done is where θ = 25◦ , and we are given |F⃗ | = 200 and |D| W = 200 · 8 cos(25◦ ) ≈ 1450 N · m.

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