Section 9.5 – The Dot Product In this section, we will explore the product of two vectors called the Dot Product or the Scalar Product. This product produces a scalar as an answer and has many practical applications in science. Let's begin with a definition: Objective #1:

The Dot Product of Two Vectors.

Dot Product Definition Let v = a1i + b1j and w = a2i + b2j. Then, the dot product v•w is: v•w = a1a2 + b1b2 Find the following if v = 2i + 3j and w = – 3i + 4j: Ex. 1a v•w Ex. 1b w•v Ex. 1c v•v Ex. 1d w•w Ex. 1e ||v|| Ex. 1f ||w|| Solution: a) v•w = 2(– 3) + 3(4) = – 6 + 12 = 6 b) w•v = – 3(2) + 4(3) = – 6 + 12 = 6 c) v•v = 2(2) + 3(3) = 4 + 9 = 13 d) w•w = (– 3)(– 3) + 4(4) = 9 + 16 = 25 e)

||v|| =

(2)2 +(3)2 =

f)

||w|| =

(− 3)2 +(4)2 =

13

Properties of the Dot Product If u, v, and w are vectors, then a) u•v = v•u b) u•(v + w) = u•v + u•w c) v•v = ||v||2 d) 0•v = 0 Objective #2

25 = 5

Commutative Property Distributive Property

Find the Angle between Two Vectors.

We can use the dot product to calculate the angle between two vectors. If we let u and v be two vectors with the same initial point, then the vectors u, v, u – v form a triangle. Let θ be the angle between u and v.

u

u–v θ v

219 The lengths of the sides of the triangle are equal to ||u||, ||v||, and ||u – v|| respectively. Recall that the law of cosines says that: c2 = a2 + b2 – 2abcos(C) But, c = ||u – v||, a = ||u||, b = ||v||, and C = θ: ||u – v||2 = ||u||2 + ||v||2 – 2||u||||v||cos(θ) Now, substitute ||u||2 = u•u, ||v||2 = v•v, and ||u – v||2 = (u – v)•(u – v) (u – v)•(u – v) = u•u + v•v – 2||u||||v||cos(θ) "FOIL" the left side: u•u – 2u•v + v•v = u•u + v•v – 2||u||||v||cos(θ) Subtract u•u + v•v from both sides and then solve for cos(θ): – 2u•v = – 2||u||||v||cos(θ) u•v = ||u||||v||cos(θ) cos(θ) =

u•v u v

Angle Between Two Vectors Theorem Let u and v be two non-zero vectors and let θ be the angle between the € where 0 ≤ θ ≤ π. Then vectors, cos(θ) =

u•v u v

If v = 2i + 3j and w = – 3i + 4j, find: Ex. 2 The angle between v and w € Solution: Recall from example #1 that: v•w = 6, ||v|| = 13 , and ||w|| = 5. Plugging in, we get: cos(θ) =

6 ( 13 )(5)

= 0.3328… (use the inverse cosine)

θ = cos – 1(0.3328…) = 70.559…˚ ≈ 70.56˚ Recall example 9 from section 9.4: € Ex. 3 Rowing across a lake, Roger maintains a constant speed of 8 miles per hour due north. If the current is 3 miles per hour in a southeasterly direction, find the actual speed & direction of the boat. Solution: We found that vn = 1.5 i + (8 – 1.5 )j and that the speed was ≈ 6.25 mph.

220 We can now use the Angle Between Two Vectors Theorem to find the angle between vb and vn directly: vb•vn = 0(1.5 ) + 8(8 – 1.5 ) = 47.029… ||vb|| = 8 and ||vn|| = 6.2497… 47.029... cos(θ) = = 0.9406… (use the inverse cosine) 8(6.2497...)

θ = 19.8419…˚ ≈ 19.84˚ The boat is moving at ≈ 6.25 mph with a bearing of N19.84˚E.

€ #3: Objective

Determine if Two Vectors are Parallel.

Two vectors are parallel if one vector is a non-zero scalar multiple of the other vector (i.e., v and w are parallel if v = αw) Since parallel vectors either have the same direction or opposite directions, then the angle between the vectors is 0 or π. Determine if the following vectors are parallel: 14 Ex. 4 v = 9i – 6j and w = – 7i + j 3

Solution: 14 v•w = 9(– 7) – 6( ) = – 63 – 28 = – 91 ||v|| =

(9)2 +(− 6)

||w|| =

(− 7) +(

cos(θ) =

2

3 2

14 2 ) 3 −91

(3 13 )(

7 13 3

117 = 3 13

=

)

= =

49 + −91 91

196 9

=

637 9

=

7 13 3

=–1

Thus, θ = π so the lines are parallel. Objective #4: €

Determine if Two Vectors are Orthogonal.

Definition Two non-zero vectors are said to be orthogonal if they are perpendicular. π This means the angle between the vectors is . 2

Theorem Two non-zero vectors are orthogonal if and only if dot product is 0.

221 Proof: We need to prove both directions. First, we will prove that if the non-zero vectors v and w are orthogonal, then v•w = 0. Then, we will prove that if v•w = 0, then the non-zero vectors are v and w orthogonal. (⇒) Assume that the two nonzero vectors v and w are orthogonal. This implies that the angle between the vectors is Thus, cos( v•w v w

π 2

)=

v•w v w

. But cos(

π 2

π 2

by definition.

) = 0, so

= 0.

Since v and w are non-zero vectors, then ||v|| ≠ 0 and ||w|| ≠ 0, so we € can multiply both sides by ||v||||w||: v•w (||v||||w||) = 0(||v||||w||)

€

v w

v•w = 0. (⇐) Assume for the two non-zero vectors v and w, v•w = 0. € Since v and w are non-zero vectors, then ||v|| ≠ 0 and ||w|| ≠ 0, so we can divide both sides by ||v||||w||: v•w = 0 ||v||||w|| ||v||||w|| v•w =0 v w

But cos(θ) =

€

v•w v w

, so

cos(θ) = 0 θ=

(use the inverse cosine)

π 2

€ the angle between the two vectors is Since

π 2

, then, by definition, the

vectors are orthogonal. Determine if the following vectors are orthogonal: 14 Ex. 5 v = 6i + 9j and w = – 7i + j 3

Solution: 14 v•w = 6(– 7) + 9( ) = – 42 + 42 = 0 3

By our theorem that we just proved, the vectors are orthogonal.

222 Objective #5:

Decompose a Vector into Two Orthogonal Vectors.

We have already seem how to write a vector v in the form ai + bj. Notice that the vectors i and j are orthogonal so we can say the v1 = ai and v2 = bj is decomposition of the v. Here v1 is a vector parallel to the positive x-axis and v2 is a vector orthogonal to the positive x-axis. Now, suppose instead of using the positive x-axis, we want to decompose v into vectors v1 and v2 such that v1 is a vector parallel a vector w and v2 is a vector orthogonal to the w. The parallel vector v1 is called the projection of v onto w. To find the v1, we draw a line from the terminal point of v to the vector w so that the line is perpendicular to the v vector w. v1 is the vector from the initial point of v to this intersection point. The v2 v1 vector v2 is v – v1 and it will be orthogonal to w by the way we constructed v1. Thus, v = v1 + v2, so v•w = (v1 + v2)•w = v1•w + v2•w But, v1 is parallel to w, so v1 = αw for some real number α and v2 is orthogonal to w, meaning v2•w = 0. Hence, v•w = v1•w + v2•w = αw•w + 0 = α||w||2 Solving for α yields: v•w = α||w||2 α=

v•w w

2

This means that v1 = αw =

v•w w

2

w

Theorem € If v and w are two non-zero vectors, then the projection of v onto w is v•w v1 = w € 2 w

and the decomposition of v into v1 and v2, where v1 is parallel to w and v2 is orthogonal to w, is

€

€

v1 =

v•w w

2

w

v2 = v – v1

w

223 Decompose v into two vectors v1 and v2 where v1 || w and v2 ⊥ w: Ex. 6

v = 2i – 3j and w = 2i – j Solution: ||w|| = (2)2 +(−1)2 = 5 v•w = 2(2) + (– 3)(– 1) = 4 + 3 = 7 v1 =

v•w

v1 =

7 5

w

2

w=

7 ( 5)

2

w=

7 5

w

(replace w with 2i – j)

(2i – j) = 2.8i – 1.4j

v2 = v – v1 = 2i – 3j – (2.8i – 1.4j) = – 0.8i – 1.6j

€ Ex. 7

A car weighing 4200 pounds is parked on a street with a slope of 8˚. Find the force required to keep the car from rolling down the hill. Solution: 8˚ The force of gravity is pulling the car straight down towards the center of w the Earth with a magnitude of 4200 pounds. Let w be the unit vector 4200 lb along the incline of the hill. The angle between the positive x-axis and the vector labeled w is 180˚ + 8˚ = 188˚. Thus, the unit vector is w = cos(188˚)i + sin(188˚)j with ||w|| = 1 The force vector due to gravity is F = – 4200j Thus, the projection of F onto w is F•w −4200j•[cos(188°)i+sin(188°)j] w= [cos(188˚)i + sin(188˚)j] 2 2 w

( 1)

= – 4200sin(188˚)j[cos(188˚)i + sin(188˚)j] = 584.527…[cos(188˚)i + sin(188˚)j] ≈ 584.53[cos(188˚)i + sin(188˚)j] € € Hence, the magnitude of the force is 584.53 pounds so 584.53 pounds of force is needed to keep the car from rolling down the hill. Objective #6:

Compute Work

In physics, when a constant force F is applied that moves an object from point A to point B, we say that the work W done is the product of the magnitude of the force and the length of the distance d the object is moved:

224 W = ||F|| ||d||, assuming the force is applied along the line of motion If the force is measured in pounds and the distance in feet, then the work will have foot-pounds as its units. If the force is measured in newtons and the distance in meters, then the work will have newton-meters or joules as its units. If the force is not along the line of motion, then the work done will be the projection of the force along the line of motion times the distance d from point A to point B: W = ||amount of force F projected on d||(distance d from A to B) F•d But the amount of force F projected on d = 2 d, so d

W = ||

F•d d

2

d||||d|| =

F•d d

2

||d||||d|| = F•d

€ Solve the following: Ex. 8 Find the work done by a force of 8 pounds acting in the € direction 50˚ to € the horizontal moving an object 7 feet from (0, 0) to (7, 0). Solution: The movement from point A, (0, 0), to point B (7, 0) is represented as d = 7i + 0j. The force vector is F = 8[cos(50˚)i + sin(50˚)j] ≈ 5.142…i + 6.128…j So, the work done is: W = F•d = (5.142…i + 6.128…j)•(7i + 0j) = 5.142…(7) + 6.128…(0) ≈ 36.0 ft-lb Ex. 9

A wagon is pulled horizontally by an exerting force of 30 pounds on the handle at a 25˚ angle with the horizontal. How much work is done in moving the wagon 200 feet? Solution: Let (0, 0) be the initial point and let (200, 0) be the terminal point the distance the wagon has been moved. The distance vector is: d = 200i + 0j. The force vector F is given by: F = ||F||cos(25˚)i + ||F||sin(25˚)j = 30cos(25˚)i + 30sin(25˚)j = 27.189…i + 12.678…j Thus, W = F•d = 27.189…(200) + (0) = 5437.846… + 0 ≈ 5437.85 So, ≈ 5,437.85 foot-pounds or work was done.