. Using our polar coordinate conversions, we could write a

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1 504 Chapte 8 Section Dot Poduct Now that we can add, sutact, and scale vectos, you might e wondeing whethe we can multiply vectos. It tuns out thee ae two diffeent ways to multiply vectos, one which esults in a nume, and one which esults in a vecto. In this section, we'll focus on the fist, called the dot poduct o scala poduct, since it poduces a single numeic value (a scala). We'll egin with some motivation. In physics, we often want to know how much of a foce is acting in the diection of motion. To detemine this, we need to know the angle etween diection of foce and the diection of motion. Likewise, in compute gaphics, the lighting system detemines how ight a tiangle on the oject should e ased on the angle etween oject and the diection of the light. In oth applications, we'e inteested in the angle etween the vectos, so let's stat thee. Suppose we have two vectos, a = a, a 1 and =, 1. Using ou pola coodinate convesions, we could wite a = a cos( α), a sin( α) and = cos( β ), sin( β ). Now, if we knew the angles α and β, we wouldn't have much wok to do - the angle etween the vectos would e θ = α β. While we cetainly could using some invese tangents to find the two angles, it would e geat if we could find a way to detemine the angle etween the vecto just fom the vecto components. To help us manipulate θ = α β, we might ty intoducing a tigonometic function: cos θ = cos α β ( ) ( ) Now we can apply the diffeence of angles identity cos θ = cos α cos β + sin α sin β ( ) ( ) ( ) ( ) ( ) Now a = a cos( ), so 1 α cos( α ) = a1 a Making those sustitutions, a1 1 a a11 + a cos( θ ) = + = a a a a cos θ = a + a ( ) 1 1, and likewise fo the othe thee components. Notice the expession on the ight is a vey simple calculation ased on the components of the vectos. It comes up so fequently we define it to e the dot poduct of the two vectos, notated y a dot. This gives us two definitions of the dot poduct.

2 Section Dot Poduct 505 Definitions of the Dot Poduct a = a1 1 + a a = a cos( θ ) The fist definition, a = a1 1 + a, gives us a simple way to calculate the dot poduct a = a cos θ, gives us a geometic fom components. The second definition, ( ) intepetation of the dot poduct, and gives us a way to find the angle etween two vectos, as we desied. Example 11 Find the dot poduct 3, 5, 1. Using the fist definition, we can calculate the dot poduct y multiplying the x components and adding that to the poduct of the y components. 3, 5,1 = (3)(5) + ( )(1) = 15 = 13 Example 1 Find the dot poduct of the two vectos shown. We can immediately see that the magnitudes of the 30 two vectos ae 7 and 6. We can quickly calculate that the angle etween the vectos is 150. Using the geometic definition of the dot poduct, 3 a = a cos ( θ ) = (6)(7) cos(150 ) = 4 = Ty it Now 1. Calculate the dot poduct 7,3, 6 Now we can etun to ou goal of finding the angle etween vectos.

3 506 Chapte 8 Example 13 An oject is eing pulled up a amp in the diection 5, 1 y a ope pulling in the diection 4,. What is the angle etween the ope and the amp? Using the component fom, we can easily calculate the dot poduct. a = 5,1 4, = (5)(4) + (1)() = 0 + = We can also calculate the magnitude of each vecto. a = = 6, = 4 + = 0 Using the geometic definition, we can solve fo the angle etween the vectos. a = a = 6 cos( θ ) 0 cos( θ ) θ = cos Example 14 Calculate the angle etween the vectos 6, 4 and, 3. Calculating the dot poduct, 6,4,3 = (6)( ) + (4)(3) = = 0 We don't even need to calculate the magnitudes in this case since the dot poduct is 0. a = a cos( θ ) 0 = a θ = cos cos( θ ) 1 0 a = cos 1 ( 0) = 90 Ty it Now. Ae the vectos 7, 3 and, 6 othogonal? If not, find the angle etween them. With the dot poduct equaling zeo, as in the last example, the angle etween the vectos will always e 90, indicating that the vectos ae othogonal, a moe geneal way of saying pependicula. This gives us a quick way to check if vectos ae othogonal.

4 Section Dot Poduct 507 Also, if the dot poduct is positive, then the inside of the invese cosine will e positive, giving an angle less than 90. A negative dot poduct will then lead to an angle lage than 90 Sign of the Dot Poduct If the dot poduct is: Zeo The vectos ae othogonal (pependicula). Positive The angle etween the vectos is less than 90 Negative The angle etween the vectos is geate than 90 Pojections In addition to finding the angle etween vectos, sometimes we want to know how much one vecto points in the diection of anothe. Fo example, when pulling an oject up a amp, we might want to know how much of the foce is exeted in the diection of motion. To detemine this we can use the idea of a pojection. a a u v In the pictue aove, u is a pojection of a onto. In othe wods, it is the potion of a that points in the same diection as. To find the length of u, we could notice that it is one side of a ight tiangle. If we define θ to e the angle etween a and u u, then cos(θ ) =, so a cos(θ ) = u. a While we could find the angle etween the vectos to detemine this magnitude, we could skip some steps y using the dot poduct diectly. Since a = a cos(θ ), a a a cos(θ ) =. Using this, we can ewite u = a cos(θ ) as u =. This gives us a the length of the pojection, sometimes denoted as comp a = u =. To find the vecto u itself, we could fist scale to a unit vecto with length 1:. Multiplying this y the length of the pojection will give a vecto in the diection of ut with the coect length.

5 508 Chapte 8 poj a a a u = = = Pojection Vecto The pojection of vecto a onto a is poj a = a The magnitude of the pojection is comp a = Example 15 Find the pojection of the vecto 3, onto the vecto 8, 6. We will need to know the dot poduct of the vectos and the magnitude of the vecto we ae pojecting onto. 3, 8,6 = (3)(8) + ( )(6) = 4 1 = 1 8,6 = = = 100 = 10 The magnitude of the pojection will e 3, 8,6 8,6 1 = = To find the pojection vecto itself, we would multiply that magnitude y 8, 6 scaled to a unit vecto. 6 8,6 6 8, = = 8,6 =, =,. 5 8, Based on the sketch aove, this answe seems easonale. Ty it Now 3. Find the component of the vecto 3, 4 that is othogonal to the vecto 8, 4 Wok In physics, when a constant foce causes an oject to move, the mechanical wok done y that foce is the poduct of the foce and the distance the oject is moved. Howeve, we only conside the potion of foce that is acting in the diection of motion.

6 Section Dot Poduct 509 This is simply the magnitude of the pojection of the foce F d vecto onto the distance vecto,. The wok done is the d poduct of that component of foce times the distance moved, the magnitude of the distance vecto. F d Wok = d = F d d F u d It tuns out that wok is simply the dot poduct of the foce vecto and the distance vecto. Wok When a foce F causes an oject to move some distance d, the wok done is Wok = F d Example 16 A cat is pulled 0 feet y applying a foce of 30 pounds on a ope held at a 30 degee angle. How much wok is done? Since wok is simply the dot poduct, we can take advantage of the geometic definition of the dot poduct in this case. Wok = F d = F d cos( θ ) = (30)(0) cos(30 ) ft-ls. 30 pounds 30 0 feet Ty it Now 4. Find the wok down moving an oject fom the point (1, 5) to (9, 14) y the foce vecto F = 3, Impotant Topics of This Section Calculate Dot Poduct Using component defintion Using geometic definition Find the angle etween two vectos Sign of the dot poduct Pojections Wok

7 510 Chapte 8 Ty it Now Answes 1. 7,3, 6 = ( 7)( ) + (3)( 6) = = 4. In the pevious Ty it Now, we found the dot poduct was -4, so the vectos ae not othogonal. The magnitudes of the vectos ae ( 7) + 3 = 58 ( ) + 6 = 40 4 θ = cos The angle etween the vectos will e and 3. We want to find the component of 3, 4 that is othogonal to the vecto 8, 4. In the pictue to the ight, that component is vecto v. Notice that u + v = a, so if we can find the pojection vecto, we can find v. 3,4 8,4 a 40 u = poj a = = 8,4 = 8,4 = 4, ( ( 8) 4 ) v u a Now we can solve u + v = a fo v. v = a u = 3,4 4, = 1, 4. The distance vecto is 9 1,14 5 = 8, 9. The wok is the dot poduct: Wok = F d = 3, 8,9 = = 4

8 Section Dot Poduct 511 Section Execises Two vectos ae descied y thei magnitude and diection in standad position. Find the dot poduct of the vectos. 1. Magnitude: 6, Diection: 45 ; Magnitude: 10, Diection: 10. Magnitude: 8, Diection: 0 ; Magnitude: 7, Diection: 305 Find the dot poduct of each pai of vectos. 3. 0, 4 ; 3, , 5 ; 3, 7 5., 1 ; 10, 13 6., 5 ; 8, 4 Find the angle etween the vectos 7. 0, 4 ; 3, , 5 ; 3, 7 9., 4 ; 1, , 1 ; 8, 11. 4, ; 8, , 3 ; 6, Find a value fo k so that, 7 and k, 4 will e othogonal. 14. Find a value fo k so that 3, 5 and, k will e othogonal. 15. Find the magnitude of the pojection of 8, 4 onto 1, Find the magnitude of the pojection of, 7 onto 4, Find the pojection of 6, 10 onto 1, Find the pojection of 0, 4 onto 3, A scientist needs to detemine the angle of eflection when a lase hits a mio. The pictue shows the vecto epesenting the lase eam, and a vecto that is othogonal to the mio. Find the acute angle etween these, the angle of eflection. 0. A tiangle has coodinates at A: (1,4), B: (,7), and C: (4,). Find the angle at point B. 1. A oat is tapped ehind a log lying paallel to the dock. It only equies 10 pounds of foce to pull the oat diectly towads you, ut ecause of the log, you'll have to pull at a 45 angle. How much foce will you have to pull with? (We'e going to assume that the log is vey slimy and doesn't contiute any additional esistance.)

9 51 Chapte 8 1. A lage oulde needs to e dagged to a new position. If pulled diectly hoizontally, the oulde would 15 equie 400 pounds of pulling foce to move. We need to pull the oulde using a ope tied to the ack of a lage tuck, foming a 15 angle fom the gound. How much foce will the tuck need to pull with? 1. Find the wok done against gavity y pushing a 0 pound cat 10 feet up a amp that is 10 aove hoizontal. Assume thee is no fiction, so the only foce is 0 pounds downwads due to gavity.. Find the wok done against gavity y pushing a 30 pound cat 15 feet up a amp that is 8 aove hoizontal. Assume thee is no fiction, so the only foce is 30 pounds downwads due to gavity. 3. An oject is pulled to the top of a 40 foot amp that foms a 10 angle with the gound. It is pulled y ope exeting a foce of 10 pounds at a 35 angle elative to the gound. Find the wok done An oject is pulled to the top of a 30 foot amp that foms a 0 angle with the gound. It is pulled y ope exeting a foce of 80 pounds at a 30 angle elative to the gound. Find the wok done.