6.4 VECTORS AND DOT PRODUCTS

Transcription

1 458 Chapter 6 Additional Topics in Trigonometry 6.4 VECTORS AND DOT PRODUCTS What yo shold learn ind the dot prodct of two ectors and se the properties of the dot prodct. ind the angle between two ectors and determine whether two ectors are orthogonal. Write a ector as the sm of two ector components. Use ectors to find the work done by a force. Why yo shold learn it Yo can se the dot prodct of two ectors to sole real-life problems inoling two ector qantities. or instance, in Exercise 76 on page 466, yo can se the dot prodct to find the force necessary to keep a sport tility ehicle from rolling down a hill. The Dot Prodct of Two Vectors So far yo hae stdied two ector operations ector addition and mltiplication by a scalar each of which yields another ector. In this section, yo will stdy a third ector operation, the dot prodct. This prodct yields a scalar, rather than a ector. Definition of the Dot Prodct The dot prodct of, and, is. Properties of the Dot Prodct Let,, and w be ectors in the plane or in space and let c be a scalar w w c c c or proofs of the properties of the dot prodct, see Proofs in Mathematics on page 490. Example inding Dot Prodcts Joe Raedle/Getty Images ind each dot prodct. a. 4, 5, 3 b.,, c. a. 4, 5, , 3 4, b.,, 0 c. 0, 3 4, Now try Exercise 7. In Example, be sre yo see that the dot prodct of two ectors is a scalar (a real nmber), not a ector. Moreoer, notice that the dot prodct can be positie, zero, or negatie.

2 Section 6.4 Vectors and Dot Prodcts 459 Example Using Properties of Dot Prodcts Let, 3,, 4, and w,. ind each dot prodct. a. w b. Begin by finding the dot prodct of and. a. 4, b., 3, , Notice that the prodct in part (a) is a ector, whereas the prodct in part (b) is a scalar. Can yo see why? Now try Exercise 7. Example 3 Dot Prodct and Magnitde Origin IGURE 6.33 The dot prodct of with itself is 5. What is the magnitde of? Becase and 5, it follows that 5. Now try Exercise 5. The Angle Between Two Vectors The angle between two nonzero ectors is the angle, 0, between their respectie standard position ectors, as shown in igre This angle can be fond sing the dot prodct. Angle Between Two Vectors If is the angle between two nonzero ectors and, then cos. or a proof of the angle between two ectors, see Proofs in Mathematics on page 490.

3 460 Chapter 6 Additional Topics in Trigonometry Example 4 inding the Angle Between Two Vectors y 6 = 3, = 4, IGURE 6.34 x ind the angle between 4, 3 and 3, 5. The two ectors and are shown in igre cos This implies that the angle between the two ectors is arccos 4, 3 3, 5 4, 3 3, Now try Exercise 35. Rewriting the expression for the angle between two ectors in the form cos Alternatie form of dot prodct prodces an alternatie way to calclate the dot prodct. rom this form, yo can see that becase and are always positie, and cos will always hae the same sign. igre 6.35 shows the fie possible orientations of two ectors. cos Opposite Direction IGURE 6.35 < < < cos < 0 Obtse Angle cos 0 90 Angle 0 < < 0 < cos < Acte Angle 0 cos Same Direction Definition of Orthogonal Vectors The ectors and are orthogonal if 0. The terms orthogonal and perpendiclar mean essentially the same thing meeting at right angles. Note that the zero ector is orthogonal to eery ector, becase 0 0.

4 Section 6.4 Vectors and Dot Prodcts 46 TECHNOLOGY The graphing tility program, inding the Angle Between Two Vectors, fond on the website for this text at academic.cengage.com, graphs two ectors a, b and c, d in standard position and finds the measre of the angle between them. Use the program to erify the soltions for Examples 4 and 5. Example 5 Determining Orthogonal Vectors Are the ectors, 3 and 6, 4 orthogonal? ind the dot prodct of the two ectors., 3 6, Becase the dot prodct is 0, the two ectors are orthogonal (see igre 6.36). y 4 = 6, =, 3 x IGURE 6.36 Now try Exercise 53. inding Vector Components Yo hae already seen applications in which two ectors are added to prodce a resltant ector. Many applications in physics and engineering pose the reerse problem decomposing a gien ector into the sm of two ector components. Consider a boat on an inclined ramp, as shown in igre The force de to graity plls the boat down the ramp and against the ramp. These two orthogonal forces, and w, are ector components of. That is, w. Vector components of The negatie of component represents the force needed to keep the boat from rolling down the ramp, whereas w represents the force that the tires mst withstand against the ramp. A procedre for finding and is shown on the following page. w IGURE 6.37 w

5 46 Chapter 6 Additional Topics in Trigonometry Definition of Vector Components Let and be nonzero ectors sch that w where and w are orthogonal and is parallel to (or a scalar mltiple of), as shown in igre The ectors and w are called ector components of. The ector is the projection of onto and is denoted by proj. The ector w is gien by w. w w is acte. IGURE 6.38 is obtse. rom the definition of ector components, yo can see that it is easy to find the component w once yo hae fond the projection of onto. To find the projection, yo can se the dot prodct, as follows. So, and w c w c w c c w c 0 proj c. is a scalar mltiple of. Take dot prodct of each side with. w and are orthogonal. Projection of onto Let and be nonzero ectors. The projection of onto is proj.

6 Section 6.4 Vectors and Dot Prodcts 463 Example 6 Decomposing a Vector into Components y = 6, w = 3, 5 IGURE 6.39 x ind the projection of 3, 5 onto 6,. Then write as the sm of two orthogonal ectors, one of which is proj. The projection of onto is as shown in igre The other component, w, is So, proj w 3, 5 6 5, 5 9 7, 5 5. w 6 5, 5 9 7, 5 5 3, 5. Now try Exercise , 6 5, 5 Example 7 inding a orce 30 IGURE 6.40 A 00-pond cart sits on a ramp inclined at 30, as shown in igre What force is reqired to keep the cart from rolling down the ramp? Becase the force de to graity is ertical and downward, yo can represent the graitational force by the ector 00j. orce de to graity To find the force reqired to keep the cart from rolling down the ramp, project onto a nit ector in the direction of the ramp, as follows. cos 30i sin 30j 3 i j Therefore, the projection of onto is proj i j. Unit ector along ramp The magnitde of this force is 00, and so a force of 00 ponds is reqired to keep the cart from rolling down the ramp. Now try Exercise 75.

7 464 Chapter 6 Additional Topics in Trigonometry Work The work W done by a constant force acting along the line of motion of an object is gien by W magnitde of forcedistance PQ \ as shown in igre 6.4. If the constant force is not directed along the line of motion, as shown in igre 6.4, the work W done by the force is gien by \ W proj PQ PQ \ cos PQ \ PQ \. Projection form for work \ proj PQ cos Alternatie form of dot prodct proj PQ P Q P Q orce acts along the line of motion. orce acts at angle with the line of motion. IGURE 6.4 IGURE 6.4 This notion of work is smmarized in the following definition. Definition of Work The work W done by a constant force as its point of application moes along the ector PQ \ is gien by either of the following.. W proj PQ \PQ \ Projection form. W PQ \ Dot prodct form Example 8 inding Work To close a sliding barn door, a person plls on a rope with a constant force of 50 ponds at a constant angle of 60, as shown in igre ind the work done in moing the barn door feet to its closed position. P ft proj PQ Q Using a projection, yo can calclate the work as follows. W proj PQ \ PQ \ Projection form for work 60 cos 60PQ \ 300 foot-ponds 50 IGURE 6.43 ft So, the work done is 300 foot-ponds. Yo can erify this reslt by finding the ectors and PQ \ and calclating their dot prodct. Now try Exercise 79.