12.1 Three Dimensional Coordinate Systems (Review) Equation of a sphere

Transcription

1 12.2 Vectors

2 12.1 Three Dimensional Coordinate Systems (Reiew) Equation of a sphere x a 2 + y b 2 + (z c) 2 = r 2 Center (a,b,c) radius r

3 12.2 Vectors Quantities like displacement, elocity, and force inole both magnitude and direction (unlike quantities like mass or time). To represent these quantities we use a ector represented by a directed line segment (arrow) Terminal point The magnitude of a ector is represented by or. or sometimes by AB Initial point We also call it the length of

4 Any other ector that has the same magnitude and direction is called an equialent or equal ector AB CD OP EF A ector in standard position has its initial point at the origin. The directed line segment are equialent (same). PQ and

5 We can multiply a ector by a real number c. This is called scalar multiplication and denoted by c. c has magnitude c times the magnitude of : c c and points in the same direction as if c 0 or opposite direction if c

6 We can add a ector to another ector u. This is called ector addition, u Vector subtraction u addition in disguise + is just ector u

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8 It is much simpler to study ectors algebraically: a, a, a a1, a2 or a a 1, 2 a 2 Standard unit (i.e. length 1) ectors a 1 i 1,0,0 j 0,1,0 k 0,0,1 a, a, a a i a j a k a a a i component of jcomponent of k component of

9 Scalar Multiplication: a1, a2, a3 scaled by a factor c ca, ca, ca multiply each component by c Vector Addition: a, a, a added to u b, b, b u a1 b1, a2 b2, a3 b 3 add componentwise Length of the ector : Unit ector u in the direction of : c a a a u a1, a2, a a 3, a, a

10 Vector u = PP from a point P ( x, y, z ) to a point P ( x, y, z ) : P P P P x x, y y,z z ( x x ) i (y y ) j (z z ) k

11 Find the component form and magnitude of the ector initial point 3, 2, 0 and terminal point 4, 4, 2. Find a unit ector in the direction of. with the 4 3,4 2,2 0 1,2,2 u ,2,2,, check: u

12 A 200 lb. traffic light supported by two cables hangs in equilibrium. As shown in figure (b), let the weight of the light be represented by w and the forces in the two cables by and. Determine the F1 F2 F1 F2 magnitude of and. F1 F2 w 0 As shown in figure (c), the forces can be arranged to form a triangle. Equilibrium implies that the sum of the forces is 0 : F F w 0 F F cos 20, F sin F F cos 15, F sin w 0, 200 x coordinates cancel: F cos 20 F cos 15 0 or F F y coordinates cancel: F sin 20 sin F cos 15 tan 20 sin F lbs. F lbs. cos 15 tan 20 sin 15 F 2 cos 15 cos 20

13 12.3 The Dot Product

14 We can add two ectors, what about multiplying two ectors? Since adding two ectors yields another ector where the corresponding components are added, will the same work for multiplication? No, the product of two ectors yielding another ector where the corresponding components are multiplied is meaningless. There are actually two ector products that yield meaningful results but neither of these gie a new ector using component-wise multiplication Dot Product 12.4 Cross Product

15 The dot product of u u, u, u and,, is u u u u Note: The dot product of two ectors is a number (scalar) not a ector. Example:

16 Check property 4: u u, u, u u u u u u u u, u, u u, u, u u u u u or u u u

17 Are the following expression meaningful? a uw scalar ector scalar ector b u w c u w scalar scalar d u w ector ector e u w scalar ector f u w scalar scalar

18 Find the angle between two ectors : Let u and be nonzero ectors, then Law of Cosines: u u u u 2 u cos 2 u u u u u u 2 2 u u u u u 2u hence u u cos cos u u

19 cos u 0 u Alternate form of dot product: u u cos u and cos will always hae the same sign If 0 acute then cos 0 or u 0 2 If obtuse then cos 0 or u 0 2 right angle if cos( )=0 or u 0 2 u and are orthogonal if u 0

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21 ector projection of u onto u proj u comp u The component of u in the direction of is u cos, which is (up to sign) the length of proj u recall u u cos hence u cos u comp u u u the ector projection of u onto has as its magnitude and goes in the same direction as proj u u proj u u

22 Work = Force x Distance W F d If force and motion hae the same direction If a constant force F applied to a body acts at an angle to the direction of motion, then the work done W is: W F cos D where D is the displacement ector Using the dot product we hae: W FD

23 A toy wagon is pulled by exerting a force of 25 pounds on a handle that makes a 20 angle with the horizontal. Find the work done in Pulling the wagon 50 feet. F 25cos 20,25sin 20 D 50,0 W FD 1250cos ft. lbs. or 1593 joules