Vector Algebra. Addition: (A + B) + C = A + (B + C) (associative) Subtraction: A B = A + (-B)



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Vector Algebra When dealing with scalars, the usual math operations (+, -, ) are sufficient to obtain any information needed. When dealing with ectors, the magnitudes can be operated on as scalars, but we must account for doing operations on or between ector directions. Addition: A + B = B + A (commutatie) (A + B) + C = A + (B + C) (associatie) Subtraction: A B = A + (-B) NOTE: When adding ectors, you can only combine components that are in the same direction! Ex. 3,2 + 4, 1 = (3 + 4) x + (2 1) y = 7x + 1y Visual Representation B -B A -A A A + B B + A A A - B A B < Tail to Tip Method Handout > Magnitude or length of a ector : Let = ax + by + cz The length or magnitude of is the distance from the origin 0,0,0 to point abc,,. Using the distance formula: = ( a 0) + ( b 0) + ( c 0) = a + b + c 2 2 2 2 2 2 1

Multiplication by a scalar: a(a + B) = aa + ab (distributie) A= ax + a y + a z, then If ka= kax + ka y + ka z NOTE: Multiplying a ector by a positie scalar does NOT effect its direction, only its magnitude. Multiplying by a negatie scalar reerses the ector s direction. Ex. 3(2x 4y + 1 z) = 6x 12y + 3z Visual Representation A 3A Dot Product: A B = ab + ab + ab where A= a 1x+ a2y+ a3z & B= b 1x+ b2y+ b3z I 1 1 2 2 3 3 II A B = ABcos where is the angle formed between the 2 ectors. NOTE: This process is also referred to as the scalar product because the final result is a scalar, not a ector. In the second expression, A and B represent the magnitudes of the ectors A and B. This method is only used when you are looking for the angle between 2 ectors or when the angle between 2 ectors is known. < Visual Representation of the Dot Product > NOTE: The dot product is commutatie, meaning A B= B A. The dot product is distributie, meaning ( A+ B) C= A C+ B C Ex. (3x + 4 y) ( 1x + 2y + 5 z ) = (3 1) + (4 2) + (0 5) = 5 2

Ex. Gien A= 3x + 2 y and B= 4x 1y, find the angle between. cos = A B AB A B AB 1 = cos 1 3 4+ 2 ( 1) = cos 9+ 4 16+ 1 1 10 = cos = 0.83298 221 rad (47.7 o ) Ex. What happens when you take the dot product of a ector with unit ectors? = x + y + z Let x = 1, 2, 3 1,0,0 = 1 y = 1, 2, 3 0,1,0 = 2 z = 1, 2, 3 0,0,1 = 3 Ex. What happens when you take the dot product between unit ectors? x x = 1,0,0 1,0,0 = 1 Likewise y y = z z = 1 x y = 1, 0,0 0,1, 0 = 0 this means they are to each other likewise x z = y z = 0 NOTE: This property is the same for other coordinate systems. * The dot product for any set of orthogonal unit ectors can be summarized by using the Kronecker delta ( δ ij ): e e =δ i j ij 3

Multiplication by a ector: (Cross Product) Let A= a1e 1+ a2e2+ a3e 3 & B= b1e 1+ b2e2+ b3e 3 I A B= ( ab 2 3 ab 3 2) e1 ( ab 1 3 ab 3 1) e2+ ( ab 1 2 ab 2 1) e 3 II e1 e2 e3 A B = det a1 a2 a 3 or b1 b2 b 3 e e e a a a b b b III A B= ABsinn where n represents the direction perp. To both A & B (n = the normal unit ector, meaning perpendicular to) NOTE: This definition is also referred to as the ector product because the final result is a ector, not a scalar. * Important features of the cross product: 1) It is only defined for 3-D (for a 2-D ector, add a 0 for the missing dimension) 2) It yields a ector that is perpendicular to both original ectors (its direction is gien by the right-hand-rule) 3) The cross product obeys the following algebraic properties A B= ( B A ) (not commutatie) A ( B+ C) = A B+ A C (distributie) c( A B) = ( ca) B= A ( cb ) (distributie) ( A B) C= B( A C) A( B C ) (ector triple product) A ( B C) = B( A C) C( A B ) (ector triple product) A ( B C) = ( A B) C (scalar triple product) A A =0 A 0= 0 4

* Geometric features of the cross product: 1) u & u are orthogonal (perp.) to both u and u u ( u) = ( u ) The magnitude (length) of u x is a measure or reflection of how perpendicular u and are. Max length is when u is to 0 length when u & are or anti- 2) u =usin 3) u = the area of a parallelogram haing u & for adjacent sides u u sin 4) 1 2 u = the area of a triangle haing u & for adjacent sides u u sin 5) u = 0 iff u & are scalar multiples of each other (parallel) 6) u ( w ) = the olume of a parallelepiped haing u, & w as adjacent edges The triple scalar product can be found using: u ( w ) = u u u w w w NOTE: This alue could be negatie! 5

APPLICATIONS: The primary applications in physics inoling cross products are: Torque Angular Momentum Magnetic Force Ex. Find the cross product A B for A= 1x + 3y + 2z & B= 3x + 4y z x y z A B= 1 3 2 = ( 3 8) x + (6 ( 1)) y + (4 9) z 3 4 1 = 11x + 7y 5z Ex. Find the cross product x y for unit ectors x = 1, 0,0 & y = 0,1,0 What do you expect the answer to be? ẑ ẑ x ŷ x y z x y = 1 0 0 = (0 0) x + (0 0) y + (1 0) z 0 1 0 = ẑ Ex. Find the cross product y x. x y z y x = 0 1 0 = (0 0) x + (0 0) y + (0 1) z 1 0 0 = z ** These last 2 examples illustrate the identity A B= ( B A ). 6

Ex. Find the area of a parallelogram haing u = 3,2, 1 & = 1,3,3 as adjacent sides. x y z u = 3 2 1 = (6 ( 3)) x + (( 1) 9) y + (9 2) z 1 3 3 = 9x 10y + 7z 2 2 2 Area = u = 9 + ( 10) + 7 = 15.2 Ex. Find the area of the parallelogram in the preious example by finding the angle between the ectors and then using u = usin. To find the angle, we use the dot product: u u 1 = cos 1 31 + 23 + ( 1)3 = cos 9+ 4+ 1 1+ 9+ 9 = cos = 68.4 266 1 6 o or (.38π) Therefore: Area = o 266 sin(68.4 ) = 15.2 7

Applications of Dot and Cross-products Work Work done by a constant force is defined as W = F d = Fdcos. In other words, how much work is done on an object is equal to the magnitude of the applied force in the direction of motion. The work done on an object is also a measure of the amount of energy the object has gained (W > 0) or lost (W < 0). Ex. How much work is required to moe a 200 kg crate 4 m if it is being dragged by a steel cord under a force F = (15x + 12 y ) N? F = (15x + 12 y ) N 200 kg 4 m Method I: (direct dot product) W = F d = ( 15x + 12y) 4x = ( 15x 4x) + ( 12y 4x ) = 60( ) + 48( ) = 60 J x x y x NOTE: x x = 1 & y x = 0 Method II: (using magnitudes and angles) W = Fdcos o = ( 369 )( 4) cos( 38.66 ) = 60 J 2 2 F = 15 + 12 = 369 d = 4 1 12 = tan = 38.66 15 o 8

Angular Momentum The angular momentum of a rotating object about a fixed point is gien by L = r p = r m, where r is the displacement from the fixed point to a point on the object and p is the linear momentum of the point located at r (or m is the mass and is the elocity of a point located at r). The physical interpretation of angular momentum is two-fold: 1) L is an indicator of the direction an object is rotating. L > 0 Counter-Clockwise L < 0 Clockwise L r p - L r p 2) L is a measure of how hard it is to stop an object that is rotating. Special Case: Circular Motion Since the angular momentum is a cross-product, the magnitude is gien by: L = rpsin = mrsin Since sin can range from -1 to 1, this means the alue of L can range from mr to mr. π π 2 2 1 sin 1 mr L mr The maximum alues (1 or -1) occur when = 90 o or 90 o. These special angles refer to an object experiencing circular motion. 9

Ex. Diatomic Molecule m r r m = 90 o Find L L = L + L 1 2 L = L + L 1 2 L = mrsin + mrsin L = mr(1) + mr(1) L = 2mr 10