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Vector Algebra When dealing with scalars, the usual math operations (+, -, …) are sufficient to obtain any information needed. When dealing with vectors, the magnitudes can be operated on as scalars, but we must account for doing operations on or between vector directions. Addition: A+B=B+A (commutative) (A + B) + C = A + (B + C) (associative) Subtraction: A – B = A + (-B) NOTE: When adding vectors, you can only combine components that are in the same direction! Ex. 3,2 + 4, −1 = (3 + 4)xˆ + (2 − 1)yˆ = 7 xˆ + 1yˆ Visual Representation B A A -B B+A -A A A+B A-B A B < Tail to Tip Method Handout > Magnitude or length of a vector |v|: Let v = axˆ + byˆ + czˆ The length or magnitude of v is the distance from the origin 0,0,0 to point a, b, c . Using the distance formula: v = (a − 0) 2 + (b − 0) 2 + (c − 0) 2 = a 2 + b 2 + c 2 1 Multiplication by a scalar: a(A + B) = aA + aB (distributive) If A = a1xˆ + a2 yˆ + a3zˆ , then kA = ka1xˆ + ka2 yˆ + ka3zˆ NOTE: Multiplying a vector by a positive scalar does NOT effect its direction, only its magnitude. Multiplying by a negative scalar reverses the vector’s direction. Ex. 3(2xˆ − 4yˆ + 1zˆ ) = 6xˆ − 12yˆ + 3zˆ Visual Representation A 3A Dot Product: I A ⋅ B = a1b1 + a2b2 + a3b3 II A ⋅ B = AB cosθ where A = a1xˆ + a2 yˆ + a3zˆ & B = b1xˆ + b2 yˆ + b3zˆ where θ is the angle formed between the 2 vectors. NOTE: This process is also referred to as the scalar product because the final result is a scalar, not a vector. In the second expression, A and B represent the magnitudes of the vectors A and B. This method is only used when you are looking for the angle between 2 vectors or when the angle between 2 vectors is known. < Visual Representation of the Dot Product > NOTE: The dot product is commutative, meaning A ⋅ B = B ⋅ A . The dot product is distributive, meaning ( A + B) ⋅ C = A ⋅ C + B ⋅ C Ex. (3xˆ + 4yˆ ) ⋅ (−1xˆ + 2yˆ + 5zˆ ) = (3 ⋅ −1) + (4 ⋅ 2) + (0 ⋅ 5) =5 2 Ex. Given A = 3xˆ + 2yˆ and B = 4xˆ − 1yˆ , find the angle between. cosθ = A⋅B Î AB ⎛ A⋅B ⎞ ⎟ ⎝ AB ⎠ θ = cos −1 ⎜ ⎛ 3 ⋅ 4 + 2 ⋅ (−1) ⎞ ⎟ ⎝ 9 + 4 16 + 1 ⎠ θ = cos −1 ⎜ ⎛ 10 ⎞ ⎟ = 0.83298 rad 221 ⎝ ⎠ θ = cos −1 ⎜ (47.7o) Ex. What happens when you take the dot product of a vector with unit vectors? Let v = v1xˆ + v2 yˆ + v3zˆ v ⋅ xˆ = v1 , v2 , v3 ⋅ 1,0,0 = v1 v ⋅ yˆ = v1 , v2 , v3 ⋅ 0,1,0 = v2 v ⋅ zˆ = v1 , v2 , v3 ⋅ 0,0,1 = v3 Ex. What happens when you take the dot product between unit vectors? xˆ ⋅ xˆ = 1,0,0 ⋅ 1,0,0 = 1 xˆ ⋅ yˆ = 1,0,0 ⋅ 0,1,0 = 0 Likewise yˆ ⋅ yˆ = zˆ ⋅ zˆ = 1 this means they are to each other likewise xˆ ⋅ zˆ = yˆ ⋅ zˆ = 0 NOTE: This property is the same for other coordinate systems. * The dot product for any set of orthogonal unit vectors can be summarized by using the Kronecker delta ( δ ij ): eˆ i ⋅ eˆ j = δ ij 3 Multiplication by a vector: (Cross Product) Let A = a1eˆ1 + a2eˆ 2 + a3eˆ 3 & B = b1eˆ1 + b2eˆ 2 + b3eˆ 3 I A × B = (a2b3 − a3b2 )eˆ1 − (a1b3 − a3b1 )eˆ 2 + (a1b2 − a2b1 )eˆ 3 II ⎛ eˆ1 eˆ 2 A × B = det ⎜⎜ a1 a2 ⎜b b ⎝ 1 2 III A × B = AB sin θ nˆ eˆ 3 ⎞ a3 ⎟⎟ b3 ⎟⎠ or eˆ1 eˆ 2 a1 a2 b1 b2 eˆ 3 a3 b3 where n̂ represents the direction perp. To both A & B ( n̂ = the normal unit vector, meaning perpendicular to) NOTE: This definition is also referred to as the vector product because the final result is a vector, not a scalar. * Important features of the cross product: 1) It is only defined for 3-D (for a 2-D vector, add a 0 for the missing dimension) 2) It yields a vector that is perpendicular to both original vectors (its direction is given by the right-hand-rule) 3) The cross product obeys the following algebraic properties A × B = − (B × A ) (not commutative) A × (B + C) = A × B + A × C (distributive) c( A × B) = (cA) × B = A × (cB) (distributive) ( A × B) × C = B( A ⋅ C) − A(B ⋅ C) (vector triple product) A × (B × C) = B( A ⋅ C) − C( A ⋅ B) (vector 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 × v & v × u are orthogonal (perp.) to both u and v u×v The magnitude (length) of u x v is a measure or reflection of how perpendicular u and v are. v u ( v × u) = −(u × v) Max length is when u is ⊥ to v 0 length when u & v are | | or anti- | | 2) u × v = uv sin θ 3) u × v = the area of a parallelogram having u & v for adjacent sides u u sinθ θ v 4) 1 u × v = the area of a triangle having u & v for adjacent sides 2 u θ u sinθ v 5) u × v = 0 iff u & v are scalar multiples of each other (parallel) 6) u ⋅ ( v × w ) = the volume of a parallelepiped having u, v & w as adjacent edges The triple scalar product can be found using: u1 u ⋅ ( v × w ) = v1 w1 u2 v2 w2 u3 v3 w3 NOTE: This value could be negative! 5 APPLICATIONS: The primary applications in physics involving cross products are: Torque Angular Momentum Magnetic Force Ex. Find the cross product A × B for A = 1xˆ + 3yˆ + 2zˆ & B = 3xˆ + 4yˆ − zˆ xˆ yˆ zˆ A × B = 1 3 2 = (−3 − 8)xˆ + (6 − (−1))yˆ + (4 − 9)zˆ 3 4 −1 = −11xˆ + 7yˆ − 5zˆ Ex. Find the cross product xˆ × yˆ for unit vectors xˆ = 1,0,0 & yˆ = 0,1,0 What do you expect the answer to be? ẑ ẑ x̂ ŷ xˆ yˆ zˆ xˆ × yˆ = 1 0 0 = (0 − 0)xˆ + (0 − 0)yˆ + (1 − 0)zˆ 0 1 0 = ẑ Ex. Find the cross product yˆ × xˆ . xˆ yˆ zˆ yˆ × xˆ = 0 1 0 = (0 − 0)xˆ + (0 − 0)yˆ + (0 − 1)zˆ 1 0 0 = −zˆ ** These last 2 examples illustrate the identity A × B = −(B × A) . 6 Ex. Find the area of a parallelogram having u = 3,2, −1 adjacent sides. & v = 1,3,3 as xˆ yˆ zˆ u × v = 3 2 −1 = (6 − (−3))xˆ + ((−1) − 9)yˆ + (9 − 2)zˆ 1 3 3 = 9xˆ − 10yˆ + 7 zˆ Area = u × v = 92 + (−10) 2 + 7 2 = 15.2 Ex. Find the area of the parallelogram in the previous example by finding the angle between the vectors and then using u × v = uv sin θ . To find the angle, we use the dot product: ⎛u⋅v⎞ ⎟ ⎝ uv ⎠ θ = cos −1 ⎜ ⎛ 3 ⋅ 1 + 2 ⋅ 3 + (−1) ⋅ 3 ⎞ ⎟ ⎝ 9 + 4 +1 1+ 9 + 9 ⎠ θ = cos −1 ⎜ ⎛ 6 ⎞ o ⎟ = 68.4 or (.38π) ⎝ 266 ⎠ θ = cos −1 ⎜ Therefore: Area = 266 sin(68.4o ) = 15.2 7 Applications of Dot and Cross-products Work Work done by a constant force is defined as W = F ⋅ d = Fd cos θ . 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 move a 200 kg crate 4 m if it is being dragged by a steel cord under a force F = (15xˆ + 12 yˆ ) N ? F = (15xˆ + 12 yˆ ) N θ 200 kg 4m Method I: (direct dot product) W = F⋅d = (15xˆ + 12yˆ ) ⋅ 4xˆ = (15xˆ ⋅ 4xˆ ) + (12yˆ ⋅ 4xˆ ) = 60 ( xˆ ⋅ xˆ ) + 48 ( yˆ ⋅ xˆ ) NOTE: xˆ ⋅ xˆ = 1 & yˆ ⋅ xˆ = 0 = 60 J Method II: (using magnitudes and angles) W = Fd cos θ = ( 369 = 60 J F = 15 + 12 = 2 ) ( 4 ) cos ( 38.66 ) o 2 369 d =4 ⎛ 12 ⎞ = 38.66o ⎟ ⎝ 15 ⎠ θ = tan −1 ⎜ 8 Angular Momentum The angular momentum of a rotating object about a fixed point is given by L = r × p = r × mv , 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 v is the velocity of a point located at r). The physical interpretation of angular momentum is two-fold: L is an indicator of the direction an object is rotating. L > 0 Counter-Clockwise L < 0 Clockwise 1) L θ r p -L 2) r p −θ |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 given by: L = rp sin θ = mvr sin θ Since sin θ can range from -1 to 1, this means the value of L can range from –mvr to mvr. Î − π 2 ≤θ ≤ π 2 −1 ≤ sin θ ≤ 1 − mvr ≤ L ≤ mvr The maximum values (1 or -1) occur when θ = 90o or – 90o. These special angles refer to an object experiencing circular motion. 9 Ex. Diatomic Molecule r θ m r m θ θ = 90o v Find |L| L = L1 + L 2 L = L1 + L 2 L = mvr sin θ + mvr sin θ L = mvr (1) + mvr (1) L = 2 mvr 10