Vector has a magnitude and a direction Scalar has a magnitude
Vector has a magnitude and a direction Scalar has a magnitude a brick on a table
Vector has a magnitude and a direction Scalar has a magnitude brick moves (2 meters)
Vector has a magnitude and a direction Scalar has a magnitude brick moves (2 meters) Brick moved 2 meters.
Vector has a magnitude and a direction Scalar has a magnitude brick moves (2 meters) Brick moved 2 meters to the right. Brick moved 2 meters.
Vector has a magnitude and a direction Scalar has a magnitude brick moves (2 meters) Brick moved 2 meters to the right. direction specified Brick moved 2 meters.
Vector has a magnitude and a direction Scalar has a magnitude brick moves (2 meters) Brick moved 2 meters to the right. s = 2 m to the right displacement direction specified Brick moved 2 meters. s = 2 m distance
Vector has a magnitude and a direction Scalar has a magnitude brick moves (2 meters) Brick moved 2 meters to the right. s = 2 m to the right direction specified Brick moved 2 meters. s = 2 m a vector displacement a scalar distance
Some quantities in physics Vector displacement s or s Scalar distance s different notations
Some quantities in physics Vector displacement s or s velocity v or v Scalar distance s speed, velocity v
Some quantities in physics Vector displacement s or s velocity v or v acceleration a or a Scalar distance s speed, velocity v acceleration a
Some quantities in physics Vector displacement s or s velocity v or v acceleration a or a force F or F Scalar distance s speed, velocity v acceleration a force F
Some quantities in physics Vector displacement s or s velocity v or v acceleration a or a force F or F Scalar distance s speed, velocity v acceleration a force F time t mass m
Working in 3D The moving brick was a 1D situation brick moves (2 meters) We will work primarily in 3D in this class.
Working in 3D The moving brick was a 1D situation brick moves (2 meters) We will work primarily in 3D in this class. Set up a right-handed coordinate system: y y z x x z right-handed system left-handed system
Working in 3D The moving brick was a 1D situation brick moves (2 meters) We will work primarily in 3D in this class. Set up a right-handed coordinate system: y y x The convention is to use a righthanded system. z x z right-handed system left-handed system
Displacement in 3D Something moves -3 m in the x direction, and -2 m in the y direction, and -6 m in the z direction
Displacement in 3D Something moves -3 m in the x direction, and -2 m in the y direction, and -6 m in the z direction = s a displacement in 3D
Displacement in 3D Something moves -3 m in the x direction, and -2 m in the y direction, and -6 m in the z direction = s a displacement in 3D Much better notation s = (3 m, 2 m, -6 m)
Displacement in 3D Something moves -3 m in the x direction, and -2 m in the y direction, and -6 m in the z direction = s a displacement in 3D Much better notation s = (3 m, 2 m, -6 m) More generally s = (s x, s y, s z ) components of the vector s
Visualizing the components s = (s x, s y, s z ) y s x z
Visualizing the components s = (s x, s y, s z ) s y y s z s s x (cf. actual physical model of this set up in class) x z
Visualizing the components s = (s x, s y, s z ) s y y s z s s x (cf. actual physical model of this set up in class) x z Magnitude of the vector? Pythagorean theorem gives it to us s = s = s2 x + s2 y + s2 z
A scalar times a vector Multiplying a vector by a scalar just scales the length of the vector. s = (s x, s y, s z ) as = (as x, as y, as z )
A scalar times a vector Multiplying a vector by a scalar just scales the length of the vector. s = (s x, s y, s z ) as = (as x, as y, as z ) y s 2s x z
Adding vectors Just add components. obvious for something like displacement s = (s x, s y, s z ) t = (t x, t y, t z ) s + t = (s x +t x, s y +t y, s z +t z )
Adding vectors Just add components. obvious for something like displacement s = (s x, s y, s z ) t = (t x, t y, t z ) Graphically + = s + t = (s x +t x, s y +t y, s z +t z )
Unit vectors Using two preceding rules (for as and s 1 +s 2 ), we can make a new useful notation using unit vectors. Define three unit vectors: x = (1, 0, 0) y = (0, 1, 0) z = (0, 0, 1) hat says that magnitude = 1
Unit vectors Using two preceding rules (for as and s 1 +s 2 ), we can make a new useful notation using unit vectors. Define three unit vectors: x = (1, 0, 0) y = (0, 1, 0) z = (0, 0, 1) hat says that magnitude = 1 Can assemble any vector by multiplying and adding these three vectors together
Unit vectors Using two preceding rules (for as and s 1 +s 2 ), we can make a new useful notation using unit vectors. Define three unit vectors: x = (1, 0, 0) y = (0, 1, 0) z = (0, 0, 1) hat says that magnitude = 1 s = (s x, s y, s z ) Can assemble any vector by multiplying and adding these three vectors together s = s x x + s y y + s z z equivalent
Representation is arbitrary Or: coordinate system is arbitrary Three identical vectors r 1 r 2 r 3
Representation is arbitrary Or: coordinate system is arbitrary Three identical vectors r 1 r 2 r 3 r 1 = r 2 = r 3 This equality is true.
Representation is arbitrary Or: coordinate system is arbitrary Three identical vectors No particular representation (coordinate system) has been chosen for any of these yet. r 1 r 2 r 3 r 1 = r 2 = r 3 This equality is true.
Representation is arbitrary Or: coordinate system is arbitrary Three identical vectors y r 1 r 2 r 3 x r 1 = r 2 = r 3 This equality is still true.
Choosing a particular representation is often convenient. s = (s x, s y, s z )
Choosing a particular representation is often convenient. s = (s x, s y, s z ) Also, thinking of the tail of the vector as being at the origin lets you think of vectors as points in 3D space. y s x z
Choosing a particular representation is often convenient. s = (s x, s y, s z ) Also, thinking of the tail of the vector as being at the origin lets you think of vectors as points in 3D space. y s x z But, take care! Vectors are fundamentally just lengths and directions and aren t tied to a representation. s
Derivative of a vector s = s x x + s y y + s z z ds ds x + ds y + ds z dt = x + y + z dt dt dt
Derivative of a vector s = s x x + s y y + s z z ds ds x + ds y + ds z dt = x + y + z dt dt dt If the position of an object is: Then the velocity is: x(t) v(t) = dx dt (example on blackboard)
Multiplying vectors (sort of) Given these objects called vectors, we can define various useful operations with them.
Multiplying vectors (sort of) Given these objects called vectors, we can define various useful operations with them. Two useful operations are multiplication-like in appearance. dot product and cross product
Multiplying vectors (sort of) Given these objects called vectors, we can define various useful operations with them. Two useful operations are multiplication-like in appearance. dot product and cross product For each: First, the definition. Then, some intuition.
Dot product s = s x x + s y y + s z z t = t x x + t y y + t z z s t = s x t x + s y t y + s z t z A dot product yields a scalar
Dot product s = s x x + s y y + s z z t = t x x + t y y + t z z s t = s x t x + s y t y + s z t z A dot product yields a scalar s t = s t cos θ st Alternative form: product of the magnitudes times cosine of the angle between the vectors
Dot product s = s x x + s y y + s z z t = t x x + t y y + t z z s t = s x t x + s y t y + s z t z A dot product yields a scalar s t = s t cos θ st Alternative form: product of the magnitudes times cosine of the angle between the vectors Clearly: s s = s 2 and s t = t s
Dot product Why? Gives the product of the magnitudes, with the modification that it only counts the components that are parallel. a b
Dot product Why? Gives the product of the magnitudes, with the modification that it only counts the components that are parallel. a θ b
Dot product Why? Gives the product of the magnitudes, with the modification that it only counts the components that are parallel. a θ b a cos θ b a b = a b cos θ
Dot product Why? Can look at it in reverse a θ b a b cos θ a b = a b cos θ
Dot product Bonus Can get angle between two vectors knowing only the components without any trouble Consider: s 1 = (1, 3, -4) s 2 = (2, -6, 1) What s the angle between these vectors?
Dot product Bonus Can get angle between two vectors knowing only the components without any trouble Consider: s 1 = (1, 3, -4) s 2 = (2, -6, 1) What s the angle between these vectors? Equate the two expressions for dot product and solve for cos θ (1)(2) + (3)(-6) + (-4)(1) = [ (1) 2 + (3) 2 + (-4) 2 ] [ (2) 2 + (-6) 2 + (1) 2 ] cos θ
Cross product s = s x x + s y y + s z z t = t x x + t y y + t z z s t = (s y t z - s z t y ) x + (s z t x - s x t z ) y + (s x t y - s y t x ) z A cross product yields another vector, perpendicular to the original two.
Cross product s = s x x + s y y + s z z t = t x x + t y y + t z z s t = (s y t z - s z t y ) x + (s z t x - s x t z ) y + (s x t y - s y t x ) z A cross product yields another vector, perpendicular to the original two. s t = s t sin θ st u Alternative form. Here, the vector u points perpendicular to the plane of s and t, according to a right-hand rule
Cross product s = s x x + s y y + s z z t = t x x + t y y + t z z s t = (s y t z - s z t y ) x + (s z t x - s x t z ) y + (s x t y - s y t x ) z A cross product yields another vector, perpendicular to the original two. s t = s t sin θ st u Not commutative! Alternative form. Here, the vector u points perpendicular to the plane of s and t, according to a right-hand rule s t = -t s
Cross product Why? Gives the product of the magnitudes with the modification that it neglects any parallel components. a θ b vector pointing out of page vector pointing into page a b a sin θ b a b = a b sin θ
Cross product Bonus If you are familiar with matrices and determinants a b = x y z a x a y a z b x b y b z
Cross product Bonus If you are familiar with matrices and determinants a b = x y z a x a y a z b x b y b z For the curious: The result of a cross product is subtly different from a regular vector and is sometimes called a pseudovector or axial vector. (further discussion in class)