2/7/2003, CROSS PRODUCT Math 21a, O. KnillHomework for Monday: Section 9.4, Numbers 16, 18, 26, 32, 34CROSS PRODUCT. The cross product of two vectors~v = (v1, v2, v3) and ~w = (w1, w2, w3) is defined as the vector~v × ~w = (v2w3− v3w2, v3w1− v1w3, v1w2− v2w1).To compute it:multiply diagonallyat the crosses.v1v2v3v1v2X X Xw1w2w3w1w2DIRECTION OF ~v × ~w: ~v × ~w is orthogonal to ~v and orthogonal to ~w.Proof. Check that ~v · (~v × ~w) = 0.LENGTH:|~v × ~w| = |~v|| ~w| sin(α)Proof. The identity |v×w|2= |v|2|w|2−(v·w)2can be proven by direct computation. Now, |v·w| = |v||w| cos(α).AREA. The length |~v × ~w| is the area of the parallelogram spanned by v and w.Proof. Check it first for v = (1, 0, 0) and w = (cos(α), sin(α), 0), where v × w = (0, 0, sin(α)) has length | sin(α)|which is indeed the area of the parallelogram spanned by v and w. A more general case can be obtained byscaling v and w: both the area as well as the cross product behave linearly in v and w.ZERO CROSS PRODUCT. We see that ~v × ~w is zero if ~v and ~w are parallel.ORIENTATION. The vectors ~v, ~w and ~v × ~w form a right handedcoordinate system. The right hand rule is: put the first vector ~v onthe thumb, the second vector ~w on the pointing finger and the thirdvector ~v × ~w on the third middle finger.EXAMPLE.~i ×~j =~k forms a right handed coordinate system.DOT PRODUCT (is a scalar)~v · ~w = ~w · ~v commutative|~v · ~w| = |~v|| ~w| cos(α) angle(a~v) · ~w = a(~v · ~w) linearity(~u + ~v) · ~w = ~u · ~w + ~v · ~w distributivity{1, 2, 3}.{3, 4, 5} in Mathematicaddt(~v · ~w) = ˙v · ~w + ~v · ˙w product ruleCROSS PRODUCT (is a vector)~v × ~w = − ~w × ~v anti-commutative|~v × ~w| = |~v||~w| sin(α) angle(a~v) × ~w = a(~v × ~w) linearity(~u + ~v) × ~w = ~u × ~w + ~v × ~w distributivityCross[{1, 2, 3}, {3, 4, 5}] in Mathematicaddt(~v × ~w) =˙~v × ~w + ~v ×˙~w product ruleTRIPLE SCALAR PRODUCT. The scalar [~u, ~v, ~w] = ~u · ~v × ~w is called the triple scalar product of ~u,~v, ~w.PARALLELEPIPED. [~u, ~v, ~w] is the volume of the parallelepipedspanned by ~u, ~v, ~w because h = ~u ·~n/|~n| is the height of the paral-lelepiped if ~n is a normal vector to the ground parallelogram whichhas area A = |~n| = |~v × ~w|. The volume of the parallelepiped isAh = |u · (~v × ~w)|.DISTANCE POINT-LINE (3D).If P is a point in space and L is the line which contains the vector ~u, thend(P, L) = |(~P Q × ~u|/|~u|is the distance between P and the line L.PLANE THROUGH 3 POINTS P, Q, R:The vector =~P Q ×~P R is orthogonal to the plane. We will see this again next week.The rest is informal and serves only as a motivation for this course:ANGULAR MOMENTUM. If a mass point of mass m moves along a curve ~r(t), then the vector~L(t) =m~r(t) × ~r0(t) is called the angular momentum.ANGULAR MOMENTUM CONSERVATION.ddt~L(t) = m~r0(t) × ~r0(t) + m~r(t) × ~r00(t) = ~r(t) ×~F (t)In a central field, where~F (t) is parallel to ~r(t), this vanishes.TORQUE. In physics, the quantity ~r(t)×~F (t) is also called the torque. The time derivative of the momentumm~r0is the force, the time derivative of the angular momentum~L is the torque.KEPLER’S AREA LAW. (Proof by Newton)The fact that~L(t) is constant means first of allthat ~r(t) stays in a plane spanned by ~r(0) and~r0(0). The experimental fact that the vector~r(t) sweeps over equal areas in equal timesexpresses the angular momentum conservation:|~r(t) × ~r0(t)dt/2| = |~Ldt/m/2| is the area of asmall triangle. The vector ~r(t) sweeps over anareaRT0|~L|dt/(2m) = |~L|T/(2m) in time [0, T ].r(t)SunEarthr’(t)dr x r/2PLACES IN PHYSICS WHERE THE CROSS PRODUCT OCCURS: (informal)The top, the motion of a rigid body is describe by the angular momentum L and the angular velocity vectorΩ in the body. Then˙L = L × Ω + M , where M is an external torque.Electromagnetism: a particle moving along ~r(t) in a magnetic field~B forexample experiences the force~F (t) = q~r0(t) ×~B, where q is the charge of theparticle.Hurricanes are powerful storms with wind velocities of 74 miles per hour or more.On the northern hemisphere, hurricanes turn counterclockwise, on the southernhemisphere clockwise. This is a feature of all low pressure systems and can beexplained by the Coriolis force. In a rotating coordinate system a particle of massm moving along ~r(t) experience the following forces: m~ω0×~r (inertia of rotation),2m~ω × ~r0(Coriolis force) and mω × (~ω × ~r)) (Centrifugal force), a fundamentalphysical force which is also responsible for the circulation in Jupiter’s Red
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