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3: Cross Product Now, |~V · ~W| = |~V||~W| Cos(Α)

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3: Cross Product Now, |~V · ~W| = |~V||~W| Cos(Α) Math S21a: Multivariable calculus Oliver Knill, Summer 2011 The sin formula: |~v × ~w| = |~v||~w| sin(α). Proof: We verify first the Lagrange’s identity |~v× ~w|2 = |~v|2|~w|2 −(~v· ~w)2 by direct computation. 3: Cross product Now, |~v · ~w| = |~v||~w| cos(α). The absolute value respectively length |~v × ~w| defines the area of the parallelo- The cross product of two vectors ~v = hv1, v2i and ~w = hw1,w2i in the plane is the gram spanned by ~v and ~w. scalar v1w2 − v2w1. Note that this was a definition so that nothing needs to be proven. But we want to make sure To remember this, we can write it as a determinant: take the product of the diagonal entries and that the definition fits with our common intuition we have about area: |~w| sin(α) is the height of v1 v2 subtract the product of the side diagonal. the parallelogram with base length |~v|. We see from the sin-formula that the area does not change " w1 w2 # if we rotate the vectors around in space because both length and angle stay the same. Area also is linear in each of the vectors v and w. If we make v twice as long, then the area gets twice as The cross product of two vectors ~v = hv1, v2, v3i and ~w = hw1,w2,w3i in space is large. defined as the vector ~v × ~w is zero exactly if ~v and ~w are parallel, that is if ~v = λ~w for some real λ. ~v × ~w = hv2w3 − v3w2, v3w1 − v1w3, v1w2 − v2w1i . Proof. This follows immediately from the sin formula and the fact that sin(α) = 0 if α = 0 or α = π. To remember it we write the product as a ”determinant”: i j k i j k The cross product can therefore be used to check whether two vectors are parallel or not. Note that v and −v are also considered parallel evenso sometimes one calls this anti-parallel. v1 v2 v3 = v2 v3 − v1 v3 + v1 v2 w1 w2 w3 w2 w3 w1 w3 w1 w2 The trigonometric sin-formula: if a, b, c are the side lengths of a triangle and which is ~i(v2w3 − v3w2) − ~j(v1w3 − v3w1)+ ~k(v1w2 − v2w1). α,β,γ are the angles opposite to a, b, c then a/ sin(α)= b/ sin(β)= c/ sin(γ. Proof. We express the area of the triangle in three different ways: 1 The cross product of h1, 2i and h4, 5i is 5 − 8= −3. ab sin(γ)= bc sin(α)= ac sin(β) . 2 The cross product of h1, 2, 3i and h4, 5, 1i is h−13, 11, −3i. Divide the first equation by sin(γ) sin(α) to get one identity. Divide the second equation by sin(α) sin(β) to get the second identity. 3 If ~v = ha, 0, 0i and ~w = hb cos(α), b sin(α), 0i, then ~v × ~w = h0, 0, ab sin(α)i which has length |ab sin(α)|. The scalar [~u, ~v, ~w]= ~u · (~v × ~w) is called the triple scalar product of ~u, ~v, ~w. The cross product ~v × ~w is orthogonal to both ~v and ~w. The product is anti- commutative. Proof. We verify for example that ~v · (~v × ~w) = 0 and look at the definition. The absolute value of [~u, ~v, ~w] defines the volume of the parallelepiped spanned Historically, the dot product and cross product emerged about at the same time. Determinants by ~u, ~v, ~w. were studied already by Gauss in 1801, matrix multiplication in 1812 by Binet. The dot prod- uct can be seen as a special case of matrix multiplication. Only in 1844 geometry in n di- mensions started to be developed by Grassman. It was Hamilton who described in 1843 first orientation The of three vectors is the sign of [~u, ~v, ~w]. It is positive if the three a multiplication ∗ of 4 vectors. It contains intrinsically both dot and cross product because vectors form a right handed coordinate system. (0, v1, v2, v3) ∗ (0,w1,w2,w3)=(−vw,v × w). The cross product can also be realized using matrix multiplication AB − BA of skew symmetric matrices. The proper way to see this now is tensor Again, we do not have to prove anything since we have just defined volume and orientation. Let analysis which started in 1890 with Ricci which contains a tensor product which has an anticom- us still see why this fits with with our intuition about volume. The value h = |~u · ~n|/|~n| is the mutative version called exterior algebra, an example of a superalgebra. More Vector calculus height of the parallelepiped if ~n =(~v × ~w) is a normal vector to the ground parallelogram of area was also developed at the same time by Clifford, Gibbs, Heaviside. This leads to mathematics A = |~n| = |~v × ~w|. The volume of the parallelepiped is hA =(~u · ~n/|~n|)|~v × ~w| which simplifies to developed in the 20’th century developed first by Poincare an Elie Cartan about 100 years ago. ~u · ~n = |(~u · (~v × ~w)| which is indeed the absolute value of the triple scalar product. The vectors Motivated heavily by physics, this was generalized to Spinor algebras which are special cases of ~v, ~w and ~v × ~w form a right handed coordinate system. If the first vector ~v is your thumb, Clifford algebras. It is fascinating to see that these geometric constructions appear in fundamental the second vector ~w is the pointing finger then ~v × ~w is the third middle finger of the right hand. physics in the structure of elementary particles. But the dot and cross product appears in down For example, the vectors ~i,~j,~i × ~j = ~k form a right handed coordinate system. to earth physics: It appears in fundamental equations like the Lorentz force F = q(v ×B) which depends on the velocity v of a particle, the charge q and the magnetic field B. An other example Since the triple scalar product is linear with respect to each vector we also see that volume is is the angular momentum mr × v or Coriolis force − − 2mω × v, where ω is a vector in the additive. Adding two equal parallelepipeds together for example gives a parallelepiped with a rotation axes and v is the velocity. volume twice the volume. 4 Problem: Find the volume of a cuboid of width a length b and height c. Answer. The cuboid is a parallelepiped spanned by ha, 0, 0ih0, b, 0i and h0, 0,ci. The triple scalar product Homework is abc. 5 Problem Find the volume of the parallelepiped which has the vertices O = (1, 1, 0),P = 1 a) Find the volume of the parallelepiped for which the base parallelogram is given by (2, 3, 1), Q = (4, 3, 1), R = (1, 4, 1). Answer: We first see that it is spanned by the vectors the points (0, 0, 0), (1, 0, 1, (2, 2, 1), (1, 2, 0) and which has an edge connecting (0, 0, 0) with ~u = h1, 2, 1i, ~v = h3, 2, 1i, and ~w = h0, 3, 1i. We get ~v × ~w = h−1, −3, 9i and ~u · (~v × ~w) = 2. (3, 4, 5). The volume is 2. b) Find the area of the base and use a) to get the height. You can skip the following remarks: We have seen that in two dimensions, the cross product is a 2 a) Assume ~u + ~v + ~w = ~0. Verify that ~u × ~v = ~v × ~w = ~w × ~u. scalar and in three dimensions, the cross product is a vector. What happens in higher dimensions? b) Find (~u + ~v) · (~v × ~w) if ~u, ~v, ~w are unit vectors which are orthogonal to each other and There is a generalization called wedge product ~v ∧ ~w but the resulting vector is in general in ~u × ~v = ~w. a new ”super” space. In four dimensions for example, the wedge product between two vectors ~v and ~w is an object with 6 = 4(4 − 1)/2 components. In n dimensions, the product has n(n − 1)/2 3 To find the equation ax + by + cz = d for the plane which contains the point P = (1, 2, 3) as components: well as the line which passes through Q = (3, 4, 4) and R = (1, 1, 2), we find a vector ha, b, ci normal to the plane and fix d so that P is in the plane. ~v ∧ ~w = hv1w2 − v2w1, v1v3 − v3w1, v1w4 − v4w1, v2w3 − v3w2, v2v4 − v4v2, v3w4 − v4v3i . 4 Verify the Lagrange formula The length of ~v ∧ ~w is the area of the parallelepiped spanned by ~v and ~w. It is only in our three dimensional space that we can identify the wedge product between two vectors as a 3-vector again. n ~ ~ ~ The wedge product can be extended to an algebra of dimension 2 . It is called a superalgebra. ~a × (b × ~c)= b(~a · ~c) − ~c(~a · b) An other approach to generalize the product is to use linear algebra and write a vector v as a 3 × 3 matrix V , now v × w corresponds to V W − WV . For those of you who take linear algebra: for general vectors ~a,~b,~c in space. The formula can be remembered as ”BAC minus CAB”. 0 −v1 v3 0 −w1 w3 0 −(v2w3 − v3w2) v1w2 − v2w1 V = v1 0 −v2 , W = w1 0 −w2 , V W − W V = v2w3 − v3w2 0 −(v3w1 − v1w3) . 5 Assume you know that the triple scalar product [~u, ~v, ~w] = ~u · (~v × ~w) between ~u, ~v, ~w is −v3 v2 0 −w3 w2 0 v2w1 − v1w2 −v1w3 − w1v3 0 equal to 4.
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