DOCSLIB.ORG

Cross Product Review

Total Page：16

File Type：pdf, Size：1020Kb

Download full-text PDF Read full-text
Cross Product Review 12.4 Cross Product Review: The dot product of uuuu123, , and v vvv 123, , is u v uvuvuv 112233 uv u u u u v u v cos or cos uv u and v are orthogonal if and only if u v 0 u uv uv compvu projvuv v v vv projvu cross product u v uv23 uv 32 i uv 13 uv 31 j uv 12 uv 21 k u v is orthogonal to both u and v. u v u v sin Geometric description of the cross product of the vectors u and v The cross product of two vectors is a vector! • u x v is perpendicular to u and v • The length of u x v is u v u v sin • The direction is given by the right hand side rule Right hand rule Place your 4 fingers in the direction of the first vector, curl them in the direction of the second vector, Your thumb will point in the direction of the cross product Algebraic description of the cross product of the vectors u and v The cross product of uu1, u 2 , u 3 and v v 1, v 2 , v 3 is uv uv23 uvuv 3231,, uvuv 1312 uv 21 check (u v ) u 0 and ( u v ) v 0 (u v ) u uv23 uvuv 3231 , uvuv 1312 , uv 21 uuu 123 , , uvu231 uvu 321 uvu 312 uvu 132 uvu 123 uvu 213 0 similary: (u v ) v 0 length u v u v sin is a little messier : 2 2 2 2 2 2 2 2uv 2 2 2 uvuv sin22 uv 1 cos uv 1 uvuv 22 uv now need to show that u v2 u 2 v 2 u v2 (try it..) An easier way to remember the formula for the cross products is in terms of determinants: ab 12 2x2 determinant: ad bc 4 6 2 cd 34 3x3 determinants: An example Copy 1st 2 columns 1 6 2 sum of sum of 1 6 2 1 6 forward backward 3 1 3 3 1 3 3 1 diagonal diagonal 4 5 2 4 5 2 4 5 products products determinant = 2 72 30 8 15 36 40 59 19 recall: uv uv23 uvuv 3231, uvuv 1312 , uv 21 i j k i j k i j u1 u 2 u 3 u 1 u 2 now we claim that uvu1 u 2 u 3 v1 v 2 v 3 v 1 v 2 v1 v 2 v 3 iuv23 j uv 31 k uv 12 k uv 21 i uv 32 j uv 13 u v uv23 uv 32 i uv 13 uv 31 j uv 12 uv 21 k uv uv23 uvuv 3231,, uvuv 1312 uv 21 Example: Let u1, 2,1 and v 3,1, 2 Find u v. i j k i j k i j uv 1 2 1 1 2 1 1 2 3 1 2 3 1 2 3 1 u v 41 i 32 j 16k uv3,5,7 Geometric Properties of the cross product: Let u and v be nonzero vectors and let be the angle between u and v. 1. u v is orthogonal to both u and v. 2. u v u v sin 3. u v 0 if and only if u and v are scalar multiples of each other (they are parallel). u u 4. uvarea of the parallelogram h u sin v determined by uv and . v 1 5. 2 uvarea of the triangle having uv and as adjacent sides. v Problem: Compute the area of the triangle with vertices (2,3,4), (1,3,2), (3,0,-6) Two sides are: uv1,0,2 and 1,3,10 i j k i j k i j uv1 0 2 1 0 2 1 0 1 3 10 1 3 2 1 3 (0 6)i ( 2 10) j (3 0) k 6i 12 j 3 k 3 6, 12,3 |u v | = 36 144 9 189 area = 21 2 Algebraic Properties of the cross product: i j k Let u, v , and w be vectors and let c be a scalar. uvu1 u 2 u 3 v v v 1. u v v u 1 2 3 uv23 uvuv 3231,, uvuv 1312 uv 21 2. u v w u v u w 3. cu v u c v c u v 4. 0 v v 0 0 5. (cv ) v 0 6. u v w u v w 7. u v w u w v u v w Volume of the parallelepiped determined by the vectors a, b , and c. Area of the base bc Height compbc a a cos Volumeb c a cos this stands Volume a b c for absolute value a b c is called the scalar triple product The vectors are in the same plane coplanar if the scalar triple product is 0. Problem: Compute the volume of the parallelepiped spanned by the 3 vectors u1,0,2 , v 1,3, 2 and w 1,3, 4 Solution: 6,0,3 1,3, 4 From slide 10: uv 6,0,3 6 12 6 Volume = 6 Quicker: 1 3 4 i j k (uv ) w 1 0 2 (0 6 12) ( 6 6 0) (uv ) w 1 3 2 u1 u 2 u 3 w 1, w 2 , w 3 6 v1 v 2 v 3 Triple scalar product i j k i j w1 w 2 w 3 u u u u u w1 w 2 w 3 1 2 3 1 2 w1, w 2 , w 3 u1 u 2 u 3 v1 v 2 v 3 v 1 v 2 (wu v) u u u u v w v1 v 2 v 3 1 2 3 v1 v 2 v 3 In physics, the cross product is used to measure torque. Q Consider a force Fr acting on a rigid body at a point given by a position vector . The torque measures the tendency of the body to rotate about the origin point P rF rF rFsin r is the angle between the force and position vectors F 12.5 Lines and Planes Recall how to describe lines in the plane (e.g. tangent lines to a graph): y mx b m is the slope by is the intercept yy 0 Point slope formula: m (xy00 , ) is on the line xx 0 y y0 y 1 y 0 (x , y ) and ( x , y ) Two point formula: 0 0 1 1 are on the line x x0 x 1 x 0 Equations of Lines and Planes A a point on the line P0 x 0, y 0 , z 0 In order to find the equation of a line, we need : B a direction vector for the line v abc, , r r0 t v vector equation of line L z Here r0 is the vector from the origin P0 P tv P x, y , z to a specific point P0 on the line P0 x 0, y 0 , z 0 r L r is the vector from the origin to r0 a general point P (x , y , z ) on the line v y v is a vector which is parallel to a vector r0 x 0,, y 0 z 0 x r x,, y z that lies on the line v is not unique: 2v , or v will also do P0 P tv t a, b , c for some t vector equation of the line L or x,,,,,, y z x0 y 0 z 0 t a b c r r0 t v equating components we get the parametric equations of the line L xx0 aty, y 0 btz, z 0 ct Solving for t we get the symmetric equations of the line L x x y y z z 0 0 0 a b c Problem: A a point on the line P0 x 0, y 0 , z 0 Find the parametric equations of the line containing P01 5,1,3 and P 3, 2,4 . B a direction vector for the line v abc, , (could also choose x,,,,,, y z x0 y 0choose z 0 t aP0 b c 5,1,3 P0 3, 2,4 ) v PPPP0 1 1 0 3 5, 2 1,4 3 2, 3,1 or v 2,3, 1 5,1,3 t 2,3, 1 The line is: x5 2 t , y 1 3 t , z 3 t Two lines in 3 space can interact in 3 ways: A) Parallel Lines - their direction vectors are scalar multiples of each other B) Intersecting Lines - there is a specific t and s, so that the lines share the same point. C) Skew Lines - their direction vectors are not parallel and there is no values of t and s that make the lines share the same point. Problem: Determine whether the lines LL12 and are parallel, skew or intersecting. If they intersect, find the point of intersection. L1 : x3 t , y 5 3 t , z 1 4 t L2 : x8 2 s , y 6 4 s , z 5 s Set the x coordinate equal to each other: 3ts 8 2 , or 2st 5 Set the y coordinate equal to each other: 5 3ts 6 4 , or 4st 3 11 We get a system of equations: 2st 5 or 4st 2 10 t 1 4st 3 11 4 st 3 11 s 2 1 4ts 5 Find the point of intersection using L1 : Check to make sure that the z 1 41 5 2 values are equal for this ts and .
Load more

Recommended publications
  • A Parallelepiped Based Approach
    3D Modelling Using Geometric Constraints: A Parallelepiped Based Approach Marta Wilczkowiak, Edmond Boyer, and Peter Sturm MOVI–GRAVIR–INRIA Rhˆone-Alpes, 38330 Montbonnot, France, [email protected], http://www.inrialpes.fr/movi/people/Surname Abstract. In this paper, efficient and generic tools for calibration and 3D reconstruction are presented. These tools exploit geometric con- straints frequently present in man-made environments and allow cam- era calibration as well as scene structure to be estimated with a small amount of user interactions and little a priori knowledge. The proposed approach is based on primitives that naturally characterize rigidity con- straints: parallelepipeds. It has been shown previously that the intrinsic metric characteristics of a parallelepiped are dual to the intrinsic charac- teristics of a perspective camera. Here, we generalize this idea by taking into account additional redundancies between multiple images of multiple parallelepipeds. We propose a method for the estimation of camera and scene parameters that bears strongsimilarities with some self-calibration approaches. Takinginto account prior knowledgeon scene primitives or cameras, leads to simpler equations than for standard self-calibration, and is expected to improve results, as well as to allow structure and mo- tion recovery in situations that are otherwise under-constrained. These principles are illustrated by experimental calibration results and several reconstructions from uncalibrated images. 1 Introduction This paper is about using partial information on camera parameters and scene structure, to simplify and enhance structure from motion and (self-) calibration. We are especially interested in reconstructing man-made environments for which constraints on the scene structure are usually easy to provide.
    [Show full text]
  • Area, Volume and Surface Area
    The Improving Mathematics Education in Schools (TIMES) Project MEASUREMENT AND GEOMETRY Module 11 AREA, VOLUME AND SURFACE AREA A guide for teachers - Years 8–10 June 2011 YEARS 810 Area, Volume and Surface Area (Measurement and Geometry: Module 11) For teachers of Primary and Secondary Mathematics 510 Cover design, Layout design and Typesetting by Claire Ho The Improving Mathematics Education in Schools (TIMES) Project 2009‑2011 was funded by the Australian Government Department of Education, Employment and Workplace Relations. The views expressed here are those of the author and do not necessarily represent the views of the Australian Government Department of Education, Employment and Workplace Relations. © The University of Melbourne on behalf of the international Centre of Excellence for Education in Mathematics (ICE‑EM), the education division of the Australian Mathematical Sciences Institute (AMSI), 2010 (except where otherwise indicated). This work is licensed under the Creative Commons Attribution‑NonCommercial‑NoDerivs 3.0 Unported License. http://creativecommons.org/licenses/by‑nc‑nd/3.0/ The Improving Mathematics Education in Schools (TIMES) Project MEASUREMENT AND GEOMETRY Module 11 AREA, VOLUME AND SURFACE AREA A guide for teachers - Years 8–10 June 2011 Peter Brown Michael Evans David Hunt Janine McIntosh Bill Pender Jacqui Ramagge YEARS 810 {4} A guide for teachers AREA, VOLUME AND SURFACE AREA ASSUMED KNOWLEDGE • Knowledge of the areas of rectangles, triangles, circles and composite figures. • The definitions of a parallelogram and a rhombus. • Familiarity with the basic properties of parallel lines. • Familiarity with the volume of a rectangular prism. • Basic knowledge of congruence and similarity. • Since some formulas will be involved, the students will need some experience with substitution and also with the distributive law.
    [Show full text]
  • Evaporationin Icparticles
    The JapaneseAssociationJapanese Association for Crystal Growth (JACG){JACG) Small Metallic ParticlesProduced by Evaporation in lnert Gas at Low Pressure Size distributions, crystal morphologyand crystal structures KazuoKimoto physiosLaboratory, Department of General Educatien, Nago)ra University Various experimental results ef the studies on fine 1. Introduction particles produced by evaperation and subsequent conden$ation in inert gas at low pressure are reviewed. Small particles of metals and semi-metals can A brief historical survey is given and experimental be produced by evaporation and subsequent arrangements for the production of the particles are condensation in the free space of an inert gas at described. The structure of the stnoke, the qualitative low pressure, a very simple technique, recently particle size clistributions, small particle statistios and "gas often referred to as evaporation technique"i) the crystallographic aspccts ofthe particles are consid- ered in some detail. Emphasis is laid on the crystal (GET). When the pressure of an inert is in the inorphology and the related crystal structures ef the gas range from about one to several tens of Torr, the particles efsome 24 elements. size of the particles produced by GET is in thc range from several to several thousand nm, de- pending on the materials evaporated, the naturc of the inert gas and various other evaporation conditions. One of the most characteristic fea- tures of the particles thus produced is that the particles have, generally speaking, very well- defined crystal habits when the particle size is in the range from about ten to several hundred nm. The crystal morphology and the relevant crystal structures of these particles greatly interested sDme invcstigators in Japan, 4nd encouraged them to study these propenies by means of elec- tron microscopy and electron difliraction.
    [Show full text]
  • Vectors, Matrices and Coordinate Transformations
    S. Widnall 16.07 Dynamics Fall 2009 Lecture notes based on J. Peraire Version 2.0 Lecture L3 - Vectors, Matrices and Coordinate Transformations By using vectors and defining appropriate operations between them, physical laws can often be written in a simple form. Since we will making extensive use of vectors in Dynamics, we will summarize some of their important properties. Vectors For our purposes we will think of a vector as a mathematical representation of a physical entity which has both magnitude and direction in a 3D space. Examples of physical vectors are forces, moments, and velocities. Geometrically, a vector can be represented as arrows. The length of the arrow represents its magnitude. Unless indicated otherwise, we shall assume that parallel translation does not change a vector, and we shall call the vectors satisfying this property, free vectors. Thus, two vectors are equal if and only if they are parallel, point in the same direction, and have equal length. Vectors are usually typed in boldface and scalar quantities appear in lightface italic type, e.g. the vector quantity A has magnitude, or modulus, A = |A|. In handwritten text, vectors are often expressed using the −→ arrow, or underbar notation, e.g. A , A. Vector Algebra Here, we introduce a few useful operations which are defined for free vectors. Multiplication by a scalar If we multiply a vector A by a scalar α, the result is a vector B = αA, which has magnitude B = |α|A. The vector B, is parallel to A and points in the same direction if α > 0.
    [Show full text]
  • Vector Analysis
    DOING PHYSICS WITH MATLAB VECTOR ANANYSIS Ian Cooper School of Physics, University of Sydney [email protected] DOWNLOAD DIRECTORY FOR MATLAB SCRIPTS cemVectorsA.m Inputs: Cartesian components of the vector V Outputs: cylindrical and spherical components and [3D] plot of vector cemVectorsB.m Inputs: Cartesian components of the vectors A B C Outputs: dot products, cross products and triple products cemVectorsC.m Rotation of XY axes around Z axis to give new of reference X’Y’Z’. Inputs: rotation angle and vector (Cartesian components) in XYZ frame Outputs: Cartesian components of vector in X’Y’Z’ frame mscript can be modified to calculate the rotation matrix for a [3D] rotation and give the Cartesian components of the vector in the X’Y’Z’ frame of reference. Doing Physics with Matlab 1 SPECIFYING a [3D] VECTOR A scalar is completely characterised by its magnitude, and has no associated direction (mass, time, direction, work). A scalar is given by a simple number. A vector has both a magnitude and direction (force, electric field, magnetic field). A vector can be specified in terms of its Cartesian or cylindrical (polar in [2D]) or spherical coordinates. Cartesian coordinate system (XYZ right-handed rectangular: if we curl our fingers on the right hand so they rotate from the X axis to the Y axis then the Z axis is in the direction of the thumb). A vector V in specified in terms of its X, Y and Z Cartesian components ˆˆˆ VVVV x,, y z VViVjVk x y z where iˆˆ,, j kˆ are unit vectors parallel to the X, Y and Z axes respectively.
    [Show full text]
  • Crystallography Ll Lattice N-Dimensional, Infinite, Periodic Array of Points, Each of Which Has Identical Surroundings
    crystallography ll Lattice n-dimensional, infinite, periodic array of points, each of which has identical surroundings. use this as test for lattice points A2 ("bcc") structure lattice points Lattice n-dimensional, infinite, periodic array of points, each of which has identical surroundings. use this as test for lattice points CsCl structure lattice points Choosing unit cells in a lattice Want very small unit cell - least complicated, fewer atoms Prefer cell with 90° or 120°angles - visualization & geometrical calculations easier Choose cell which reflects symmetry of lattice & crystal structure Choosing unit cells in a lattice Sometimes, a good unit cell has more than one lattice point 2-D example: Primitive cell (one lattice pt./cell) has End-centered cell (two strange angle lattice pts./cell) has 90° angle Choosing unit cells in a lattice Sometimes, a good unit cell has more than one lattice point 3-D example: body-centered cubic (bcc, or I cubic) (two lattice pts./cell) The primitive unit cell is not a cube 14 Bravais lattices Allowed centering types: P I F C primitive body-centered face-centered C end-centered Primitive R - rhombohedral rhombohedral cell (trigonal) centering of trigonal cell 14 Bravais lattices Combine P, I, F, C (A, B), R centering with 7 crystal systems Some combinations don't work, some don't give new lattices - C tetragonal C-centering destroys cubic = P tetragonal symmetry 14 Bravais lattices Only 14 possible (Bravais, 1848) System Allowed centering Triclinic P (primitive) Monoclinic P, I (innerzentiert) Orthorhombic P, I, F (flächenzentiert), A (end centered) Tetragonal P, I Cubic P, I, F Hexagonal P Trigonal P, R (rhombohedral centered) Choosing unit cells in a lattice Unit cell shape must be: 2-D - parallelogram (4 sides) 3-D - parallelepiped (6 faces) Not a unit cell: Choosing unit cells in a lattice Unit cell shape must be: 2-D - parallelogram (4 sides) 3-D - parallelepiped (6 faces) Not a unit cell: correct cell Stereographic projections Show or represent 3-D object in 2-D Procedure: 1.
    [Show full text]
  • When Does a Cross Product on R^{N} Exist?
    WHEN DOES A CROSS PRODUCT ON Rn EXIST? PETER F. MCLOUGHLIN It is probably safe to say that just about everyone reading this article is familiar with the cross product and the dot product. However, what many readers may not be aware of is that the familiar properties of the cross product in three space can only be extended to R7. Students are usually first exposed to the cross and dot products in a multivariable calculus or linear algebra course. Let u = 0, v = 0, v, 3 6 6 e and we be vectors in R and let a, b, c, and d be real numbers. For review, here are some of the basic properties of the dot and cross products: (i) (u·v) = cosθ (where θ is the angle formed by the vectors) √(u·u)(v·v) (ii) ||(u×v)|| = sinθ √(u·u)(v·v) (iii) u (u v) = 0 and v (u v)=0. (perpendicularproperty) (iv) (u· v×) (u v) + (u · v)2×= (u u)(v v) (Pythagorean property) (v) ((au×+ bu· ) ×(cv + dv))· = ac(u · v)+·ad(u v)+ bc(u v)+ bd(u v) e × e × × e e × e × e We will refer to property (v) as the bilinear property. The reader should note that properties (i) and (ii) imply the Pythagorean property. Recall, if A is a square matrix then A denotes the determinant of A. If we let u = (x , x , x ) and | | 1 2 3 v = (y1,y2,y3) then we have: e1 e2 e3 u v = x1y1 + x2y2 + x3y3 and (u v) = x1 x2 x3 · × y1 y2 y3 It should be clear that the dot product can easily be extended to Rn, however, arXiv:1212.3515v7 [math.HO] 30 Oct 2013 it is not so obvious how the cross product could be extended.
    [Show full text]
  • 15 BASIC PROPERTIES of CONVEX POLYTOPES Martin Henk, J¨Urgenrichter-Gebert, and G¨Unterm
    15 BASIC PROPERTIES OF CONVEX POLYTOPES Martin Henk, J¨urgenRichter-Gebert, and G¨unterM. Ziegler INTRODUCTION Convex polytopes are fundamental geometric objects that have been investigated since antiquity. The beauty of their theory is nowadays complemented by their im- portance for many other mathematical subjects, ranging from integration theory, algebraic topology, and algebraic geometry to linear and combinatorial optimiza- tion. In this chapter we try to give a short introduction, provide a sketch of \what polytopes look like" and \how they behave," with many explicit examples, and briefly state some main results (where further details are given in subsequent chap- ters of this Handbook). We concentrate on two main topics: • Combinatorial properties: faces (vertices, edges, . , facets) of polytopes and their relations, with special treatments of the classes of low-dimensional poly- topes and of polytopes \with few vertices;" • Geometric properties: volume and surface area, mixed volumes, and quer- massintegrals, including explicit formulas for the cases of the regular simplices, cubes, and cross-polytopes. We refer to Gr¨unbaum [Gr¨u67]for a comprehensive view of polytope theory, and to Ziegler [Zie95] respectively to Gruber [Gru07] and Schneider [Sch14] for detailed treatments of the combinatorial and of the convex geometric aspects of polytope theory. 15.1 COMBINATORIAL STRUCTURE GLOSSARY d V-polytope: The convex hull of a finite set X = fx1; : : : ; xng of points in R , n n X i X P = conv(X) := λix λ1; : : : ; λn ≥ 0; λi = 1 : i=1 i=1 H-polytope: The solution set of a finite system of linear inequalities, d T P = P (A; b) := x 2 R j ai x ≤ bi for 1 ≤ i ≤ m ; with the extra condition that the set of solutions is bounded, that is, such that m×d there is a constant N such that jjxjj ≤ N holds for all x 2 P .
    [Show full text]
  • Ε and Cross Products in 3-D Euclidean Space
    Last Latexed: September 8, 2005 at 10:14 1 Last Latexed: September 8, 2005 at 10:14 2 It is easy to evaluate the 27 coefficients kij, because the cross product of two ijk and cross products in 3-D orthogonal unit vectors is a unit vector orthogonal to both of them. Thus Euclidean space eˆ1 eˆ2 =ˆe3,so312 =1andk12 =0ifk = 1 or 2. Applying the same argument× toe ˆ eˆ ande ˆ eˆ , and using the antisymmetry of the cross 2 × 3 3 × 1 These are some notes on the use of the antisymmetric symbol for ex- product, A~ B~ = B~ A~,weseethat ijk × − × pressing cross products. This is an extremely powerful tool for manipulating = = =1; = = = 1, cross products and their generalizations in higher dimensions, and although 123 231 312 132 213 321 − many low level courses avoid the use of ,IthinkthisisamistakeandIwant and = 0 for all other values of the indices, i.e. = 0 whenever any you to become proficient with it. ijk ijk two of the indices are equal. Note that changes sign not only when the last In a cartesian coordinate system a vector ~ has components along each V Vi two indices are interchanged (a consequence of the antisymmetry of the cross of the three orthonormal basis vectorse ˆ ,orV~ = V eˆ . The dot product i Pi i i product), but whenever any two of its indices are interchanged. Thus is of two vectors, A~ B~ , is bilinear and can therefore be written as ijk · zero unless (1, 2, 3) (i, j, k) is a permutation, and is equal to the sign of the permutation if it→ exists.
    [Show full text]
  • Crystal Structure
    Physics 927 E.Y.Tsymbal Section 1: Crystal Structure A solid is said to be a crystal if atoms are arranged in such a way that their positions are exactly periodic. This concept is illustrated in Fig.1 using a two-dimensional (2D) structure. y T C Fig.1 A B a x 1 A perfect crystal maintains this periodicity in both the x and y directions from -∞ to +∞. As follows from this periodicity, the atoms A, B, C, etc. are equivalent. In other words, for an observer located at any of these atomic sites, the crystal appears exactly the same. The same idea can be expressed by saying that a crystal possesses a translational symmetry. The translational symmetry means that if the crystal is translated by any vector joining two atoms, say T in Fig.1, the crystal appears exactly the same as it did before the translation. In other words the crystal remains invariant under any such translation. The structure of all crystals can be described in terms of a lattice, with a group of atoms attached to every lattice point. For example, in the case of structure shown in Fig.1, if we replace each atom by a geometrical point located at the equilibrium position of that atom, we obtain a crystal lattice. The crystal lattice has the same geometrical properties as the crystal, but it is devoid of any physical contents. There are two classes of lattices: the Bravais and the non-Bravais. In a Bravais lattice all lattice points are equivalent and hence by necessity all atoms in the crystal are of the same kind.
    [Show full text]
  • Tutorial 3 Using MATLAB in Linear Algebra
    Edward Neuman Department of Mathematics Southern Illinois University at Carbondale [email protected] One of the nice features of MATLAB is its ease of computations with vectors and matrices. In this tutorial the following topics are discussed: vectors and matrices in MATLAB, solving systems of linear equations, the inverse of a matrix, determinants, vectors in n-dimensional Euclidean space, linear transformations, real vector spaces and the matrix eigenvalue problem. Applications of linear algebra to the curve fitting, message coding and computer graphics are also included. For the reader's convenience we include lists of special characters and MATLAB functions that are used in this tutorial. Special characters ; Semicolon operator ' Conjugated transpose .' Transpose * Times . Dot operator ^ Power operator [ ] Emty vector operator : Colon operator = Assignment == Equality \ Backslash or left division / Right division i, j Imaginary unit ~ Logical not ~= Logical not equal & Logical and | Logical or { } Cell 2 Function Description acos Inverse cosine axis Control axis scaling and appearance char Create character array chol Cholesky factorization cos Cosine function cross Vector cross product det Determinant diag Diagonal matrices and diagonals of a matrix double Convert to double precision eig Eigenvalues and eigenvectors eye Identity matrix fill Filled 2-D polygons fix Round towards zero fliplr Flip matrix in left/right direction flops Floating point operation count grid Grid lines hadamard Hadamard matrix hilb Hilbert
    [Show full text]
  • Recitation Video Transcript (PDF)
    MITOCW | MIT18_06SC_110609_L4_300k-mp4 LINAN CHEN: Hello. Welcome back to recitation. I'm sure you are becoming more and more familiar with the determinants of matrices. In the lecture, we also learned the geometric interpretation of the determinant. The absolute value of the determinant of a matrix is simply equal to the volume of the parallelepiped spanned by the row vectors of that matrix. So today, we're going to apply this fact to solve the following problem. I have a tetrahedron, T, in this 3D space. And the vertices of T are given by O, which is the origin, A_1, A_2, and A_3. So I have highlighted this tetrahedron using the blue chalk. So this is T. And our first goal is to compute the volume of T using the determinant. And the second part is: if I fix A_1 and A_2, but move A_3 to another point, A_3 prime, which is given by this coordinate, I ask you to compute the volume again. OK. So since we want to use the fact that the determinant is related to the volume, we have to figure out which volume we should be looking at. We know that the determinant is related to the volume of a parallelepiped. But here, we only have a tetrahedron. So the first goal should be to find out which parallelepiped you should be working with. OK, why don't you hit pause and try to work it out yourself. You can sketch the parallelepiped on this picture. And I will return in a while and continue working with you.
    [Show full text]