{"id":34243,"date":"2022-03-29T13:00:59","date_gmt":"2022-03-29T13:00:59","guid":{"rendered":"https:\/\/toposuranos.com\/material\/?p=34243"},"modified":"2025-08-27T21:04:54","modified_gmt":"2025-08-27T21:04:54","slug":"algebra-and-projections-in-rn-cross-product-in-r3","status":"publish","type":"post","link":"https:\/\/toposuranos.com\/material\/en\/algebra-and-projections-in-rn-cross-product-in-r3\/","title":{"rendered":"Algebra and Projections in Rn, Cross Product in R3"},"content":{"rendered":"<style>\np, ul, ol{\ntext-align: justify;\n}\nh1{\ntext-align:center;\ntext-transform: uppercase;\n}\nh2{\ntext-align:center;\ntext-transform: uppercase;\nfont-size:24pt;\n}\nh3 { \n    text-align: center;\n    text-transform: uppercase;\n    font-size: 24px !important;\n}\n<\/style>\n<h1>Algebra and Projections in Rn, Cross Product in <span class=\"katex-eq\" data-katex-display=\"false\">{\\mathbb{R}^3}<\/span><\/h1>\n<p style=\"text-align:center;\"><em><strong>Abstract:<\/strong><\/br>This series is the direct continuation of the series on the Euclidean Space of n dimensions. Here we will review some linear algebra concepts that help to better understand the n-dimensional Euclidean space, examine the concepts of projections of one vector onto another, demonstrate the Pythagorean theorem, and conclude with a review of the cross product in <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3<\/span> and its relationship with the other products of the 3-dimensional Euclidean space. <\/p>\n<p style=\"text-align:center;\"><strong>INDEX<\/strong><br \/>\n<a href=\"#Independencia-Lineal-Ortogonalidad-y-Proyecciones\">Linear Independence, Orthogonality, and Projections<\/a><br \/>\n<a href=\"#El-Teorema-de-Pitagoras-y-la-Proyecci\u00f3n-sobre-un-Subespacio\">The Pythagorean Theorem and Projection onto a Subspace<\/a><br \/>\n<a href=\"#El-Producto-Escalar-y-Vectorial-en-R3\">The Dot and Cross Product in <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3<\/span><\/a>\n<\/p>\n<p><a name=\"Independencia-Lineal-Ortogonalidad-y-Proyecciones\"><\/a><br \/>\n<center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/vtNHkaHD3aA\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/p>\n<h2>Linear Independence, Orthogonal, and Projections<\/h2>\n<h3>Linear combination and linear independence<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=vtNHkaHD3aA&#038;t=138s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">A nonzero vector<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z}<\/span> can be constructed as a <strong>linear combination<\/strong> with respect to other nonzero vectors <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> if there exists a pair of real numbers <span class=\"katex-eq\" data-katex-display=\"false\">\\alpha<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\beta<\/span>, not both zero simultaneously, such that:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z} = \\alpha \\vec{x} + \\beta\\vec{y}<\/span>\n<p>That is, the vector <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z}<\/span> can be constructed as a weighted sum of the vectors <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}.<\/span>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=vtNHkaHD3aA&#038;t=609s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Analogously, it is said<\/span><\/strong><\/a> that the vectors <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> are <strong>linearly independent<\/strong> if <\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">(\\alpha \\vec{x} + \\beta\\vec{y} = \\vec{0} ) \\longleftrightarrow (\\alpha=0 \\wedge \\beta=0 )<\/span>\n<p>Linear independence between the vectors <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> tells us that <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> cannot be obtained as a (nonzero) scalar multiple of <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> and vice versa.<\/p>\n<p>The concept of linear independence we have just reviewed can be extended to larger sets of vectors. The set of nonzero vectors <span class=\"katex-eq\" data-katex-display=\"false\">\\{\\vec{x}_1, \\cdots, \\vec{x}_n\\}<\/span> is said to be linearly independent when<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\displaystyle \\left[\\left(\\sum_{i=1}^n \\alpha_i \\vec{x}_i \\right) = \\vec{0} \\right] \\longleftrightarrow \\left[\\bigwedge_{i=1}^n (\\alpha_i = 0) \\right]<\/span>\n<h3>The angle formed by two vectors and orthogonality<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=vtNHkaHD3aA&#038;t=1289s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">If we recall the Cauchy-Schwarz inequality,<\/span><\/strong><\/a> it tells us that <span class=\"katex-eq\" data-katex-display=\"false\">(\\forall \\vec{x},\\vec{y}\\in\\mathbb{R}^n)(|\\vec{x}\\cdot\\vec{y}| \\leq \\|\\vec{x}\\| \\|\\vec{y}\\|).<\/span> Keeping this in mind, it is easy to verify that for any pair of vectors <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x},\\vec{y}\\in\\mathbb{R}^n\\setminus\\{\\vec{0}\\}<\/span> the following relation holds:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle -1 \\leq \\frac{\\vec{x}\\cdot\\vec{y}}{\\|\\vec{x}\\|\\|\\vec{y}\\|}\\leq 1<\/span>\n<p>We can now intuit a relationship between the dot product and the angle formed by the vectors <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span>, because they generate a plane isometric to <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^2<\/span>. Therefore, without loss of generality, we can imagine them as being elements of <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^2<\/span> with angles with respect to the <span class=\"katex-eq\" data-katex-display=\"false\">\\hat{x}<\/span> axis of <span class=\"katex-eq\" data-katex-display=\"false\">\\theta_x<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\theta_y,<\/span> respectively, so that the vectors are written in polar form as:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n\\vec{x} &amp;= \\|\\vec{x}\\|(\\cos(\\theta_x) , \\sin(\\theta_x)) \\\\ \\\\ \\vec{y} &amp;= \\|\\vec{y}\\|(\\cos(\\theta_y) , \\sin(\\theta_y))\n\n\\end{array}<\/span>\n<p>Thus we can assume (without loss of generality, again) that <span class=\"katex-eq\" data-katex-display=\"false\">\\theta_x \\lt \\theta_y,<\/span> and then calculate the dot product <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\cdot\\vec{y}.<\/span> Doing so, we obtain the following result:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\\vec{x}\\cdot \\vec{y} &amp;=  \\|\\vec{x}\\|  \\|\\vec{y}\\| (\\cos(\\theta_x)\\cos(\\theta_y) + \\sin(\\theta_x)\\sin(\\theta_y)) \\\\ \\\\ &amp;=  \\|\\vec{x}\\|  \\|\\vec{y}\\| \\cos(\\theta_y-\\theta_x)\n\n\\end{array}<\/span>\n<p>Now, by taking the difference between the greater and smaller angular positions, we obtain the angle between the vectors, <span class=\"katex-eq\" data-katex-display=\"false\">\\angle(\\vec{x},\\vec{y})=\\theta_y - \\theta_x.<\/span> And with this we can now write:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\displaystyle \\cos\\left(\\angle(\\vec{x},\\vec{y}) \\right) = \\frac{\\vec{x} \\cdot \\vec{y}}{\\|\\vec{x}\\|\\|\\vec{y}\\|}\n\n<\/span>\n<p>Here we must emphasize that <span class=\"katex-eq\" data-katex-display=\"false\">\\angle(\\vec{x},\\vec{y})\\in [0, \\pi]<\/span>\n<p>From this we can connect the Cauchy-Schwarz inequality with the geometry of angles, and it also allows us to obtain a rigorous notion of orthogonality. Two vectors are said to be <strong>Orthogonal<\/strong> when they form an angle of <span class=\"katex-eq\" data-katex-display=\"false\">\\pi\/2<\/span> radians between them, in the sense explained in the previous paragraph. This is equivalent to saying that <span class=\"katex-eq\" data-katex-display=\"false\">\\cos\\left(\\angle(\\vec{x},\\vec{y})\\right) = 0,<\/span> which in turn is equivalent to saying that <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\cdot\\vec{y} = 0.<\/span> For this reason, stating the orthogonality of the vectors <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> is equivalent to saying that <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\cdot\\vec{y}=0.<\/span>\n<h4>If two nonzero vectors are orthogonal, then they are linearly independent<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=vtNHkaHD3aA&#038;t=2365s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">This is a somewhat intuitive property of vectors<\/span><\/strong><\/a> in <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> whose formal proof is not so direct, and it is also a property that can sometimes cause some confusion: The orthogonality of two vectors implies their linear independence, but the linear independence of two vectors does not necessarily imply their orthogonality. To see the latter, a simple counterexample suffices:<\/p>\n<p>If we take the vectors <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{A}=(1,0)<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{B}=(1,1),<\/span> which are clearly not orthogonal because <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{A}\\cdot\\vec{B}=1,<\/span> we see that if we do<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\alpha\\vec{A} + \\beta\\vec{B} = \\vec{0}\n\n<\/span>\n<p>Then we have<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n\\alpha + \\beta &amp;= 0 \\\\ \\beta &amp;= 0\n\n\\end{array}<\/span>\n<p>and therefore: <span class=\"katex-eq\" data-katex-display=\"false\">\\alpha = 0  \\wedge \\beta=0.<\/span> And with this we conclude that:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\alpha\\vec{A} + \\beta\\vec{B} = \\vec{0} \\longleftrightarrow  \\alpha = 0  \\wedge \\beta=0\n\n<\/span>\n<p>Which is equivalent to saying that <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{A}<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{B}<\/span> are linearly independent. This makes it very explicit that it is not true that linear independence implies orthogonality. However, orthogonality does imply linear independence, and this is what I will formally demonstrate below. For this, let us consider the following set of premises:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\mathcal{H}= \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\vec{x}\\cdot\\vec{y}=0, \\alpha\\vec{x}+\\beta\\vec{y} = \\vec{0}\\}<\/span>\n<p>From this we can produce the following reasoning:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n(1) &amp;\\mathcal{H}\\vdash \\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\} &amp;{;\\;Assumption}\\\\ \\\\\n\n(2) &amp;\\mathcal{H}\\vdash \\vec{x}\\cdot\\vec{y}=0 &amp;{\\;Assumption} \\\\ \\\\\n\n(3) &amp;\\mathcal{H}\\vdash \\alpha\\vec{x} + \\beta\\vec{y} = \\vec{0} &amp;{\\;Assumption} \\\\ \\\\\n\n(4) &amp;\\mathcal{H}\\vdash (\\alpha\\vec{x} + \\beta\\vec{y})\\cdot\\vec{x} = \\alpha\\|\\vec{x}\\|^2 + \\beta(\\vec{x}\\cdot\\vec{y}) &amp;{;\\; Bilinearity} \\\\ \\\\\n\n(5) &amp;\\mathcal{H}\\vdash  \\alpha\\|\\vec{x}\\|^2 = 0 &amp; {;\\; From(2,3,4)} \\\\ \\\\\n\n(6) &amp;\\mathcal{H}\\vdash  \\alpha  = 0 &amp; {;\\; From(1,5)} \\\\ \\\\\n\n(7) &amp;\\mathcal{H}\\vdash (\\alpha\\vec{x} + \\beta\\vec{y})\\cdot\\vec{y} = \\alpha(\\vec{x}\\cdot\\vec{y}) + \\beta\\|\\vec{y}\\|^2 &amp; {;\\;Bilinearity} \\\\ \\\\\n\n(8) &amp;\\mathcal{H}\\vdash \\beta\\|\\vec{y}\\|^2 = 0 &amp;{;\\;From(2,3,7)} \\\\ \\\\\n\n(9) &amp;\\mathcal{H}\\vdash \\beta = 0 &amp;{;\\;From(1,8)} \\\\ \\\\\n\n(10) &amp;\\mathcal{H}\\vdash \\alpha= 0 \\wedge \\beta = 0 &amp;{;\\;\\wedge-int(6,9)}\n\n\\end{array}<\/span>\n<p>With this we conclude that<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\n\\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\vec{x}\\cdot\\vec{y}=0, \\alpha\\vec{x}+\\beta\\vec{y} = \\vec{0}\\} \\vdash \\alpha= 0 \\wedge \\beta = 0  <\/span>\n<p>Finally, by applying the deduction theorem to this last expression we have:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\vec{x}\\cdot\\vec{y}=0\\} \\vdash (\\alpha\\vec{x}+\\beta\\vec{y} = \\vec{0}) \\rightarrow (\\alpha= 0 \\wedge \\beta = 0)<\/span>\n<p>The proof that yields the arrow in the opposite direction is trivial.<\/p>\n<p>That is: if <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> are nonzero and orthogonal vectors, then they are linearly independent.<\/p>\n<h3>The projection of one vector onto another<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=vtNHkaHD3aA&#038;t=3055s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Suppose we have two nonzero vectors<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> that form between them an angle <span class=\"katex-eq\" data-katex-display=\"false\">\\angle(\\vec{x},\\vec{y})<\/span> and we ask ourselves \u00abTo what extent is the vector <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> found along the vector <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span>?\u00bb or \u00abWhat is the size of the shadow of the vector <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> when projected onto the direction of the vector <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span>?\u00bb. This question can be resolved through trigonometry, and with this we define the projection of a vector <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> onto another <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y},<\/span> <span class=\"katex-eq\" data-katex-display=\"false\">Proy_{\\vec{y}}(\\vec{x}),<\/span> through the expression:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">Proy_{\\vec{y}}(\\vec{x}) = \\| \\vec{x}\\| \\cos(\\angle(\\vec{x},\\vec{y})) \\hat{y}<\/span>\n<p>If we combine this with what was seen in previous paragraphs, we can write:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle Proy_{\\vec{y}}(\\vec{x}) = {\\| \\vec{x}\\|} \\left(\\frac{\\vec{x}\\cdot\\vec{y}}{{\\|\\vec{x}\\|} \\|\\vec{y}\\|}\\right)\\color{red}{\\hat{y}} =  \\left(\\frac{\\vec{x}\\cdot\\vec{y}}{\\|\\vec{y}\\|} \\right)\\color{red}{\\frac{\\vec{y}}{\\|\\vec{y}\\|}} = \\left(\\frac{\\vec{x}\\cdot\\vec{y}}{\\|\\vec{y}\\|^2}\\right)\\vec{y} = \\left(\\frac{\\vec{x}\\cdot\\vec{y}}{\\vec{y}\\cdot\\vec{y}}\\right)\\vec{y}<\/span>\n<p>since, let us recall<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\displaystyle \\cos(\\angle(\\vec{x},\\vec{y}))  = \\frac{\\vec{x}\\cdot\\vec{y}}{\\|\\vec{x}\\| \\|\\vec{y}\\|}<\/span>\n<p>Projections are important because they allow us to express vectors in terms of any basis as the sum of their projections:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x} = \\displaystyle \\sum_{i=1}^n \\alpha_i \\hat{u}_i<\/span>\n<p>Where <span class=\"katex-eq\" data-katex-display=\"false\">\\{\\vec{u}_i\\}_{i=1,\\cdots, n}<\/span> is a basis of linearly independent vectors of <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> and the coefficients <span class=\"katex-eq\" data-katex-display=\"false\">\\alpha_i = (\\vec{x}\\cdot\\vec{u}_i)\/\\|\\vec{u}_i\\|<\/span> are precisely the projections onto each element of the basis, which constitute the coordinates of <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> with respect to the basis <span class=\"katex-eq\" data-katex-display=\"false\">\\{\\hat{u}_i\\}_{i=1,\\cdots, n}<\/span> of <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n.<\/span>\n<p><a name=\"El-Teorema-de-Pitagoras-y-la-Proyecci\u00f3n-sobre-un-Subespacio\"><\/a><br \/>\n<center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/CGrr6IDnvjs\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/p>\n<h2>The Pythagorean Theorem and Projection onto a Subspace<\/h2>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=CGrr6IDnvjs&#038;t=254s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">The Pythagorean theorem is a result<\/span><\/strong><\/a> well known to all, with countless proofs. One possible proof of this theorem emerges precisely from the topics we have developed for the Euclidean space, with the added value of being valid for any number of dimensions.<\/p>\n<h3>Proving the Pythagorean Theorem<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=CGrr6IDnvjs&#038;t=533s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">If we have a right triangle with legs<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">a<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">b,<\/span> and hypotenuse <span class=\"katex-eq\" data-katex-display=\"false\">c,<\/span> the Pythagorean theorem tells us that <span class=\"katex-eq\" data-katex-display=\"false\">a^2+b^2=c^2.<\/span> With this understood, we can represent each leg through a pair of orthogonal vectors <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}<\/span> and write the Pythagorean theorem as follows:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}\\} \\vdash\n\n \\vec{x}\\bot\\vec{y} \\leftrightarrow (\\|\\vec{x} + \\vec{y}\\|^2 = \\|\\vec{x}\\|^2 + \\|\\vec{y}\\|^2)<\/span>\n<p>Where the expression <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\bot\\vec{y}<\/span> indicates that both vectors are orthogonal, that is: nonzero and such that <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\cdot\\vec{y}=0.<\/span> In this way, a biconditional relationship is established between orthogonality and the sum of the squared magnitudes of two vectors.<\/p>\n<p>This vector form of representing the Pythagorean theorem can be proved through the following two reasonings:<\/p>\n<p><strong>First, in the forward direction:<\/strong><\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n(1) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\vec{x}\\bot\\vec{y}\\} \\vdash \\vec{x}\\bot\\vec{y} &amp; {;\\;Assumption} \\\\ \\\\\n\n(2) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\vec{x}\\bot\\vec{y}\\} \\vdash \\vec{x}\\cdot\\vec{y}= 0 &amp; {;\\;From(1)} \\\\ \\\\\n\n(3) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\vec{x}\\bot\\vec{y}\\} \\vdash \\|\\vec{x} + \\vec{y}\\|^2 = (\\vec{x} + \\vec{y})\\cdot(\\vec{x} + \\vec{y}) = \\|\\vec{x}\\|^2 + 2(\\vec{x}\\cdot\\vec{y}) + \\|\\vec{y}\\|^2 &amp; \\\\\n\n&amp;;\\; Property\\;of\\;the\\;Euclidean\\;norm\\;and\\;the\\;dot\\;product &amp; \\\\ \\\\\n\n(4) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\vec{x}\\bot\\vec{y}\\} \\vdash \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2 &amp; {;\\;From(2,3)} \\\\ \\\\\n\n(5) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}\\} \\vdash \\vec{x}\\bot\\vec{y} \\rightarrow ( \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2) &amp; {;\\;DT(4)} \\end{array}<\/span>\n<p><strong>And now in the reverse direction:<\/strong><\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rll}\n\n(1) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2\\} \\vdash \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2 &amp; {;\\;Assumption} \\\\ \\\\\n\n(2) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2\\} \\vdash \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2 +2(\\vec{x}\\cdot\\vec{y}) + \\|\\vec{y}\\|^2 &amp;  \\\\\n\n&amp;;\\; Property\\;of\\;the\\;Euclidean\\;norm\\;and\\;the\\;dot\\;product &amp;\\\\ \\\\\n\n(3) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2\\} \\vdash  \\vec{x}\\cdot\\vec{y}=0 &amp; {;\\;From(1,2)} \\\\ \\\\\n\n(4) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}, \\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2\\} \\vdash  \\vec{x}\\bot\\vec{y} &amp; {;\\;From(3)} \\\\ \\\\\n\n(5) &amp; \\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}\\} \\vdash (\\|\\vec{x} + \\vec{y}\\|^2 =  \\|\\vec{x}\\|^2  + \\|\\vec{y}\\|^2) \\rightarrow  \\vec{x}\\bot\\vec{y} &amp; {;\\;DT(4)} \\end{array}<\/span>\n<p><strong>And finally, by combining both reasonings we obtain what we wanted to demonstrate:<\/strong><\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\{\\vec{x},\\vec{y}\\in \\mathbb{R}^n\\setminus\\{\\vec{0}\\}\\} \\vdash   \\vec{x}\\bot\\vec{y} \\leftrightarrow (\\|\\vec{x} + \\vec{y}\\|^2 = \\|\\vec{x}\\|^2 + \\|\\vec{y}\\|^2)<\/span>\n<h3>The Projection of a Vector onto a Subspace of <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span><\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=CGrr6IDnvjs&#038;t=1545s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Let us consider a subspace<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> of <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> formed by a basis of unit vectors <span class=\"katex-eq\" data-katex-display=\"false\">\\{\\hat{v}_1, \\cdots, \\hat{v}_k\\}.<\/span> If we take a vector <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\in\\mathbb{R}^n\\setminus\\{\\vec{0}\\},<\/span> the projection of the vector <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> onto the space <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> is defined by the expression:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">Proy_{H}(\\vec{x}) = \\displaystyle \\sum_{j=1}^k (\\vec{x} \\cdot \\hat{v}_j)\\hat{v}_j<\/span>\n<p>That a set is orthonormal means that all its elements are orthogonal to each other and each one has norm equal to one.<\/p>\n<p>This is, so to speak, the shadow cast by a vector onto each of the components of the subspace <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> of <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> <\/p>\n<p><center><img decoding=\"async\" src=\"data:image\/gif;base64,R0lGODlhAQABAIAAAAAAAP\/\/\/yH5BAEAAAAALAAAAAABAAEAAAIBRAA7\" data-src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEga986LBrInk-B_9gUKPe01TF10dNECXU54KK1bSf3mAPakWE-FqdqyPbb0TVy88OfGxQmJRd-yW4dwAfcC21i2dM0KZqQjPe_Qx0M5OUz4f_P6IipJQ6PcxtkOmcO7-GqRiGZ-3StQpzy8FMIfPYE89Wae6JZIC2Jk9dSTPFTK1L4TsnpkcdpV1Dbr\" width=\"578\" height=\"591\" class=\"alignnone size-full lazyload\" \/><noscript><img decoding=\"async\" src=\"https:\/\/blogger.googleusercontent.com\/img\/a\/AVvXsEga986LBrInk-B_9gUKPe01TF10dNECXU54KK1bSf3mAPakWE-FqdqyPbb0TVy88OfGxQmJRd-yW4dwAfcC21i2dM0KZqQjPe_Qx0M5OUz4f_P6IipJQ6PcxtkOmcO7-GqRiGZ-3StQpzy8FMIfPYE89Wae6JZIC2Jk9dSTPFTK1L4TsnpkcdpV1Dbr\" width=\"578\" height=\"591\" class=\"alignnone size-full lazyload\" \/><\/noscript><\/center><\/p>\n<h3>Distance between a Point or Vector of <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> and a Subspace of <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span><\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=CGrr6IDnvjs&#038;t=1974s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">From the projection of a vector<\/span><\/strong><\/a> <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\in\\mathbb{R}^n\\setminus\\{\\vec{0}\\}<\/span> onto a subspace <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> of <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span>, we can construct a vector of the form<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x} - Proy_{H}(\\vec{x})<\/span>\n<p>The vector formed in this way will be a vector that connects a point of the subspace <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> with the point of coordinates <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x},<\/span> which emerges orthogonally to the subspace <span class=\"katex-eq\" data-katex-display=\"false\">H.<\/span> This is not difficult to prove: if we take any vector <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z}\\in H<\/span> and calculate the dot product <span class=\"katex-eq\" data-katex-display=\"false\">(\\vec{x}-Proy_{H}(\\vec{x}))\\cdot \\vec{z},<\/span> it suffices to see that the result of this operation is zero. Let us do the calculations to verify this:<\/p>\n<p>If <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z}\\in H,<\/span> then it will be of the form<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z}=\\displaystyle \\sum_{j=1}^k \\beta_j\\hat{v}_j<\/span>\n<p>Where <span class=\"katex-eq\" data-katex-display=\"false\">\\{\\hat{v}_j\\}_{j=1}^k<\/span> is an orthonormal basis of <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\beta_j \\in\\mathbb{R}<\/span> are the coefficients of <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z}<\/span> in <span class=\"katex-eq\" data-katex-display=\"false\">H.<\/span> Taking this into account, the calculation of the dot product <span class=\"katex-eq\" data-katex-display=\"false\">(\\vec{x}-Proy_{H}(\\vec{x}))\\cdot \\vec{z},<\/span> yields:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl} (\\vec{x}-Proy_{H}(\\vec{x}))\\cdot \\vec{z} &amp;= \\left(\\vec{x} - \\displaystyle \\sum_{j=1}^k (\\vec{x} \\cdot \\hat{v}_j)\\hat{v}_j \\right) \\cdot \\displaystyle \\sum_{j=1}^k \\beta_j\\hat{v}_j \\\\ \\\\ &amp;= \\vec{x} \\cdot \\displaystyle \\sum_{j=1}^k \\beta_j\\hat{v}_j - \\displaystyle \\sum_{j=1}^k (\\vec{x} \\cdot \\hat{v}_j)\\hat{v}_j \\cdot \\displaystyle \\sum_{j=1}^k \\beta_j\\hat{v}_j \\end{array}<\/span>\n<p>But since <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> is a vector of <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> of which <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> is a subspace, it is possible to find a set of <span class=\"katex-eq\" data-katex-display=\"false\">n-k<\/span> vectors orthonormal to each other and also orthonormal to all the vectors of <span class=\"katex-eq\" data-katex-display=\"false\">H,<\/span> say <span class=\"katex-eq\" data-katex-display=\"false\">\\{\\hat{v}_{k+1}, \\cdots, \\hat{v}_n\\},<\/span> such that together with the basis of <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> they form a basis for <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> and we can write<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x} = \\displaystyle  \\sum_{j=1}^k (\\vec{x}\\cdot\\hat{v}_j )\\hat{v}_j + \\sum_{j=k+1}^n \\alpha_j \\hat{v}_j <\/span>\n<p>So that the development above continues as follows:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl}\n\n(\\vec{x}-Proy_{H}(\\vec{x}))\\cdot \\vec{z} &amp;= \\displaystyle \\left( \\sum_{j=1}^k (\\vec{x}\\cdot\\hat{v}_j )\\hat{v}_j + \\sum_{j=k+1}^n \\alpha_j \\hat{v}_j\\right) \\cdot  \\sum_{j=1}^k \\beta_j\\hat{v}_j -  \\sum_{j=1}^k (\\vec{x} \\cdot \\hat{v}_j)\\hat{v}_j \\cdot  \\sum_{j=1}^k \\beta_j\\hat{v}_j \\\\ \\\\\n\n&amp;=  \\displaystyle \\sum_{j=1}^k (\\vec{x}\\cdot\\hat{v}_j )\\hat{v}_j \\cdot \\sum_{j=1}^k \\beta_j\\hat{v}_j + \\underbrace{\\color{red}{\\sum_{j=k+1}^n \\alpha_j \\hat{v}_j \\cdot \\sum_{j=1}^k \\beta_j\\hat{v}_j}}_{(*)} - \\sum_{j=1}^k (\\vec{x} \\cdot \\hat{v}_j)\\hat{v}_j \\cdot  \\sum_{j=1}^k \\beta_j\\hat{v}_j \\\\ \\\\\n\n&amp;=  \\displaystyle \\sum_{j=1}^k (\\vec{x}\\cdot\\hat{v}_j )\\hat{v}_j \\cdot \\sum_{j=1}^k \\beta_j\\hat{v}_j  - \\sum_{j=1}^k (\\vec{x} \\cdot \\hat{v}_j)\\hat{v}_j \\cdot  \\sum_{j=1}^k \\beta_j\\hat{v}_j \\\\ \\\\\n\n&amp;= 0  \\end{array}<\/span>\n<p>(*) Zero sum because <span class=\"katex-eq\" data-katex-display=\"false\">\\{v_j\\}_{j=1}^n<\/span> is an orthonormal basis of <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n.<\/span>\n<p>From this we can show that the distance between the subspace <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> and the vector <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> is given by:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x} - Proy_{H}(\\vec{x})\\|<\/span>\n<h4>Proof<\/h4>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=CGrr6IDnvjs&#038;t=2995s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">To prove this result it will be shown<\/span><\/strong><\/a> that for all <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{z}\\in H<\/span> it will always hold that <span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x} - Proy_{H}(\\vec{x})\\| \\leq \\|\\vec{x} - \\vec{z}\\|,<\/span> for this we will use the Pythagorean theorem in the following way:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl} \\|\\vec{x} - \\vec{z}\\|^2 &amp;= \\| \\left(\\vec{x} -Proy_{H}(\\vec{x}) \\right) + \\left(Proy_{H}(\\vec{x}) - \\vec{z}\\right)\\|^2 \\\\ \\\\ &amp;= \\| \\vec{x} -Proy_{H}(\\vec{x}) \\|^2 + \\|Proy_{H}(\\vec{x}) - \\vec{z}\\|^2 \\\\ \\\\ \\end{array}<\/span>\n<p>This last equality is obtained because the vectors <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x} -Proy_{H}(\\vec{x})<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">Proy_{H}(\\vec{x}) - \\vec{z}<\/span> are orthogonal. And therefore:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x} - Proy_{H}(\\vec{x})\\|^2 \\leq \\|\\vec{x} - \\vec{z}\\|^2<\/span>\n<p>which is what we wanted to demonstrate.<\/p>\n<p>With this result in hand, we can say that the distance between a point <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\in\\mathbb{R}^n<\/span> and a subspace <span class=\"katex-eq\" data-katex-display=\"false\">H<\/span> of <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n<\/span> generated by the orthonormal vectors <span class=\"katex-eq\" data-katex-display=\"false\">\\{\\hat{v}_1, \\cdots, \\hat{v}_k\\}<\/span> is given by:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">dist(\\vec{x},H) =\\left\\|\\vec{x} - Proy_{H}(\\vec{x})\\right\\|= \\left\\|\\vec{x} - \\displaystyle \\sum_{j=1}^k (\\vec{x} \\cdot \\hat{v}_j)\\hat{v}_j\\right\\|<\/span>\n<p><a name=\"El-Producto-Escalar-y-Vectorial-en-R3\"><\/a><br \/>\n<center><iframe class=\"lazyload\" width=\"560\" height=\"315\" data-src=\"https:\/\/www.youtube.com\/embed\/uei6y2tniOc\" title=\"YouTube video player\" frameborder=\"0\" allow=\"accelerometer; autoplay; clipboard-write; encrypted-media; gyroscope; picture-in-picture\" allowfullscreen><\/iframe><\/center><\/p>\n<h2>The Dot and Cross Product in <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3<\/span><\/h2>\n<p><strong><a href=\"https:\/\/www.youtube.com\/watch?v=uei6y2tniOc&#038;t=242s\" rel=\"noopener\" target=\"_blank\"><span style=\"color: #ff0000;\">We will now change our focus slightly<\/span><\/strong><\/a> to concentrate on vectors in <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3.<\/span> Here, in addition to the operations we have already reviewed in general for <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^n,<\/span> it is also possible to define the cross product, which takes two vectors and produces another vector as a result. This is a product exclusive to <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3<\/span> (and possibly to <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^7<\/span>, whose case we will not analyze here). Generally, the vectors of the canonical basis of <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3<\/span> are represented by the letters <span class=\"katex-eq\" data-katex-display=\"false\">\\hat{x}, \\hat{y}, \\hat{z}<\/span> or as <span class=\"katex-eq\" data-katex-display=\"false\">\\hat{\\imath}, \\hat{\\jmath}, \\hat{k}<\/span>. The preference for one or the other is personal.<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl} \\hat{\\imath} = \\hat{x}&amp;=(1,0,0)\\\\ \\hat{\\jmath} =\\hat{y}&amp;=(0,1,0)\\\\ \\hat{k} =\\hat{z}&amp;=(0,0,1)\\\\ \\end{array}<\/span>\n<p>Thus, if we have a vector of the form <span class=\"katex-eq\" data-katex-display=\"false\">(a,b,c),<\/span> it can be written algebraically as follows:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">(a,b,c) = a\\hat{x} + b\\hat{y} + c\\hat{z}<\/span>\n<h3>The Cross Product in <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3<\/span><\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=uei6y2tniOc&#038;t=330s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Let <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}=(x_1,x_2,x_3)<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}=(y_1,y_2,y_3)<\/span> be vectors in <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3.<\/span><\/span><\/strong><\/a> The cross product of <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> with <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y},<\/span> <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\times\\vec{y}<\/span> is defined by:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\begin{array}{rl} \\vec{x}\\times\\vec{y} &amp;= \\left|\\begin{array}{ccc} \\hat{x} &amp; \\hat{y} &amp; \\hat{z} \\\\ x_1 &amp; x_2 &amp; x_3 \\\\ y_1 &amp; y_2 &amp; y_3 \\end{array}\\right| \\\\ \\\\ &amp;=\\hat{x}x_2y_3 + \\hat{y}x_3y_1 + \\hat{z} x_1y_2 - \\left( \\hat{z} x_2 y_1 + \\hat{y} x_1 y_3 + \\hat{x}x_3y_2\\right) \\\\ \\\\ &amp;=\\hat{x}(x_2y_3 - x_3y_2) + \\hat{y}(x_3y_1 - x_1y_3) + \\hat{z}(x_1y_2 - x_2y_1) \\end{array}<\/span>\n<h3>Lagrange&#8217;s Identity<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=uei6y2tniOc&#038;t=1399s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">For the case of vectors in <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3<\/span><\/span><\/strong><\/a> we can recognize three types of \u00abproducts\u00bb: the dot product <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\cdot\\vec{y},<\/span> the cross product <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\times\\vec{y},<\/span> and the product of the norms <span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x}\\|\\|\\vec{y}\\|.<\/span> These three products are related to each other through Lagrange&#8217;s identity<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x}\\times\\vec{y}\\|^2  = \\|\\vec{x}\\|^2\\|\\vec{y}\\|^2- (\\vec{x}\\cdot\\vec{y})^2 <\/span>\n<h4>Proof of Lagrange&#8217;s Identity<\/h4>\n<p>Let <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}=(x_1,x_2,x_3)<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}=(y_1,y_2,y_3)<\/span> be vectors in <span class=\"katex-eq\" data-katex-display=\"false\">\\mathbb{R}^3,<\/span> then we have:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\begin{array}{rl} \\vec{x}\\times\\vec{y} &amp;=(x_2y_3 - x_3y_2) \\hat{x} + (x_3y_1 - x_1y_3)\\hat{y} + (x_1y_2 - x_2y_1)\\hat{z} \\end{array}<\/span>\n<p>So that:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\begin{array}{rl}\n\n\\|\\vec{x}\\times\\vec{y}\\|^2 &amp;=(x_2y_3 - x_3y_2)^2 + (x_3y_1 - x_1y_3)^2 + (x_1y_2 - x_2y_1)^2 \\\\ \\\\\n\n&amp;= \\color{green}{x_2^2y_3^2 - 2x_2x_3y_3y_2 + x_3^2y_2^2} + \\cdots\\\\ \\\\\n\n&amp;\\cdots + \\color{blue}{x_3^2y_1^2 - 2x_3x_1y_1y_3 + x_1^2y_3^2} + \\cdots \\\\ \\\\\n\n&amp;\\cdots + \\color{red}{x_1^2y_2^2 - 2x_1x_2y_2y_1 + x_2^2y_1^2} \\end{array}<\/span>\n<p>On the other hand:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\"> \\begin{array}{rl}\n\n\\|\\vec{x}\\|^2 \\|\\vec{y}\\|^2 - (\\vec{x}\\cdot\\vec{y})^2 &amp;= (x_1^2 + x_2^2 + x_3^2)(y_1^2+y_2^2 + y_3^2) - (x_1y_1 + x_2y_2 + x_3 y_3)^2 \\\\ \\\\ \\\\\n\n&amp;=  {x_1^2y_1^2} + \\color{red}{x_1^2y_2^2} + \\color{blue}{x_1^2y_3^2} + \\cdots \\\\ \\\\\n\n&amp;\\cdots + \\color{red}{x_2^2y_1^2} +  {x_2^2y_2^2} + \\color{green}{x_2^2y_3^2} + \\cdots \\\\ \\\\\n\n&amp;\\cdots + \\color{blue}{x_3^2y_1^2} + \\color{green}{x_3^2y_2^2} +  {x_3^2y_3^2} + \\cdots \\\\ \\\\\n\n&amp;\\cdots - \\left[ {x_1^2y_1^2} +  {x_2^2y_2^2} +  {x_3^2y_3^2} + \\right. \\cdots \\\\ \\\\\n\n&amp;\\cdots + 2\\left(\\color{red}{x_1x_2y_1y_2} + \\color{blue}{x_1x_3y_1y_3} + \\color{green}{x_2x_3y_2y_3} \\right)\\left.\\right] \\\\ \\\\ \\\\\n\n&amp;= \\color{red}{x_1^2y_2^2 - 2x_1x_2y_2y_1 + x_2^2y_1^2} + \\cdots \\\\ \\\\\n\n&amp; \\cdots + \\color{blue}{x_1^2y_3^2 - 2x_1x_3y_3y_1 + x_3^2y_1^2} + \\cdots \\\\ \\\\\n\n&amp; \\cdots + \\color{green}{x_2^2y_3^2 - 2x_2x_3y_3y_2 + x_3^2y_2^2}\n\n\\end{array}<\/span>\n<p>Finally, by comparing the colored expressions we obtain what we wanted to demonstrate.<\/p>\n<h3>The Cross Product and the Angle Between Vectors<\/h3>\n<p><a href=\"https:\/\/www.youtube.com\/watch?v=uei6y2tniOc&#038;t=1954s\" rel=\"noopener\" target=\"_blank\"><strong><span style=\"color: #ff0000;\">Earlier we saw that there is a close relationship<\/span><\/strong><\/a> between the angle sustained by two vectors and the result of the dot product, which is given by the relation <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}\\cdot\\vec{y} = \\|\\vec{x}\\|\\|\\vec{y}\\|\\cos(\\angle(\\vec{x},\\vec{y})).<\/span> It turns out that a similar thing occurs with the cross product and it is given by the following relation:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x}\\times\\vec{y}\\| = \\|\\vec{x}\\|\\|\\vec{y}\\| \\sin(\\angle(\\vec{x},\\vec{y}))<\/span>\n<p>This expression is a direct result of Lagrange&#8217;s identity demonstrated above, and the proof goes more or less as follows:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\begin{array}{rl} \\|\\vec{x}\\times\\vec{y}\\|^2 &amp;= \\|\\vec{x}\\|^2\\|\\vec{y}\\|^2 - (\\vec{x}\\cdot\\vec{y})^2 \\\\ \\\\ &amp;= \\|\\vec{x}\\|^2\\|\\vec{y}\\|^2 - (\\|\\vec{x}\\|\\|\\vec{y}\\|\\cos(\\angle(\\vec{x},\\vec{y})))^2 \\\\ \\\\ &amp;= \\|\\vec{x}\\|^2\\|\\vec{y}\\|^2 - \\|\\vec{x}\\|^2\\|\\vec{y}\\|^2\\cos^2(\\angle(\\vec{x},\\vec{y})) \\\\ \\\\ &amp;= \\|\\vec{x}\\|^2\\|\\vec{y}\\|^2 (1 - \\cos^2(\\angle(\\vec{x},\\vec{y}))) \\\\ \\\\ &amp;= \\|\\vec{x}\\|^2\\|\\vec{y}\\|^2 \\sin^2(\\angle(\\vec{x},\\vec{y})) \\end{array}<\/span>\n<p>Finally, taking square roots we arrive at:<\/p>\n<p style=\"text-align:center;\"><span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x}\\times\\vec{y}\\| = \\|\\vec{x}\\|\\|\\vec{y}\\|\\; |\\sin(\\angle(\\vec{x},\\vec{y}))|<\/span>\n<p>But recall that <span class=\"katex-eq\" data-katex-display=\"false\">\\angle(\\vec{x},\\vec{y})\\in[0,\\pi],<\/span> and in that range of values the sine function is always non-negative, so we can remove the absolute value and arrive at what we wanted to demonstrate.<\/p>\n<p>From this expression we can infer that the result of the operation <span class=\"katex-eq\" data-katex-display=\"false\">\\|\\vec{x}\\times\\vec{y}\\|<\/span> gives us the area generated by the vectors <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{x}<\/span> and <span class=\"katex-eq\" data-katex-display=\"false\">\\vec{y}.<\/span>\n","protected":false},"excerpt":{"rendered":"<p>Algebra and Projections in Rn, Cross Product in Abstract:This series is the direct continuation of the series on the Euclidean Space of n dimensions. Here we will review some linear algebra concepts that help to better understand the n-dimensional Euclidean space, examine the concepts of projections of one vector onto another, demonstrate the Pythagorean theorem, [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":34241,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"iawp_total_views":1,"footnotes":""},"categories":[567,1116],"tags":[],"class_list":["post-34243","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-mathematics","category-multivariable-calculus"],"yoast_head":"<!-- This site is optimized with the Yoast SEO plugin v26.7 - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>Algebra and Projections in Rn, Cross Product in R3 - toposuranos.com\/material<\/title>\n<meta name=\"description\" content=\"Explore projections and vectors in Rn: linear independence, orthogonality, Pythagorean theorem, and cross product in Euclidean spaces.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/toposuranos.com\/material\/en\/algebra-and-projections-in-rn-cross-product-in-r3\/\" \/>\n<meta property=\"og:locale\" content=\"es_ES\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Algebra and Projections in Rn, Cross Product in R3\" \/>\n<meta property=\"og:description\" content=\"Explore projections and vectors in Rn: linear independence, orthogonality, Pythagorean theorem, and cross product in Euclidean spaces.\" \/>\n<meta property=\"og:url\" content=\"https:\/\/toposuranos.com\/material\/en\/algebra-and-projections-in-rn-cross-product-in-r3\/\" \/>\n<meta property=\"og:site_name\" content=\"toposuranos.com\/material\" \/>\n<meta property=\"article:publisher\" content=\"https:\/\/www.facebook.com\/groups\/toposuranos\" \/>\n<meta property=\"article:published_time\" content=\"2022-03-29T13:00:59+00:00\" \/>\n<meta property=\"article:modified_time\" content=\"2025-08-27T21:04:54+00:00\" \/>\n<meta property=\"og:image\" content=\"http:\/\/toposuranos.com\/material\/wp-content\/uploads\/2022\/03\/vectoresyproyeccionesrn-1024x498.jpg\" \/>\n<meta name=\"author\" content=\"giorgio.reveco\" \/>\n<meta name=\"twitter:card\" content=\"summary_large_image\" \/>\n<meta name=\"twitter:title\" content=\"Algebra and Projections in Rn, Cross Product in R3\" \/>\n<meta name=\"twitter:description\" content=\"Explore projections and vectors in Rn: linear independence, orthogonality, Pythagorean theorem, and cross product in Euclidean spaces.\" \/>\n<meta name=\"twitter:image\" content=\"https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2022\/03\/vectoresyproyeccionesrn.jpg\" \/>\n<meta name=\"twitter:creator\" content=\"@topuranos\" \/>\n<meta name=\"twitter:site\" content=\"@topuranos\" \/>\n<meta name=\"twitter:label1\" content=\"Escrito por\" \/>\n\t<meta name=\"twitter:data1\" content=\"giorgio.reveco\" \/>\n\t<meta name=\"twitter:label2\" content=\"Tiempo de lectura\" \/>\n\t<meta name=\"twitter:data2\" content=\"1 minuto\" \/>\n<script type=\"application\/ld+json\" class=\"yoast-schema-graph\">{\"@context\":\"https:\/\/schema.org\",\"@graph\":[{\"@type\":\"Article\",\"@id\":\"https:\/\/toposuranos.com\/material\/en\/algebra-and-projections-in-rn-cross-product-in-r3\/#article\",\"isPartOf\":{\"@id\":\"https:\/\/toposuranos.com\/material\/en\/algebra-and-projections-in-rn-cross-product-in-r3\/\"},\"author\":{\"name\":\"giorgio.reveco\",\"@id\":\"https:\/\/toposuranos.com\/material\/#\/schema\/person\/e15164361c3f9a2a02cf6c234cf7fdc1\"},\"headline\":\"Algebra and Projections in Rn, Cross Product in R3\",\"datePublished\":\"2022-03-29T13:00:59+00:00\",\"dateModified\":\"2025-08-27T21:04:54+00:00\",\"mainEntityOfPage\":{\"@id\":\"https:\/\/toposuranos.com\/material\/en\/algebra-and-projections-in-rn-cross-product-in-r3\/\"},\"wordCount\":4206,\"commentCount\":0,\"publisher\":{\"@id\":\"https:\/\/toposuranos.com\/material\/#organization\"},\"image\":{\"@id\":\"https:\/\/toposuranos.com\/material\/en\/algebra-and-projections-in-rn-cross-product-in-r3\/#primaryimage\"},\"thumbnailUrl\":\"https:\/\/toposuranos.com\/material\/wp-content\/uploads\/2022\/03\/vectoresyproyeccionesrn.jpg\",\"articleSection\":[\"Mathematics\",\"Multivariable Calculus\"],\"inLanguage\":\"es\",\"potentialAction\":[{\"@type\":\"CommentAction\",\"name\":\"Comment\",\"target\":[\"https:\/\/toposuranos.com\/material\/en\/algebra-and-projections-in-rn-cross-product-in-r3\/#respond\"]}]},{\"@type\":\"WebPage\",\"@id\":\"https:\/\/toposuranos.com\/material\/en\/algebra-and-projections-in-rn-cross-product-in-r3\/\",\"url\":\"https:\/\/toposuranos.com\/material\/en\/algebra-and-projections-in-rn-cross-product-in-r3\/\",\"name\":\"Algebra and Projections in Rn, Cross Product in R3 - 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