{"id":281,"date":"2021-04-18T15:06:09","date_gmt":"2021-04-18T19:06:09","guid":{"rendered":"http:\/\/pressbooks.library.upei.ca\/statics\/?post_type=chapter&#038;p=281"},"modified":"2025-07-31T18:19:04","modified_gmt":"2025-07-31T22:19:04","slug":"cross-products","status":"publish","type":"chapter","link":"https:\/\/pressbooks.library.upei.ca\/statics\/chapter\/cross-products\/","title":{"raw":"1.5 Cross Products","rendered":"1.5 Cross Products"},"content":{"raw":"This is a second way to calculate the product of two vectors. It creates a third vector that is perpendicular to the plane made from the two vectors, as shown in the figure [footnote]<a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Cross-product-with-area.svg\">https:\/\/commons.wikimedia.org\/wiki\/File:Cross-product-with-area.svg<\/a> [\/footnote].\r\n\r\n<img src=\"https:\/\/upload.wikimedia.org\/wikipedia\/commons\/5\/5f\/Cross-product-with-area.svg\" alt=\"File:Cross-product-with-area.svg - Wikimedia Commons\" class=\"n3VNCb alignright\" data-noaft=\"1\" width=\"289\" height=\"251\" \/>The black arrow is perpendicular to the grey plane made from blue and red vectors. This is how you will find the amount of torque created from a force, which we will do many times. Also, unlike the dot product, <span>a<\/span> x <span>b<\/span>\u00a0 is a different direction than <span>b<\/span> x <span>a<\/span>.\r\n\r\n[latex]\\vec A\\times\\vec B=\\begin{bmatrix} \\underline{\\hat{i}} &amp; \\underline{\\hat{j}} &amp; \\underline{\\hat{k}} \\\\ A_x &amp; A_y &amp; A_z \\\\ B_x &amp; B_y &amp; B_z \\end{bmatrix}[\/latex]\r\n\r\n&nbsp;\r\n\r\n[latex]\\vec {\\textbf{A}}\\times\\vec{\\textbf{B}}=(A_yB_z-A_zB_y)\\underline{\\hat{\\textbf{i}}}+(A_zB_x-A_xB_z)\\underline{\\hat{\\textbf{j}}}+(A_xB_y-A_yB_x)\\underline{\\hat{\\textbf{k}}}[\/latex]\r\n<p style=\"text-align: center\">[latex]|\\vec {\\textbf{A}}\\times\\vec {\\textbf{B}}|=|\\vec A||\\vec B|\\sin\\theta[\/latex]<\/p>\r\n\r\n<div class=\"textbox\">\r\n\r\n<span>The <\/span><strong>vector product<\/strong><span> of two vectors [latex]\\vec A \\text{\u00a0 and\u00a0 }\\vec B[\/latex]<\/span><span>\u00a0<\/span><span>is denoted by [latex]\\vec A \\text{ x }\\vec B[\/latex]<\/span><span>\u00a0and is often referred to as a\u00a0<\/span><strong>cross product<\/strong><span>. The vector product is a vector that has its direction perpendicular to both vectors [latex]\\vec A \\text{\u00a0 and\u00a0 }\\vec B[\/latex]<\/span><span>. In other words, vector [latex]\\vec A \\text{ x }\\vec B[\/latex]<\/span><span> is perpendicular to the plane that contains vectors [latex]\\vec A \\text{\u00a0 and\u00a0 }\\vec B[\/latex]<\/span><span>.<\/span><span><\/span>\r\n\r\nSource: University Physics Volume 1, OpenStax CNX, <a href=\"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/chapter\/2-4-products-of-vectors\/\">https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/chapter\/2-4-products-of-vectors\/<\/a>\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n\r\nUnit vectors allow for a straightforward calculation of the cross product of two vectors under even the most general circumstances, e.g. circumstances in which each of the vectors is pointing in an arbitrary direction in a three-dimensional space. To take advantage of the method, we need to know the cross product of the Cartesian coordinate axis unit vectors i, j, and k with each other. First off, we should note that any vector crossed with itself gives zero. This is evident from the\u00a0 equation:\r\n<p style=\"text-align: center\">[latex]|\\vec A\\times\\vec B|=|\\vec A||\\vec B|\\sin\\theta[\/latex].<\/p>\r\nbecause if A and B are in the same direction, then \u03b8 = 0\u00b0, and since sin 0\u00b0 = 0, we have [latex]|\\vec A\\times\\vec B|=0[\/latex].\u00a0Regarding the unit vectors, this means that:\r\n<p style=\"text-align: center\">[latex]\\underline{\\hat{i}}\\times\\underline{\\hat{i}}=0\\\\\\underline{\\hat{j}}\\times\\underline{\\hat{j}}=0\\\\\\underline{\\hat{k}}\\times\\underline{\\hat{k}}=0[\/latex]<\/p>\r\nNext, we note that the magnitude of the cross product of two vectors that are perpendicular to each other is just the ordinary product of the magnitudes of the vectors. This is also evident from the equation:\r\n<p style=\"text-align: center\">[latex]|\\vec {A}\\times\\vec{B}|=|\\vec A||\\vec B|\\sin\\theta[\/latex]<\/p>\r\nbecause if [latex]\\vec A[\/latex] is perpendicular to [latex]\\vec B[\/latex]\u00a0then \u03b8 = 90\u00b0 and sin90\u00b0 = 1 so:\r\n<p style=\"text-align: center\">[latex]|\\vec A\\times\\vec B|=|\\vec A||\\vec B|[\/latex]<\/p>\r\nNow, if A and B are unit vectors, then their magnitudes are both 1, so the product of their magnitudes is also 1. Furthermore, the unit vectors i, j, and k are all perpendicular to each other, so the magnitude of the cross product of any one of them with any other one of them is the product of the two magnitudes, that is, 1.\r\n\r\nNow, how about the direction? Let\u2019s use the right-hand rule to get the direction of i\u00d7j:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0584.jpg\" alt=\"A 3D coordinate system showing the x, y, and z axes with unit vectors labeled \u00ee (x-axis), \u0135 (y-axis), and k\u0302 (z-axis) pointing in their respective positive directions.\" class=\"aligncenter wp-image-410\" width=\"327\" height=\"320\" \/>\r\n\r\n<em>[To find the direction, we use the right-hand rule, which we will cover more in section 3.1. Here is an overview<\/em>.] With the fingers of the right hand pointing directly away from the right elbow, and in the same direction as i, (the first vector in \u201c[latex]\\underline{\\hat{i}}\\times\\underline{\\hat{j}}[\/latex]\u201d) to make it so that if one were to close the fingers, they would point in the same direction as [latex]\\underline{\\hat{j}}[\/latex], the palm must be facing in the +y direction. That being the case, the extended thumb must be pointing in the +z direction. Putting the magnitude (the magnitude of each unit vector is 1) and direction (+z) information together, we see that:\r\n<ul>\r\n \t<li>[latex]\\underline{\\hat{i}}\\times\\underline{\\hat{j}}=\\underline{\\hat{k}}[\/latex]<\/li>\r\n \t<li>[latex]\\underline{\\hat{j}}\\times\\underline{\\hat{k}}=\\underline{\\hat{i}}[\/latex]<\/li>\r\n \t<li>[latex]\\underline{\\hat{k}}\\times\\underline{\\hat{i}}=\\underline{\\hat{j}}[\/latex]<\/li>\r\n \t<li>[latex]\\underline{\\hat{j}}\\times\\underline{\\hat{i}}=-\\underline{\\hat{k}}[\/latex]<\/li>\r\n \t<li>[latex]\\underline{\\hat{k}}\\times\\underline{\\hat{j}}=-\\underline{\\hat{i}}[\/latex]<\/li>\r\n \t<li>[latex]\\underline{\\hat{i}}\\times\\underline{\\hat{k}}=-\\underline{\\hat{j}}[\/latex]<\/li>\r\n<\/ul>\r\nOne way of remembering this is to write [latex]\\underline{\\hat{i}},\\underline{\\hat{j}},\\underline{\\hat{k}}[\/latex] twice in succession:\r\n<p style=\"text-align: center\">[latex]\\underline{\\hat{\\textbf{i}}},\\underline{\\hat{\\textbf{j}}},\\underline{\\hat{\\textbf{k}}},\\underline{\\hat{\\textbf{i}}},\\underline{\\hat{\\textbf{j}}},\\underline{\\hat{\\textbf{k}}}[\/latex]<\/p>\r\nThen, crossing any one of the first three vectors into the vector immediately to its right yields the next vector to the right. But crossing any one of the last three vectors into the vector immediately to its left yields the negative of the next vector to the left (left-to-right \u201c+\u201c, but right-to-left \u201c\u2212\u201c).\r\n\r\nNow we\u2019re ready to look at the general case. Any vector\u00a0 [latex]\\vec A[\/latex] \u00a0can be expressed in terms of unit vectors:\r\n\r\n[latex]\\vec{\\textbf{A}}=A_x\\hat{\\underline{i}}+A_y\\hat{\\underline{j}}+A_z\\hat{\\underline{k}}[\/latex]\r\n\r\nDoing the same for a vector\u00a0 [latex]\\vec B[\/latex]\u00a0 <span><\/span> then allows us to write the cross product as:\r\n\r\n[latex]\\vec{\\textbf{A}}\\times\\vec{\\textbf{B}}=(A_x\\hat{\\underline{i}}+A_y\\hat{\\underline{j}}+A_z\\hat{\\underline{k}})\\times(B_x\\hat{\\underline{i}}+B_y\\hat{\\underline{j}}+B_z\\hat{\\underline{k}})[\/latex]\r\n\r\nUsing the distributive rule for multiplication, we can write this as:\r\n\r\n[latex]\\begin{aligned}\\vec{\\textbf{A}}\\times\\vec{\\textbf{B}}=&amp;A_x\\hat{\\underline{i}}\\times(B_x\\hat{\\underline{i}}+B_y\\hat{\\underline{j}}+B_z\\hat{\\underline{k}})+\\\\&amp;A_y\\hat{\\underline{j}}\\times(B_x\\hat{\\underline{i}}+B_y\\hat{\\underline{j}}+B_z\\hat{\\underline{k}})+\\\\&amp;A_z\\hat{\\underline{k}}\\times(B_x\\hat{\\underline{i}}+B_y\\hat{\\underline{j}}+B_z\\hat{\\underline{k}})\\end{aligned}[\/latex]\r\n\r\n[latex]\\begin{aligned}\\vec{\\textbf{A}}\\times\\vec{\\textbf{B}}=&amp;A_x\\hat{\\underline{i}}\\times B_x\\hat{\\underline{i}}+A_x\\hat{\\underline{i}}\\times B_y\\hat{\\underline{j}}+A_x\\hat{\\underline{i}}\\times\u00a0 B_z\\hat{\\underline{k}}+\\\\&amp;A_y\\hat{\\underline{j}}\\times B_x\\hat{\\underline{i}}+A_y\\hat{\\underline{j}}\\times B_y\\hat{\\underline{j}}+A_y\\hat{\\underline{j}}\\times B_z\\hat{\\underline{k}}+\\\\&amp;A_z\\hat{\\underline{k}}\\times B_x\\hat{\\underline{i}}+A_z\\hat{\\underline{k}}\\times B_y\\hat{\\underline{j}}+A_z\\hat{\\underline{k}}\\times B_z\\hat{\\underline{k}}\\end{aligned}[\/latex]\r\n\r\nUsing, in each term, the commutative rule and the associative rule for multiplication, we can write this as:\r\n\r\n[latex]\\begin{aligned}\\vec{\\textbf{A}}\\times\\vec{\\textbf{B}}=&amp;A_xB_x(\\hat{\\underline{i}}\\times\\hat{\\underline{i}})+A_xB_y(\\hat{\\underline{i}}\\times\\hat{\\underline{j}})+A_xB_z(\\hat{\\underline{i}}\\times\\hat{\\underline{k}})+\\\\&amp;A_yB_x(\\hat{\\underline{j}}\\times\\hat{\\underline{i}})+A_yB_y(\\hat{\\underline{j}}\\times\\hat{\\underline{j}})+A_yB_z(\\hat{\\underline{j}}\\times\\hat{\\underline{k}})+\\\\&amp;A_zB_x(\\hat{\\underline{k}}\\times\\hat{\\underline{i}})+A_zB_y(\\hat{\\underline{k}}\\times\\hat{\\underline{j}})+A_zB_z(\\hat{\\underline{k}}\\times\\hat{\\underline{k}})\\end{aligned}[\/latex]\r\n\r\nNow we evaluate the cross product that appears in each term:\r\n\r\n[latex]\\begin{aligned}\\vec A\\times\\vec B=&amp;A_xB_x(0)+A_xB_y(\\underline{\\hat{k}})+A_xB_z(-\\underline{\\hat{j}})+\\\\&amp;A_yB_x(-\\underline{\\hat{k}})+A_yB_y(0)+A_yB_z(\\underline{\\hat{i}})+\\\\&amp;A_zB_x(\\underline{\\hat{j}})+A_zB_y(-\\underline{\\hat{i}})+A_zB_z(0)\\end{aligned}[\/latex]\r\n\r\nEliminating the zero terms and grouping the terms with i together, the terms with j together, and the terms with k together yields:\r\n\r\n[latex]\\begin{aligned}\\vec A\\times\\vec B=&amp;A_yB_z(\\underline{\\hat{i}})+A_zB_y(-\\underline{\\hat{i}})+\\\\&amp;A_zB_x(\\underline{\\hat{j}})+A_xB_z(-\\underline{\\hat{j}})+\\\\&amp;A_xB_y(\\underline{\\hat{k}})+A_yB_x(-\\underline{\\hat{k}})\\end{aligned}[\/latex]\r\n\r\nFactoring out the unit vectors yields:\r\n\r\n[latex]\\begin{aligned}\\vec A\\times\\vec B=&amp;(A_yB_z-A_zB_y)\\underline{\\hat{i}}+\\\\&amp;(A_zB_x-A_xB_z)\\underline{\\hat{j}}+\\\\&amp;(A_xB_y-A_yB_x)\\underline{\\hat{k}}\\end{aligned}[\/latex]\r\n\r\nwhich can be written on one line as:\r\n\r\n[latex]\\vec A\\times\\vec B=(A_yB_z-A_zB_y)\\underline{\\hat{i}}+(A_zB_x-A_xB_z)\\underline{\\hat{j}}+(A_xB_y-A_yB_x)\\underline{\\hat{k}}[\/latex]\r\n\r\nThis is our end result. We can arrive at this result much more quickly if we borrow a tool from that branch of mathematics known as linear algebra (the mathematics of matrices).\r\n\r\nWe form the 3\u00d73 matrix:\r\n<p style=\"text-align: center\">[latex]\\begin{bmatrix} \\underline{\\hat{i}} &amp; \\underline{\\hat{j}} &amp; \\underline{\\hat{k}} \\\\ A_x &amp; A_y &amp; A_z \\\\ B_x &amp; B_y &amp; B_z \\end{bmatrix}[\/latex]<\/p>\r\nby writing i, j, k as the first row, then the components of the first vector that appear in the cross product as the second row, and finally the components of the second vector that appear in the cross product as the last row. It turns out that the cross product is equal to the determinant of that matrix. We use absolute value signs on the entire matrix to signify \u201cthe determinant of the matrix.\u201d So we have:\r\n<p style=\"text-align: center\">[latex]\\vec A\\times\\vec B=\\begin{bmatrix} \\underline{\\hat{i}} &amp; \\underline{\\hat{j}} &amp; \\underline{\\hat{k}} \\\\ A_x &amp; A_y &amp; A_z \\\\ B_x &amp; B_y &amp; B_z \\end{bmatrix}[\/latex]<\/p>\r\nTo take the determinant of a 3\u00d73 matrix, you work your way across the top row. For each element in that row, you take the product of the elements along the diagonal that extends down and to the right, minus the product of the elements down and to the left, and you add the three results (one result for each element in the top row) together. If there are no elements down and to the appropriate side, you move over to the other side of the matrix (see below) to complete the diagonal.\r\n<p style=\"text-align: center\">For the first element of the first row, the i, take the product down and to the right,<\/p>\r\n<p style=\"text-align: center\"><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0597.jpg\" alt=\"Diagram showing vectors i, j, k above columns labeled Ax, Ay, Az, and Bx, By, Bz, with two red arrows pointing from Ax to Ay and from Ay to Bz.\" class=\"aligncenter wp-image-422\" width=\"139\" height=\"132\" \/><\/p>\r\n<p style=\"text-align: center\">( this yields iA<sub>y<\/sub>B<sub>z<\/sub>) minus the product down and to the left<\/p>\r\n<p style=\"text-align: center\"><img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0598.jpg\" alt=\"Three columns with unit vectors \u00ee, \u0135, and k\u0302 on top. Below each vector, arrows point to Ax, By, and Az in the first, second, and third columns, respectively. Bx, By, and Bz are below them.\" class=\"aligncenter wp-image-423\" width=\"163\" height=\"135\" \/><\/p>\r\n<p style=\"text-align: center\">( the product down-and-to-the-left is iA<sub>z<\/sub>B<sub>y<\/sub>).<\/p>\r\nFor the first element in the first row, we thus have: [latex]\\underline{\\hat{i}}[\/latex]A<sub>y<\/sub>B<sub>z<\/sub> \u2212 [latex]\\underline{\\hat{i}}[\/latex]A<sub>z<\/sub>B<sub>y<\/sub> which can be written as: (A<sub>y<\/sub>B<sub>z<\/sub> \u2212 A<sub>z<\/sub>B<sub>y<\/sub>)[latex]\\underline{\\hat{i}}[\/latex]. Repeating the process for the second and third elements in the first row (the j and the k) we get (A<sub>z<\/sub>B<sub>x<\/sub> \u2212 A<sub>x<\/sub>B<sub>z<\/sub>)[latex]\\underline{\\hat{j}}[\/latex] and (A<sub>x<\/sub>B<sub>y<\/sub> \u2212 A<code>y<\/code>B<sub>x<\/sub>)[latex]\\underline{\\hat{k}}[\/latex] respectively. Adding the three results to form the determinant of the matrix results in:\r\n<p style=\"text-align: center\">[latex]\\vec A\\times\\vec B=(A_yB_z-A_zB_y)\\underline{\\hat{i}}+(A_zB_x-A_xB_z)\\underline{\\hat{j}}+(A_xB_y-A_yB_x)\\underline{\\hat{k}}[\/latex]<\/p>\r\nAs we found before, \u201cthe hard way.\u201d Below, the diagram shows the direction of each part of the cross product:\r\n\r\n<img src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/CNX_UPhysics_02_04_uniprod.jpg\" alt=\"Four diagrams of coordinate axes in 3D boxes illustrate the right-hand rule for cross products, labeling directions for x (\u00ee), y (\u0135), and z (k\u0302) with curved arrows and vector equations.\" class=\"aligncenter wp-image-550\" width=\"471\" height=\"447\" \/>\r\n\r\nSource: Calculus Based Physics, <span>Jeffrey W. Schnick,<\/span> <a href=\"https:\/\/openlibrary.ecampusontario.ca\/catalogue\/item\/?id=ce74a181-ccde-491c-848d-05489ed182e7\">https:\/\/openlibrary.ecampusontario.ca\/catalogue\/item\/?id=ce74a181-ccde-491c-848d-05489ed182e7<\/a> page 136\u2013141\r\n\r\n<\/div>\r\n<div class=\"textbox textbox--key-takeaways\"><header class=\"textbox__header\">\r\n<p class=\"textbox__title\">Key Takeaways<\/p>\r\n\r\n<\/header>\r\n<div class=\"textbox__content\">\r\n<div class=\"textbox__content\">\r\n\r\n<em>Basically: <\/em>Cross product is a method to find a vector that is a product of two vectors, perpendicular to the plane created from the two vectors.\r\n\r\n<i>Application: While bicycling, the force of my foot pushing on the pedal produces the most rotation when the force is 90 degrees (perpendicular) from the surface of the pedal.<\/i> The pedals themselves rotate, so I can change the angle I am pushing with, otherwise, when the pedal is at the bottom, it would be very difficult to produce forward motion.\r\n\r\n<em>Looking ahead: <\/em>In the next section, 1.6 on Torque, we will use the cross product to find the moment produced by forces.\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>","rendered":"<p>This is a second way to calculate the product of two vectors. It creates a third vector that is perpendicular to the plane made from the two vectors, as shown in the figure <a class=\"footnote\" title=\"https:\/\/commons.wikimedia.org\/wiki\/File:Cross-product-with-area.svg\" id=\"return-footnote-281-1\" href=\"#footnote-281-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a>.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/upload.wikimedia.org\/wikipedia\/commons\/5\/5f\/Cross-product-with-area.svg\" alt=\"File:Cross-product-with-area.svg - Wikimedia Commons\" class=\"n3VNCb alignright\" data-noaft=\"1\" width=\"289\" height=\"251\" \/>The black arrow is perpendicular to the grey plane made from blue and red vectors. This is how you will find the amount of torque created from a force, which we will do many times. Also, unlike the dot product, <span>a<\/span> x <span>b<\/span>\u00a0 is a different direction than <span>b<\/span> x <span>a<\/span>.<\/p>\n<p>[latex]\\vec A\\times\\vec B=\\begin{bmatrix} \\underline{\\hat{i}} & \\underline{\\hat{j}} & \\underline{\\hat{k}} \\\\ A_x & A_y & A_z \\\\ B_x & B_y & B_z \\end{bmatrix}[\/latex]<\/p>\n<p>&nbsp;<\/p>\n<p>[latex]\\vec {\\textbf{A}}\\times\\vec{\\textbf{B}}=(A_yB_z-A_zB_y)\\underline{\\hat{\\textbf{i}}}+(A_zB_x-A_xB_z)\\underline{\\hat{\\textbf{j}}}+(A_xB_y-A_yB_x)\\underline{\\hat{\\textbf{k}}}[\/latex]<\/p>\n<p style=\"text-align: center\">[latex]|\\vec {\\textbf{A}}\\times\\vec {\\textbf{B}}|=|\\vec A||\\vec B|\\sin\\theta[\/latex]<\/p>\n<div class=\"textbox\">\n<p><span>The <\/span><strong>vector product<\/strong><span> of two vectors [latex]\\vec A \\text{\u00a0 and\u00a0 }\\vec B[\/latex]<\/span><span>\u00a0<\/span><span>is denoted by [latex]\\vec A \\text{ x }\\vec B[\/latex]<\/span><span>\u00a0and is often referred to as a\u00a0<\/span><strong>cross product<\/strong><span>. The vector product is a vector that has its direction perpendicular to both vectors [latex]\\vec A \\text{\u00a0 and\u00a0 }\\vec B[\/latex]<\/span><span>. In other words, vector [latex]\\vec A \\text{ x }\\vec B[\/latex]<\/span><span> is perpendicular to the plane that contains vectors [latex]\\vec A \\text{\u00a0 and\u00a0 }\\vec B[\/latex]<\/span><span>.<\/span><span><\/span><\/p>\n<p>Source: University Physics Volume 1, OpenStax CNX, <a href=\"https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/chapter\/2-4-products-of-vectors\/\">https:\/\/courses.lumenlearning.com\/suny-osuniversityphysics\/chapter\/2-4-products-of-vectors\/<\/a><\/p>\n<\/div>\n<div class=\"textbox\">\n<p>Unit vectors allow for a straightforward calculation of the cross product of two vectors under even the most general circumstances, e.g. circumstances in which each of the vectors is pointing in an arbitrary direction in a three-dimensional space. To take advantage of the method, we need to know the cross product of the Cartesian coordinate axis unit vectors i, j, and k with each other. First off, we should note that any vector crossed with itself gives zero. This is evident from the\u00a0 equation:<\/p>\n<p style=\"text-align: center\">[latex]|\\vec A\\times\\vec B|=|\\vec A||\\vec B|\\sin\\theta[\/latex].<\/p>\n<p>because if A and B are in the same direction, then \u03b8 = 0\u00b0, and since sin 0\u00b0 = 0, we have [latex]|\\vec A\\times\\vec B|=0[\/latex].\u00a0Regarding the unit vectors, this means that:<\/p>\n<p style=\"text-align: center\">[latex]\\underline{\\hat{i}}\\times\\underline{\\hat{i}}=0\\\\\\underline{\\hat{j}}\\times\\underline{\\hat{j}}=0\\\\\\underline{\\hat{k}}\\times\\underline{\\hat{k}}=0[\/latex]<\/p>\n<p>Next, we note that the magnitude of the cross product of two vectors that are perpendicular to each other is just the ordinary product of the magnitudes of the vectors. This is also evident from the equation:<\/p>\n<p style=\"text-align: center\">[latex]|\\vec {A}\\times\\vec{B}|=|\\vec A||\\vec B|\\sin\\theta[\/latex]<\/p>\n<p>because if [latex]\\vec A[\/latex] is perpendicular to [latex]\\vec B[\/latex]\u00a0then \u03b8 = 90\u00b0 and sin90\u00b0 = 1 so:<\/p>\n<p style=\"text-align: center\">[latex]|\\vec A\\times\\vec B|=|\\vec A||\\vec B|[\/latex]<\/p>\n<p>Now, if A and B are unit vectors, then their magnitudes are both 1, so the product of their magnitudes is also 1. Furthermore, the unit vectors i, j, and k are all perpendicular to each other, so the magnitude of the cross product of any one of them with any other one of them is the product of the two magnitudes, that is, 1.<\/p>\n<p>Now, how about the direction? Let\u2019s use the right-hand rule to get the direction of i\u00d7j:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0584.jpg\" alt=\"A 3D coordinate system showing the x, y, and z axes with unit vectors labeled \u00ee (x-axis), \u0135 (y-axis), and k\u0302 (z-axis) pointing in their respective positive directions.\" class=\"aligncenter wp-image-410\" width=\"327\" height=\"320\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0584.jpg 1040w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0584-300x294.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0584-1024x1003.jpg 1024w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0584-768x752.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0584-65x64.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0584-225x220.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0584-350x343.jpg 350w\" sizes=\"auto, (max-width: 327px) 100vw, 327px\" \/><\/p>\n<p><em>[To find the direction, we use the right-hand rule, which we will cover more in section 3.1. Here is an overview<\/em>.] With the fingers of the right hand pointing directly away from the right elbow, and in the same direction as i, (the first vector in \u201c[latex]\\underline{\\hat{i}}\\times\\underline{\\hat{j}}[\/latex]\u201d) to make it so that if one were to close the fingers, they would point in the same direction as [latex]\\underline{\\hat{j}}[\/latex], the palm must be facing in the +y direction. That being the case, the extended thumb must be pointing in the +z direction. Putting the magnitude (the magnitude of each unit vector is 1) and direction (+z) information together, we see that:<\/p>\n<ul>\n<li>[latex]\\underline{\\hat{i}}\\times\\underline{\\hat{j}}=\\underline{\\hat{k}}[\/latex]<\/li>\n<li>[latex]\\underline{\\hat{j}}\\times\\underline{\\hat{k}}=\\underline{\\hat{i}}[\/latex]<\/li>\n<li>[latex]\\underline{\\hat{k}}\\times\\underline{\\hat{i}}=\\underline{\\hat{j}}[\/latex]<\/li>\n<li>[latex]\\underline{\\hat{j}}\\times\\underline{\\hat{i}}=-\\underline{\\hat{k}}[\/latex]<\/li>\n<li>[latex]\\underline{\\hat{k}}\\times\\underline{\\hat{j}}=-\\underline{\\hat{i}}[\/latex]<\/li>\n<li>[latex]\\underline{\\hat{i}}\\times\\underline{\\hat{k}}=-\\underline{\\hat{j}}[\/latex]<\/li>\n<\/ul>\n<p>One way of remembering this is to write [latex]\\underline{\\hat{i}},\\underline{\\hat{j}},\\underline{\\hat{k}}[\/latex] twice in succession:<\/p>\n<p style=\"text-align: center\">[latex]\\underline{\\hat{\\textbf{i}}},\\underline{\\hat{\\textbf{j}}},\\underline{\\hat{\\textbf{k}}},\\underline{\\hat{\\textbf{i}}},\\underline{\\hat{\\textbf{j}}},\\underline{\\hat{\\textbf{k}}}[\/latex]<\/p>\n<p>Then, crossing any one of the first three vectors into the vector immediately to its right yields the next vector to the right. But crossing any one of the last three vectors into the vector immediately to its left yields the negative of the next vector to the left (left-to-right \u201c+\u201c, but right-to-left \u201c\u2212\u201c).<\/p>\n<p>Now we\u2019re ready to look at the general case. Any vector\u00a0 [latex]\\vec A[\/latex] \u00a0can be expressed in terms of unit vectors:<\/p>\n<p>[latex]\\vec{\\textbf{A}}=A_x\\hat{\\underline{i}}+A_y\\hat{\\underline{j}}+A_z\\hat{\\underline{k}}[\/latex]<\/p>\n<p>Doing the same for a vector\u00a0 [latex]\\vec B[\/latex]\u00a0 <span><\/span> then allows us to write the cross product as:<\/p>\n<p>[latex]\\vec{\\textbf{A}}\\times\\vec{\\textbf{B}}=(A_x\\hat{\\underline{i}}+A_y\\hat{\\underline{j}}+A_z\\hat{\\underline{k}})\\times(B_x\\hat{\\underline{i}}+B_y\\hat{\\underline{j}}+B_z\\hat{\\underline{k}})[\/latex]<\/p>\n<p>Using the distributive rule for multiplication, we can write this as:<\/p>\n<p>[latex]\\begin{aligned}\\vec{\\textbf{A}}\\times\\vec{\\textbf{B}}=&A_x\\hat{\\underline{i}}\\times(B_x\\hat{\\underline{i}}+B_y\\hat{\\underline{j}}+B_z\\hat{\\underline{k}})+\\\\&A_y\\hat{\\underline{j}}\\times(B_x\\hat{\\underline{i}}+B_y\\hat{\\underline{j}}+B_z\\hat{\\underline{k}})+\\\\&A_z\\hat{\\underline{k}}\\times(B_x\\hat{\\underline{i}}+B_y\\hat{\\underline{j}}+B_z\\hat{\\underline{k}})\\end{aligned}[\/latex]<\/p>\n<p>[latex]\\begin{aligned}\\vec{\\textbf{A}}\\times\\vec{\\textbf{B}}=&A_x\\hat{\\underline{i}}\\times B_x\\hat{\\underline{i}}+A_x\\hat{\\underline{i}}\\times B_y\\hat{\\underline{j}}+A_x\\hat{\\underline{i}}\\times\u00a0 B_z\\hat{\\underline{k}}+\\\\&A_y\\hat{\\underline{j}}\\times B_x\\hat{\\underline{i}}+A_y\\hat{\\underline{j}}\\times B_y\\hat{\\underline{j}}+A_y\\hat{\\underline{j}}\\times B_z\\hat{\\underline{k}}+\\\\&A_z\\hat{\\underline{k}}\\times B_x\\hat{\\underline{i}}+A_z\\hat{\\underline{k}}\\times B_y\\hat{\\underline{j}}+A_z\\hat{\\underline{k}}\\times B_z\\hat{\\underline{k}}\\end{aligned}[\/latex]<\/p>\n<p>Using, in each term, the commutative rule and the associative rule for multiplication, we can write this as:<\/p>\n<p>[latex]\\begin{aligned}\\vec{\\textbf{A}}\\times\\vec{\\textbf{B}}=&A_xB_x(\\hat{\\underline{i}}\\times\\hat{\\underline{i}})+A_xB_y(\\hat{\\underline{i}}\\times\\hat{\\underline{j}})+A_xB_z(\\hat{\\underline{i}}\\times\\hat{\\underline{k}})+\\\\&A_yB_x(\\hat{\\underline{j}}\\times\\hat{\\underline{i}})+A_yB_y(\\hat{\\underline{j}}\\times\\hat{\\underline{j}})+A_yB_z(\\hat{\\underline{j}}\\times\\hat{\\underline{k}})+\\\\&A_zB_x(\\hat{\\underline{k}}\\times\\hat{\\underline{i}})+A_zB_y(\\hat{\\underline{k}}\\times\\hat{\\underline{j}})+A_zB_z(\\hat{\\underline{k}}\\times\\hat{\\underline{k}})\\end{aligned}[\/latex]<\/p>\n<p>Now we evaluate the cross product that appears in each term:<\/p>\n<p>[latex]\\begin{aligned}\\vec A\\times\\vec B=&A_xB_x(0)+A_xB_y(\\underline{\\hat{k}})+A_xB_z(-\\underline{\\hat{j}})+\\\\&A_yB_x(-\\underline{\\hat{k}})+A_yB_y(0)+A_yB_z(\\underline{\\hat{i}})+\\\\&A_zB_x(\\underline{\\hat{j}})+A_zB_y(-\\underline{\\hat{i}})+A_zB_z(0)\\end{aligned}[\/latex]<\/p>\n<p>Eliminating the zero terms and grouping the terms with i together, the terms with j together, and the terms with k together yields:<\/p>\n<p>[latex]\\begin{aligned}\\vec A\\times\\vec B=&A_yB_z(\\underline{\\hat{i}})+A_zB_y(-\\underline{\\hat{i}})+\\\\&A_zB_x(\\underline{\\hat{j}})+A_xB_z(-\\underline{\\hat{j}})+\\\\&A_xB_y(\\underline{\\hat{k}})+A_yB_x(-\\underline{\\hat{k}})\\end{aligned}[\/latex]<\/p>\n<p>Factoring out the unit vectors yields:<\/p>\n<p>[latex]\\begin{aligned}\\vec A\\times\\vec B=&(A_yB_z-A_zB_y)\\underline{\\hat{i}}+\\\\&(A_zB_x-A_xB_z)\\underline{\\hat{j}}+\\\\&(A_xB_y-A_yB_x)\\underline{\\hat{k}}\\end{aligned}[\/latex]<\/p>\n<p>which can be written on one line as:<\/p>\n<p>[latex]\\vec A\\times\\vec B=(A_yB_z-A_zB_y)\\underline{\\hat{i}}+(A_zB_x-A_xB_z)\\underline{\\hat{j}}+(A_xB_y-A_yB_x)\\underline{\\hat{k}}[\/latex]<\/p>\n<p>This is our end result. We can arrive at this result much more quickly if we borrow a tool from that branch of mathematics known as linear algebra (the mathematics of matrices).<\/p>\n<p>We form the 3\u00d73 matrix:<\/p>\n<p style=\"text-align: center\">[latex]\\begin{bmatrix} \\underline{\\hat{i}} & \\underline{\\hat{j}} & \\underline{\\hat{k}} \\\\ A_x & A_y & A_z \\\\ B_x & B_y & B_z \\end{bmatrix}[\/latex]<\/p>\n<p>by writing i, j, k as the first row, then the components of the first vector that appear in the cross product as the second row, and finally the components of the second vector that appear in the cross product as the last row. It turns out that the cross product is equal to the determinant of that matrix. We use absolute value signs on the entire matrix to signify \u201cthe determinant of the matrix.\u201d So we have:<\/p>\n<p style=\"text-align: center\">[latex]\\vec A\\times\\vec B=\\begin{bmatrix} \\underline{\\hat{i}} & \\underline{\\hat{j}} & \\underline{\\hat{k}} \\\\ A_x & A_y & A_z \\\\ B_x & B_y & B_z \\end{bmatrix}[\/latex]<\/p>\n<p>To take the determinant of a 3\u00d73 matrix, you work your way across the top row. For each element in that row, you take the product of the elements along the diagonal that extends down and to the right, minus the product of the elements down and to the left, and you add the three results (one result for each element in the top row) together. If there are no elements down and to the appropriate side, you move over to the other side of the matrix (see below) to complete the diagonal.<\/p>\n<p style=\"text-align: center\">For the first element of the first row, the i, take the product down and to the right,<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0597.jpg\" alt=\"Diagram showing vectors i, j, k above columns labeled Ax, Ay, Az, and Bx, By, Bz, with two red arrows pointing from Ax to Ay and from Ay to Bz.\" class=\"aligncenter wp-image-422\" width=\"139\" height=\"132\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0597.jpg 942w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0597-300x285.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0597-768x730.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0597-65x62.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0597-225x214.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0597-350x333.jpg 350w\" sizes=\"auto, (max-width: 139px) 100vw, 139px\" \/><\/p>\n<p style=\"text-align: center\">( this yields iA<sub>y<\/sub>B<sub>z<\/sub>) minus the product down and to the left<\/p>\n<p style=\"text-align: center\"><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0598.jpg\" alt=\"Three columns with unit vectors \u00ee, \u0135, and k\u0302 on top. Below each vector, arrows point to Ax, By, and Az in the first, second, and third columns, respectively. Bx, By, and Bz are below them.\" class=\"aligncenter wp-image-423\" width=\"163\" height=\"135\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0598.jpg 1013w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0598-300x249.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0598-768x638.jpg 768w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0598-65x54.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0598-225x187.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/IMG-0598-350x291.jpg 350w\" sizes=\"auto, (max-width: 163px) 100vw, 163px\" \/><\/p>\n<p style=\"text-align: center\">( the product down-and-to-the-left is iA<sub>z<\/sub>B<sub>y<\/sub>).<\/p>\n<p>For the first element in the first row, we thus have: [latex]\\underline{\\hat{i}}[\/latex]A<sub>y<\/sub>B<sub>z<\/sub> \u2212 [latex]\\underline{\\hat{i}}[\/latex]A<sub>z<\/sub>B<sub>y<\/sub> which can be written as: (A<sub>y<\/sub>B<sub>z<\/sub> \u2212 A<sub>z<\/sub>B<sub>y<\/sub>)[latex]\\underline{\\hat{i}}[\/latex]. Repeating the process for the second and third elements in the first row (the j and the k) we get (A<sub>z<\/sub>B<sub>x<\/sub> \u2212 A<sub>x<\/sub>B<sub>z<\/sub>)[latex]\\underline{\\hat{j}}[\/latex] and (A<sub>x<\/sub>B<sub>y<\/sub> \u2212 A<code>y<\/code>B<sub>x<\/sub>)[latex]\\underline{\\hat{k}}[\/latex] respectively. Adding the three results to form the determinant of the matrix results in:<\/p>\n<p style=\"text-align: center\">[latex]\\vec A\\times\\vec B=(A_yB_z-A_zB_y)\\underline{\\hat{i}}+(A_zB_x-A_xB_z)\\underline{\\hat{j}}+(A_xB_y-A_yB_x)\\underline{\\hat{k}}[\/latex]<\/p>\n<p>As we found before, \u201cthe hard way.\u201d Below, the diagram shows the direction of each part of the cross product:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/CNX_UPhysics_02_04_uniprod.jpg\" alt=\"Four diagrams of coordinate axes in 3D boxes illustrate the right-hand rule for cross products, labeling directions for x (\u00ee), y (\u0135), and z (k\u0302) with curved arrows and vector equations.\" class=\"aligncenter wp-image-550\" width=\"471\" height=\"447\" srcset=\"https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/CNX_UPhysics_02_04_uniprod.jpg 596w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/CNX_UPhysics_02_04_uniprod-300x285.jpg 300w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/CNX_UPhysics_02_04_uniprod-65x62.jpg 65w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/CNX_UPhysics_02_04_uniprod-225x214.jpg 225w, https:\/\/pressbooks.library.upei.ca\/statics\/wp-content\/uploads\/sites\/56\/2021\/04\/CNX_UPhysics_02_04_uniprod-350x332.jpg 350w\" sizes=\"auto, (max-width: 471px) 100vw, 471px\" \/><\/p>\n<p>Source: Calculus Based Physics, <span>Jeffrey W. Schnick,<\/span> <a href=\"https:\/\/openlibrary.ecampusontario.ca\/catalogue\/item\/?id=ce74a181-ccde-491c-848d-05489ed182e7\">https:\/\/openlibrary.ecampusontario.ca\/catalogue\/item\/?id=ce74a181-ccde-491c-848d-05489ed182e7<\/a> page 136\u2013141<\/p>\n<\/div>\n<div class=\"textbox textbox--key-takeaways\">\n<header class=\"textbox__header\">\n<p class=\"textbox__title\">Key Takeaways<\/p>\n<\/header>\n<div class=\"textbox__content\">\n<div class=\"textbox__content\">\n<p><em>Basically: <\/em>Cross product is a method to find a vector that is a product of two vectors, perpendicular to the plane created from the two vectors.<\/p>\n<p><i>Application: While bicycling, the force of my foot pushing on the pedal produces the most rotation when the force is 90 degrees (perpendicular) from the surface of the pedal.<\/i> The pedals themselves rotate, so I can change the angle I am pushing with, otherwise, when the pedal is at the bottom, it would be very difficult to produce forward motion.<\/p>\n<p><em>Looking ahead: <\/em>In the next section, 1.6 on Torque, we will use the cross product to find the moment produced by forces.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-281-1\"><a href=\"https:\/\/commons.wikimedia.org\/wiki\/File:Cross-product-with-area.svg\">https:\/\/commons.wikimedia.org\/wiki\/File:Cross-product-with-area.svg<\/a>  <a href=\"#return-footnote-281-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":60,"menu_order":5,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-281","chapter","type-chapter","status-publish","hentry"],"part":3,"_links":{"self":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/281","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/users\/60"}],"version-history":[{"count":38,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/281\/revisions"}],"predecessor-version":[{"id":2841,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/281\/revisions\/2841"}],"part":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/parts\/3"}],"metadata":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapters\/281\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/media?parent=281"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/pressbooks\/v2\/chapter-type?post=281"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/contributor?post=281"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.library.upei.ca\/statics\/wp-json\/wp\/v2\/license?post=281"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}