{"id":5451,"date":"2022-06-02T18:26:50","date_gmt":"2022-06-02T18:26:50","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=5451"},"modified":"2022-10-21T00:16:44","modified_gmt":"2022-10-21T00:16:44","slug":"the-cross-product","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/the-cross-product\/","title":{"raw":"The Cross Product","rendered":"The Cross Product"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\n<ul class=\"os-abstract\">\r\n \t<li><span class=\"os-abstract-content\">Calculate the cross product of two given vectors.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Use determinants to calculate a cross product.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<section id=\"fs-id1163724083973\" data-depth=\"1\">\r\n<h2 data-type=\"title\">The Cross Product and Its Properties<\/h2>\r\n<p id=\"fs-id1163724087473\">The dot product is a multiplication of two vectors that results in a scalar. In this section, we introduce a product of two vectors that generates a third vector orthogonal to the first two. Consider how we might find such a vector. Let [latex]\\mathbf{u} =\\langle u_1, u_2, u_3 \\rangle[\/latex] and [latex]\\mathbf{v} =\\langle v_1, v_2, v_3 \\rangle[\/latex] be nonzero vectors. We want to find a vector [latex]\\mathbf{w} =\\langle w_1, w_2, w_3 \\rangle[\/latex] orthogonal to both [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex]\u2014that is, we want to find [latex]\\mathbf{w}[\/latex] such that [latex]\\mathbf{u} \\cdot \\mathbf{w} = 0[\/latex] and [latex]\\mathbf{v} \\cdot \\mathbf{w} =0[\/latex]. Therefore, [latex]w_1[\/latex], [latex]w_2[\/latex], and [latex]w_3[\/latex] must satisfy<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{aligned}\r\nu_1w_1 + u_2w_2 + u_3w_3 &amp;= 0 \\\\\r\nv_1w_1 + v_2w_2 + v_3w_3 &amp;= 0.\\\\\r\n\\end{aligned}\r\n[\/latex]<\/center>If we multiply the top equation by [latex]v_3[\/latex] and the bottom equation by [latex]u_3[\/latex] and subtract, we can eliminate the variable [latex]w_3[\/latex], which gives\r\n\r\n<center>[latex](u_1v_3 - v_1u_3)w_1 + (u_2v_3 - v_2u_3)w_2 = 0[\/latex].<\/center>\r\n<p id=\"fs-id1163723884318\">If we select<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\nw_1 &amp;= u_2v_3 - v_2u_3\\\\\r\nw_2 &amp;= -(u_1v_3 - u_3v_1),\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>\r\n<p id=\"fs-id1163723528456\">we get a possible solution vector. Substituting these values back into the original equations gives<\/p>\r\n\r\n<center>[latex]w_3 = u_1v_2 - u_2v_1[\/latex].<\/center>\r\n<p id=\"fs-id1163723670069\">That is, vector<\/p>\r\n\r\n<center>[latex]\\mathbf{w} = \\langle u_2v_3 - u_3v_2, -(u_1v_3 - u_3v_1), u_1v_2 - u_2v_1 \\rangle[\/latex]<\/center>\r\n<p id=\"fs-id1163724043762\">is orthogonal to both [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex], which leads us to define the following operation, called the cross product.<\/p>\r\n\r\n<\/section>\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\" style=\"text-align: center;\" data-type=\"title\"><span id=\"2\" class=\"os-title-label\" data-type=\"\">DEFINITION<\/span><\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-id1163723775830\">Let [latex]\\mathbf{u} =\\langle u_1, u_2, u_3 \\rangle[\/latex] and [latex]\\mathbf{v} =\\langle v_1, v_2, v_3 \\rangle[\/latex]. Then, the\u00a0<strong>cross product<\/strong> [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex]\u00a0is vector<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{u} \\times \\mathbf{v} &amp;= (u_2v_3 - u_3v_2)\\mathbf{i} -(u_1v_3 - u_3v_1)\\mathbf{j}+ (u_1v_2 - u_2v_1)\\mathbf{k}\\\\\r\n&amp;= \\langle u_2v_3 - u_3v_2, -(u_1v_3 - u_3v_1), u_1v_2 - u_2v_1 \\rangle.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/div>\r\n<\/section><\/div>\r\n<p id=\"fs-id1163724072465\">From the way we have developed [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex], it should be clear that the cross product is orthogonal to both [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex]. However, it never hurts to check. To show that [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] is orthogonal to [latex]\\textbf u[\/latex], we calculate the dot product of [latex]\\textbf u[\/latex] and [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex].<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{u} \\cdot (\\mathbf{u} \\times \\mathbf{v}) &amp;= \\langle u_1,u_2,u_3 \\rangle \\cdot \\langle u_2v_3 - u_3v_2, -u_1v_3 - u_3v_1, u_1v_2 - u_2v_1 \\rangle\\\\\r\n&amp;= u_1(u_2v_3 - u_3v_2) + u_2 (-u_1v_3 + u_3v_1) +u_3( u_1v_2 - u_2v_1)\\\\\r\n&amp;= u_1u_2v_3 - u_1u_3v_2 -u_1u_2v_3+u_2u_3v_1 +u_1u_3v_2-u_2u_3v_1\\\\\r\n&amp;= (u_1u_2v_3 - u_1u_2v_3) +(-u_1u_3v_2+u_1u_3v_2) + (u_2u_3v_1-u_2u_3v_1)\\\\\r\n&amp;= 0\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>In a similar manner, we can show that the cross product is also orthogonal to [latex]\\textbf v[\/latex].\r\n<div id=\"fs-id1163724060614\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\r\n<div id=\"fs-id1163723497714\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding a cross product<\/h3>\r\nLet [latex]\\mathbf{p} =\\langle -1,2,5 \\rangle[\/latex] and [latex]\\mathbf{q} =\\langle 4,0,-3 \\rangle[\/latex] (Figure 1). Find [latex]\\mathbf{p} \\times \\mathbf{q}[\/latex].\r\n\r\n[caption id=\"attachment_5136\" align=\"aligncenter\" width=\"501\"]<img class=\"size-full wp-image-5136\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27155235\/Figure-2.53.jpg\" alt=\"This figure is the 3-dimensional coordinate system. It has two vectors in standard position. The first vector is labeled \u201cp = &lt;-1, 2, 5&gt;.\u201d The second vector is labeled \u201cq = &lt;4, 0, -3&gt;.\u201d\" width=\"501\" height=\"497\" \/> Figure 1. Finding a cross product to two given vectors.[\/caption]\r\n\r\n[reveal-answer q=\"934723610\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"934723610\"]\r\n<p id=\"fs-id1163723747703\">Substitute the components of the vectors into\u00a0the Cross Product Equation:<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{p} \\times \\mathbf{q} &amp;= \\langle -1,2,5 \\rangle \\times \\langle 4,0,-3 \\rangle\\\\\r\n&amp;= \\langle p_2q_3-p_3q_2, p_3q_1 - p_1q_3,p_1q_2 -p_2q_1 \\rangle \\\\\r\n&amp;= \\langle 2(-3)-5(0), -(-1)(-3)+5(4), (-1)(0)-2(4) \\rangle\\\\\r\n&amp;= \\langle -6,17,-8 \\rangle.\\\\\\end{align*}\r\n[\/latex]<\/center>[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind [latex]\\mathbf{p} \\times \\mathbf{q}[\/latex] for [latex]\\mathbf{p} =\\langle 5,1,2 \\rangle[\/latex] and [latex]\\mathbf{q} =\\langle -2,0,1 \\rangle[\/latex]. Express the answer using standard unit vectors.\r\n\r\n[reveal-answer q=\"125237156\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"125237156\"]\r\n\r\n[latex]\\mathbf{i} -9\\mathbf{j}+2\\mathbf{k}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try IT.\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7753558&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=fEMzUqQTmoo&amp;video_target=tpm-plugin-aghj6e1q-fEMzUqQTmoo\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.30_transcript.html\">\u201cCP 2.30\u201d here (opens in new window).<\/a><\/span><\/center>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]5986[\/ohm_question]\r\n\r\n<\/div>\r\nAlthough it may not be obvious from\u00a0the Cross Product Equation, the direction of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] is given by the right-hand rule. If we hold the right hand out with the fingers pointing in the direction of [latex]\\textbf u[\/latex], then curl the fingers toward vector [latex]\\textbf v[\/latex], the thumb points in the direction of the cross product, as shown.\r\n\r\n[caption id=\"attachment_5138\" align=\"aligncenter\" width=\"666\"]<img class=\"size-full wp-image-5138\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27155328\/Figure-2.54.jpg\" alt=\"This figure has two images. The first image has three vectors with the same initial point. Two of the vectors are labeled \u201cu\u201d and \u201cv.\u201d The angle between u and v is theta. The third vector is perpendicular to u and v. It is labeled \u201cu cross v.\u201d The second image has three vectors. The vectors are labeled \u201cu, v, and u cross v.\u201d \u201cu cross v\u201d is perpendicular to u and v. Also, on the image of these three vectors is a right hand. The fingers are in the direction of u. As the hand is closing, the direction of the closing fingers is the direction of v. The thumb is up and in the direction of \u201cu cross v.\u201d\" width=\"666\" height=\"416\" \/> Figure 2. The direction of [latex]{\\bf{u}}\\times{\\bf{v}}[\/latex] is determined by the right-hand rule.[\/caption]Notice what this means for the direction of [latex]\\mathbf{v} \\times \\mathbf{u}[\/latex]. If we apply the right-hand rule to [latex]\\mathbf{v} \\times \\mathbf{u}[\/latex], we start with our fingers pointed in the direction of [latex]\\textbf v[\/latex], then curl our fingers toward the vector [latex]\\textbf u[\/latex]. In this case, the thumb points in the opposite direction of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex]. (Try it!)\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: anticommutativity of the cross product<\/h3>\r\nLet [latex]\\mathbf{u} =\\langle 0,2,1 \\rangle[\/latex] and [latex]\\mathbf{v} =\\langle 3,-1,0 \\rangle[\/latex]. Calculate [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] and [latex]\\mathbf{v} \\times \\mathbf{u}[\/latex] and graph them.\r\n\r\n[caption id=\"attachment_5141\" align=\"aligncenter\" width=\"524\"]<img class=\"size-full wp-image-5141\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27155653\/Figure-2.55.jpg\" alt=\"This figure is the 3-dimensional coordinate system. It has two vectors in standard position. The first vector is labeled \u201cu = &lt;0, 2, 1&gt;.\u201d The second vector is labeled \u201cv = &lt;3, -1, 0&gt;.\u201d\" width=\"524\" height=\"497\" \/> Figure 3. Are the cross products [latex]{\\bf{u}}\\times{\\bf{v}}[\/latex] and [latex]{\\bf{v}}\\times{\\bf{u}}[\/latex] in the same direction?[\/caption][reveal-answer q=\"856612738\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"856612738\"]\r\n<p id=\"fs-id1163723474704\">We have<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{u} \\times \\mathbf{v} &amp;= \\langle (0+1),-(0-3),(0-6) \\rangle = \\langle 1,3,-6\\rangle\\\\\r\n\\mathbf{v} \\times \\mathbf{u}&amp;= \\langle (-1-0),-(3-0),(6-0) \\rangle = \\langle -1,-3,6\\rangle.\\\\\\end{align*}\r\n[\/latex]<\/center>\r\n<p id=\"fs-id1163724049208\">We see that, in this case, [latex]\\mathbf{u} \\times \\mathbf{v} = -(\\mathbf{v} \\times \\mathbf{u})[\/latex] (<a class=\"autogenerated-content\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/2-4-the-cross-product#CNX_Calc_Figure_12_04_013\">Figure 2.56<\/a>). We prove this in general later in this section.<\/p>\r\n\r\n[caption id=\"attachment_5142\" align=\"aligncenter\" width=\"501\"]<img class=\"size-full wp-image-5142\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27155808\/Figure-2.56.jpg\" alt=\"\" width=\"501\" height=\"672\" \/> Figure 4. The cross products [latex]{\\bf{u}}\\times{\\bf{v}}[\/latex] and [latex]{\\bf{v}}\\times{\\bf{u}}[\/latex] are both orthogonal to [latex]{\\bf{u}}[\/latex] and [latex]{\\bf{v}}[\/latex], but in opposite directions.[\/caption][\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nSuppose vectors [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] lie in the [latex]xy[\/latex]-plane (the [latex]z[\/latex]-component of each vector is zero). Now suppose the [latex]x[\/latex]- and [latex]y[\/latex]-components of [latex]\\textbf u[\/latex] and the [latex]y[\/latex]-component of [latex]\\textbf v[\/latex] are all positive, whereas the [latex]x[\/latex]-component of [latex]\\textbf v[\/latex] is negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] point?\r\n\r\n[reveal-answer q=\"373093789\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"373093789\"]\r\n\r\nUp (the positive [latex]z[\/latex]-direction)\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1163724097644\">The cross products of the standard unit vectors [latex]\\textbf i[\/latex], [latex]\\textbf j[\/latex], and [latex]\\textbf k[\/latex] can be useful for simplifying some calculations, so let\u2019s consider these cross products. A straightforward application of the definition shows that<\/p>\r\n\r\n<center>[latex]\\mathbf{i} \\times \\mathbf{i} = \\mathbf{j} \\times \\mathbf{j}= \\mathbf{k} \\times \\mathbf{k} = 0[\/latex].<\/center>\r\n<p id=\"fs-id1163723478160\">(The cross product of two vectors is a vector, so each of these products results in the zero vector, not the scalar 0.) It\u2019s up to you to verify the calculations on your own.<\/p>\r\n<p id=\"fs-id1163723096820\">Furthermore, because the cross product of two vectors is orthogonal to each of these vectors, we know that the cross product of [latex]\\textbf i[\/latex] and [latex]\\textbf j[\/latex] is parallel to [latex]\\textbf k[\/latex]. Similarly, the vector product of [latex]\\textbf i[\/latex] and [latex]\\textbf k[\/latex] is parallel to [latex]\\textbf j[\/latex], and the vector product of [latex]\\textbf j[\/latex] and [latex]\\textbf k[\/latex] is parallel to [latex]\\textbf i[\/latex]. We can use the right-hand rule to determine the direction of each product. Then we have<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{aligned}\r\n\\mathbf{i} \\times \\mathbf{j} &amp;= \\mathbf{k} &amp; \\mathbf{j} \\times \\mathbf{i} &amp;= -\\mathbf{k}\\\\\r\n\\mathbf{j} \\times \\mathbf{k} &amp;= \\mathbf{i} &amp; \\mathbf{k} \\times \\mathbf{j} &amp;= -\\mathbf{i} \\\\\r\n\\mathbf{k} \\times \\mathbf{i} &amp;= \\mathbf{j} &amp; \\mathbf{i} \\times \\mathbf{k} &amp;= -\\mathbf{j}.\\\\\\end{aligned}\r\n[\/latex]<\/center>\r\n<p id=\"fs-id1163723472405\">These formulas come in handy later.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: cross product of standard unit vectors<\/h3>\r\nFind [latex]\\mathbf{i} \\times (\\mathbf{j} \\times \\mathbf{k})[\/latex].\r\n\r\n[reveal-answer q=\"734661173\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"734661173\"]\r\n\r\nWe know that [latex]\\mathbf{j} \\times \\mathbf{k} = \\mathbf{i}[\/latex]. Therefore, [latex]\\mathbf{i} \\times (\\mathbf{j} \\times \\mathbf{k}) = \\mathbf{i} \\times \\mathbf{i} =0[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind [latex](\\mathbf{i} \\times \\mathbf{j}) \\times (\\mathbf{k} \\times \\mathbf{i})[\/latex].\r\n\r\n[reveal-answer q=\"092380347\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"092380347\"]\r\n\r\n[latex]-\\mathbf{i}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nAs we have seen, the dot product is often called the\u00a0<em data-effect=\"italics\">scalar product<\/em>\u00a0because it results in a scalar. The cross product results in a vector, so it is sometimes called the\u00a0<strong><span id=\"d53b2426-3a76-4d95-af4f-6532ad8e4e9a_term82\" data-type=\"term\">vector product<\/span><\/strong>. These operations are both versions of vector multiplication, but they have very different properties and applications. Let\u2019s explore some properties of the cross product. We prove only a few of them. Proofs of the other properties are left as exercises.\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: properties of the cross product<\/span><\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-id1163723277869\">Let [latex]\\textbf u[\/latex], [latex]\\textbf v[\/latex], and [latex]\\textbf w[\/latex] be vectors in space, and let [latex]c[\/latex] be a scalar.<\/p>\r\n\\[ \\begin{array}{lrl}\r\n\\mbox{i.} &amp; \\mathbf{u \\times v} &amp;= -( \\mathbf{v \\times u}) &amp; \\mbox{Anticommutative property} \\\\\r\n\\mbox{ii.} &amp; \\mathbf{u} \\times (\\mathbf{v} + \\textbf{w}) &amp;= \\mathbf{u \\times v} + \\mathbf{u \\times w} &amp; \\mbox{Distributive property} \\\\\r\n\\mbox{iii.} &amp; c(\\mathbf{u \\times v}) &amp;= (c\\mathbf{u}) \\times \\mathbf{v} = \\mathbf{u} \\times (c\\mathbf{v}) &amp; \\mbox{ Multiplication by a constant} \\\\\r\n\\mbox{iv.} &amp; \\mathbf{u} \\times \\mathbf{0} &amp;= \\mathbf{0} \\times \\mathbf{u} = \\mathbf{0} &amp; \\mbox{ Cross product of the zero vector} \\\\\r\n\\mbox{iv.} &amp; \\mathbf{v} \\times \\mathbf{v} &amp;= \\mathbf{0} &amp; \\mbox{ \u00a0Cross product of a vector with itself} \\\\\r\n\\mbox{iv.} &amp; \\mathbf{u} \\times (\\mathbf{v} \\times \\textbf{w}) &amp;= (\\mathbf{u} \\times \\textbf{v}) \\times \\mathbf{w} &amp; \\mbox{ Scalar triple product} \\\\\r\n\\end{array}\\]\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<h3 data-type=\"title\">Proof<\/h3>\r\n<p id=\"fs-id1163723338502\">For property i., we want to show [latex]\\mathbf{u \\times v} = -( \\mathbf{v \\times u})[\/latex]. We have<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{u} \\times \\mathbf{v} &amp;= \\langle u_1, u_2, u_3 \\rangle \\times \\langle v_1, v_2, v_3 \\rangle\\\\\r\n&amp;= \\langle u_2v_3 - u_3v_2, -u_1v_3 + u_3v_1, u_1v_2 - u_2v_1 \\rangle \\\\\r\n&amp;= -\\langle u_3v_2 - u_2v_3, -u_3v_1 + u_1v_3, u_2v_1 - u_1v_2 \\rangle\\\\\r\n&amp;= -\\langle v_1, v_2, v_3 \\rangle \\times \\langle u_1, u_2, u_3 \\rangle \\\\\r\n&amp;= -( \\mathbf{v \\times u}).\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>&nbsp;\r\n<p id=\"fs-id1163723523497\">Unlike most operations we\u2019ve seen, the cross product is not commutative. This makes sense if we think about the right-hand rule.<\/p>\r\n<p id=\"fs-id1163723523503\">For property iv., this follows directly from the definition of the cross product. We have<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{u} \\times \\mathbf{0} &amp;= \\langle u_2(0) - u_3(0), -(u_2(0) - u_3(0)), u_1(0) - u_2(0) \\rangle\\\\\r\n&amp;= \\langle 0,0,0\\rangle = \\mathbf{0}.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>\r\n<p id=\"fs-id1163723200827\">Then, by property i., [latex]\\mathbf{0} \\times \\mathbf{u} = \\mathbf{0}[\/latex] as well. Remember that the dot product of a vector and the zero vector is the\u00a0<em data-effect=\"italics\">scalar<\/em> [latex]0[\/latex], whereas the cross product of a vector with the zero vector is the\u00a0<em data-effect=\"italics\">vector <\/em>[latex]\\textbf 0[\/latex].<\/p>\r\n<p id=\"fs-id1163723525985\">Property vi. looks like the associative property, but note the change in operations:<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{u} \\cdot (\\mathbf{v \\times w}) &amp;= \\mathbf{u} \\cdot \\langle v_2w_3 - v_3w_2, -v_1w_3 + v_3w_1,v_1w_2 - v_2w_1 \\rangle\\\\\r\n&amp;= u_1(v_2w_3 - v_3w_2) + u_2(-v_1w_3 + v_3w_1) + u_3(v_1w_2 - v_2w_1) \\\\\r\n&amp;= u_1v_2w_3 - u_1v_3w_2 - u_2v_1w_3 + u_2v_3w_1 + u_3v_1w_2 - u_3v_2w_1 \\\\\r\n&amp;= (u_2v_3 - u_3v_2)w_1 + (u_3v_1 - u_1v_3)w_2 + (u_1v_2 - u_2v_1)w_3\\\\\r\n&amp;= \\langle u_2v_3 - u_3v_2,u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \\rangle \\times \\langle w_1, w_2, w_3 \\rangle \\\\\r\n&amp;= (\\mathbf{u \\times v}) \\cdot \\mathbf{w}.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>&nbsp;\r\n\r\n[latex]_\\blacksquare[\/latex]\r\n<div id=\"fs-id1163723525994\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\r\n<div id=\"fs-id1163723500201\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: using the properties of the cross product<\/h3>\r\nUse the cross product properties to calculate [latex](2\\mathbf{i} \\times 3\\mathbf{j}) \\times \\mathbf{j}[\/latex].\r\n\r\n[reveal-answer q=\"859323711\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"859323711\"]\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n(2\\mathbf{i} \\times 3\\mathbf{j}) \\times \\mathbf{j} &amp;= 2(\\mathbf{i} \\times 3\\mathbf{j}) \\times \\mathbf{j}\\\\\r\n&amp;= 2(3)(\\mathbf{i} \\times \\mathbf{j}) \\times \\mathbf{j} \\\\\r\n&amp;= (6\\mathbf{k}) \\times \\mathbf{j}\\\\\r\n&amp;= 6 (\\mathbf{k} \\times \\mathbf{j})\\\\\r\n&amp;= 6(-\\mathbf{i}) = -6\\mathbf{i}. \\\\\r\n\\end{align*}\r\n[\/latex]<\/center>&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nUse the cross product properties to calculate [latex](\\mathbf{i} \\times \\mathbf{k}) \\times (\\mathbf{k} \\times \\mathbf{j})[\/latex].\r\n\r\n[reveal-answer q=\"092367174\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"092367174\"]\r\n\r\n[latex]-\\mathbf{k}[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nSo far in this section, we have been concerned with the direction of the vector [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex], but we have not discussed its magnitude. It turns out there is a simple expression for the magnitude of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] involving the magnitudes of [latex]\\mathbf u[\/latex] and [latex]\\mathbf v[\/latex], and the sine of the angle between them.\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: magnitude of the cross product<\/span><\/h3>\r\n\r\n<hr \/>\r\n\r\nLet [latex]\\mathbf u[\/latex] and [latex]\\mathbf v[\/latex] be vectors, and let [latex]\\theta[\/latex] be the angle between them. Then, [latex]||\\mathbf{u} \\times \\mathbf{v}|| = ||\\mathbf{u}|| \\cdot ||\\mathbf{v}||\\cdot \\sin{\\theta}[\/latex].\r\n\r\n<\/div>\r\n<h3 data-type=\"title\">Proof<\/h3>\r\n<p id=\"fs-id1163723236451\">Let [latex]\\mathbf{u} = \\langle u_1,u_2, u_3\\rangle[\/latex] and [latex]\\mathbf{v} = \\langle v_1,v_2, v_3\\rangle[\/latex] be vectors, and let [latex]\\theta[\/latex] denote the angle between them. Then<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n||\\mathbf{u} \\times \\mathbf{v}||^2 &amp;= (u_2v_3 - u_3v_2)^2 + (u_3v_1 - u_1v_3)^2 + (u_1v_2 - u_2v_1)^2 \\\\\r\n&amp;= u_2^2v_3^2 -2u_2u_3v_2v_3 + u_3^2v_2^2 + u_3^2v_1^2 -2u_1u_3v_1v_3 + u_1^2v_3^2 + u_1^2v_2^2 -2u_1u_2v_1v_2 + u_2^2v_1^2 \\\\\r\n&amp;= u_1^2v_1^2 + u_1^2v_2^2 + u_1^2v_3^2 +u_2^2v_1^2 + u_2^2v_2^2 +u_2^2v_3^2 +u_3^2v_1^2 +u_3^2v_2^2+u_3^2v_3^2 - (u_1^2v_1^2 + u_2^2v_2^2 +u_3^2v_3^2 +2u_1u_2v_1v_2 + 2u_1u_3v_1v_3 + 2u_2u_3v_2v_3) \\\\\r\n&amp;= (u_1^2 + u_2^2 + u_3^2)(v_1^2 +v_2^2+v_3^2) - (u_1v_1 + u_2v_2 +u_3v_3)^2 \\\\\r\n&amp;= ||\\mathbf{u}||^2||\\mathbf{v}||^2 - (\\mathbf{u} \\cdot \\mathbf{v})^2 \\\\\r\n&amp;= ||\\mathbf{u}||^2||\\mathbf{v}||^2 - ||\\mathbf{u}||^2||\\mathbf{v}||^2\\cos^2{\\theta} \\\\\r\n&amp;= ||\\mathbf{u}||^2||\\mathbf{v}||^2 - (1-\\cos^2{\\theta}) \\\\\r\n&amp;= ||\\mathbf{u}||^2||\\mathbf{v}||^2 - (\\sin^2{\\theta}). \\\\\r\n\\end{align*}\r\n[\/latex]<\/center>&nbsp;\r\n<p id=\"fs-id1163724074652\">Taking square roots and noting that [latex]\\sqrt{\\sin^2{\\theta}} = \\sin{\\theta}[\/latex] for [latex]0\\geq \\theta \\geq 180[\/latex], we have the desired result:<\/p>\r\n\r\n<center>[latex]||\\mathbf{u} \\times \\mathbf{v}|| = ||\\mathbf{u}||\\;||\\mathbf{v}||\\;\\sin{\\theta}[\/latex].<\/center>\r\n<div id=\"fs-id1163724074696\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]_\\blacksquare[\/latex]<\/div>\r\n<div data-type=\"equation\" data-label=\"\"><\/div>\r\n<p id=\"fs-id1163724074769\">This definition of the cross product allows us to visualize or interpret the product geometrically. It is clear, for example, that the cross product is defined only for vectors in three dimensions, not for vectors in two dimensions. In two dimensions, it is impossible to generate a vector simultaneously orthogonal to two nonparallel vectors.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: calculating the cross product<\/h3>\r\nUse\u00a0<a class=\"autogenerated-content\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/2-4-the-cross-product#fs-id1163724130388\">Properties of the Cross Product<\/a>\u00a0to find the magnitude of the cross product of [latex]\\mathbf{u} = \\langle 0,4,0\\rangle[\/latex] and\u00a0[latex]\\mathbf{v} = \\langle 0,0,-3\\rangle[\/latex].\r\n\r\n[reveal-answer q=\"466215391\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"466215391\"]\r\n<p id=\"fs-id1163724049532\">We have<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n||\\mathbf{u} \\times \\mathbf{v}|| &amp;= ||\\mathbf{u}||\\cdot||\\mathbf{v}||\\cdot\\sin{\\theta} \\\\\r\n&amp;= \\sqrt{0^2+4^2+0^2} \\cdot \\sqrt{0^2+0^2+(-3)^2} \\cdot \\sin{\\frac{\\pi}{2}}\\\\\r\n&amp;= 4(3)(1) = 12. \\\\\r\n\\end{align*}\r\n[\/latex]<\/center>&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nUse\u00a0<a class=\"autogenerated-content\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/2-4-the-cross-product#fs-id1163724130388\">Properties of the Cross Product<\/a>\u00a0to find the magnitude of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex], where [latex]\\mathbf{u} = \\langle -8,0,0\\rangle[\/latex] and\u00a0[latex]\\mathbf{v} = \\langle 0,2,0\\rangle[\/latex].\r\n\r\n[reveal-answer q=\"635451352\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"635451352\"]\r\n\r\n[latex]16[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<h2 data-type=\"title\">Determinants and the Cross Product<\/h2>\r\n<p id=\"fs-id1163723093323\">Using\u00a0the Cross Product Equation\u00a0to find the cross product of two vectors is straightforward, and it presents the cross product in the useful component form. The formula, however, is complicated and difficult to remember. Fortunately, we have an alternative. We can calculate the cross product of two vectors using\u00a0<strong><span id=\"d53b2426-3a76-4d95-af4f-6532ad8e4e9a_term83\" data-type=\"term\">determinant<\/span><\/strong>\u00a0notation.<\/p>\r\n<p id=\"fs-id1163723093336\">A [latex]2 \\times 2[\/latex] determinant is defined by<\/p>\r\n\r\n<center>[latex]\\begin{vmatrix} a_1 &amp; a_2 \\\\ b_1 &amp; b_2 \\end{vmatrix} = a_1b_2 - b_1a_2 [\/latex].<\/center>\r\n<p id=\"fs-id1163724065992\">For example,<\/p>\r\n\r\n<center>[latex]\\begin{vmatrix} 3 &amp; -2 \\\\ 5 &amp; 1 \\end{vmatrix} = 3(1) - 5(-2) = 3+10 = 13 [\/latex].<\/center>\r\n<p id=\"fs-id1163724066065\">A [latex]3 \\times 3[\/latex] determinant is defined in terms of [latex]2 \\times 2[\/latex] determinants as follows:<\/p>\r\n\r\n<center>[latex]\\begin{vmatrix} a_1 &amp; a_2 &amp; a_3 \\\\ b_1 &amp; b_2 &amp; b_3 \\\\ c_1 &amp; c_2 &amp; c_3\\end{vmatrix} = a_1\\begin{vmatrix} b_2 &amp; b_3 \\\\ c_2 &amp; c_3 \\end{vmatrix} - a_2 \\begin{vmatrix} b_1 &amp; b_3 \\\\ c_1 &amp; c_3 \\end{vmatrix} + a_3\\begin{vmatrix} b_1 &amp; b_2 \\\\ c_1 &amp; c_2 \\end{vmatrix}[\/latex].<\/center>\r\n<p id=\"fs-id1163724050342\"><a class=\"autogenerated-content\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/2-4-the-cross-product#fs-id1163724066100\">Equation 2.10<\/a>\u00a0is referred to as the\u00a0<em data-effect=\"italics\">expansion of the determinant along the first row<\/em>. Notice that the multipliers of each of the [latex]2 \\times 2[\/latex] determinants on the right side of this expression are the entries in the first row of the [latex]3 \\times 3[\/latex] determinant. Furthermore, each of the [latex]2 \\times 2[\/latex] determinants contains the entries from the [latex]3 \\times 3[\/latex] determinant that would remain if you crossed out the row and column containing the multiplier. Thus, for the first term on the right, [latex]a_1[\/latex] is the multiplier, and the [latex]2 \\times 2[\/latex] determinant contains the entries that remain if you cross out the first row and first column of the [latex]3 \\times 3[\/latex] determinant. Similarly, for the second term, the multiplier is [latex]a_2[\/latex], and the [latex]2 \\times 2[\/latex] determinant contains the entries that remain if you cross out the first row and second column of the [latex]3 \\times 3[\/latex] determinant. Notice, however, that the coefficient of the second term is negative. The third term can be calculated in similar fashion.<\/p>\r\n\r\n<div class=\"textbox exercises\">\r\n<h3>Example: using Expansion along the first row to compute a\u00a0[latex]3 \\times 3[\/latex] determinant<\/h3>\r\nEvaluate the determinant [latex]\\begin{vmatrix} 2 &amp; 5 &amp; -1 \\\\ -1 &amp; 1 &amp; 3 \\\\ -2 &amp; 3 &amp; 4\\end{vmatrix}[\/latex].\r\n\r\n[reveal-answer q=\"157384568\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"157384568\"]\r\n<p id=\"fs-id1163723183874\">We have<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\begin{vmatrix} 2 &amp; 5 &amp; -1 \\\\ -1 &amp; 1 &amp; 3 \\\\ -2 &amp; 3 &amp; 4\\end{vmatrix} &amp;= 2\\begin{vmatrix} 1 &amp; 3 \\\\ 3 &amp; 4 \\end{vmatrix} -5 \\begin{vmatrix} -1 &amp; 3 \\\\ -2 &amp; 4 \\end{vmatrix} -1\\begin{vmatrix} -1 &amp; 1 \\\\ -2 &amp; 3 \\end{vmatrix} \\\\\r\n&amp;= 2(4-9) -5(-4+6) -1(-3+2)\\\\\r\n&amp;= 2(-5) -5(2) -1(-1) = -10 - 10 +1\\\\\r\n&amp;= -19. \\\\\r\n\\end{align*}\r\n[\/latex]<\/center>&nbsp;\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nEvaluate the determinant [latex]\\begin{vmatrix} 1 &amp; -2 &amp; -1 \\\\ 3 &amp; 2 &amp; -3 \\\\ 1 &amp; 5 &amp; 4\\end{vmatrix}[\/latex].\r\n\r\n[reveal-answer q=\"178463041\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"178463041\"]\r\n\r\n[latex]40[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1163723298997\">Technically, determinants are defined only in terms of arrays of real numbers. However, the determinant notation provides a useful mnemonic device for the cross product formula.<\/p>\r\n\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\" style=\"text-align: center;\" data-type=\"title\"><span id=\"21\" class=\"os-title-label\" data-type=\"\">RULE: CROSS PRODUCT CALCULATED BY A DETERMINANT<\/span><\/h3>\r\n\r\n<hr \/>\r\n\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n<p id=\"fs-id1163723299007\">Let [latex]\\mathbf{u} = \\langle u_1,u_2, u_3\\rangle[\/latex] and [latex]\\mathbf{v} = \\langle v_1,v_2, v_3\\rangle[\/latex] be vectors. Then the cross product [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] is given by<\/p>\r\n\r\n<center>[latex]\\mathbf{u} \\times \\mathbf{v} = \\begin{vmatrix} \\mathbf{i} &amp; \\mathbf{j} &amp; \\mathbf{k} \\\\ u_1 &amp; u_2 &amp; u_3 \\\\ v_1 &amp; v_2 &amp; v_3\\end{vmatrix} = \\begin{vmatrix} u_2 &amp; u_3 \\\\ v_2 &amp; v_3 \\end{vmatrix}\\mathbf{i}-\\begin{vmatrix} u_1 &amp; u_3 \\\\ v_1 &amp; v_3 \\end{vmatrix}\\mathbf{j} + \\begin{vmatrix} u_1 &amp; u_2 \\\\ v_1 &amp; v_2 \\end{vmatrix}\\mathbf{k}[\/latex].<\/center><\/div>\r\n<\/section><\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: using determinant notation to find\u00a0[latex]\\mathbf{p} \\times \\mathbf{q}[\/latex]<\/h3>\r\nLet [latex]\\mathbf{p} = \\langle -1,2,5\\rangle[\/latex] and [latex]\\mathbf{q} = \\langle 4,0,-3\\rangle[\/latex]. Find [latex]\\mathbf{p} \\times \\mathbf{q}[\/latex].\r\n\r\n[reveal-answer q=\"188409830\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"188409830\"]\r\n<p id=\"fs-id1163724057826\">We set up our determinant by putting the standard unit vectors across the first row, the components of [latex]\\textbf u[\/latex] in the second row, and the components of [latex]\\textbf v[\/latex] in the third row. Then, we have<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{p} \\times \\mathbf{q} &amp;= \\begin{vmatrix} \\mathbf{i} &amp; \\mathbf{j} &amp; \\mathbf{k} \\\\ -1 &amp; 2 &amp; 5 \\\\ 4 &amp; 0 &amp; -3\\end{vmatrix} = \\begin{vmatrix} 2 &amp; 5 \\\\ 0 &amp; -3 \\end{vmatrix}\\mathbf{i} - \\begin{vmatrix} -1 &amp; 5 \\\\ 4 &amp; -3 \\end{vmatrix}\\mathbf{j} +\\begin{vmatrix} -1 &amp; 2 \\\\ 4 &amp; 0 \\end{vmatrix}\\mathbf{k} \\\\\r\n&amp;=(-6-0)\\mathbf{i} -(3-20)\\mathbf{j} +(0-8)\\mathbf{k}\\\\\r\n&amp;= -6\\mathbf{i} + 17\\mathbf{j} -8\\mathbf{k}.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>&nbsp;\r\n<p id=\"fs-id1163723314873\">Notice that this answer confirms the calculation of the cross product in\u00a0<a class=\"autogenerated-content\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/2-4-the-cross-product#fs-id1163723497714\">Example 2.31<\/a>.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nUse determinant notation to find [latex]\\mathbf{a} \\times \\mathbf{b}[\/latex], where [latex]\\mathbf{a} = \\langle 8,2,3\\rangle[\/latex] and\u00a0[latex]\\mathbf{b} = \\langle -1,0,4\\rangle[\/latex].\r\n\r\n[reveal-answer q=\"457345863\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"457345863\"]\r\n\r\n[latex]8\\mathbf{i} -35\\mathbf{j} +2\\mathbf{k}[\/latex].\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to the above Try IT.\r\n\r\n<center><iframe src=\"\/\/plugin.3playmedia.com\/show?mf=7809775&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=umfZKb2xRMw&amp;video_target=tpm-plugin-ma5mr889-umfZKb2xRMw\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/center><center>You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.36_transcript.html\">\u201cCP 2.36\u201d here (opens in new window).<\/a><\/center>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Calculate the cross product of two given vectors.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Use determinants to calculate a cross product.<\/span><\/li>\n<\/ul>\n<\/div>\n<section id=\"fs-id1163724083973\" data-depth=\"1\">\n<h2 data-type=\"title\">The Cross Product and Its Properties<\/h2>\n<p id=\"fs-id1163724087473\">The dot product is a multiplication of two vectors that results in a scalar. In this section, we introduce a product of two vectors that generates a third vector orthogonal to the first two. Consider how we might find such a vector. Let [latex]\\mathbf{u} =\\langle u_1, u_2, u_3 \\rangle[\/latex] and [latex]\\mathbf{v} =\\langle v_1, v_2, v_3 \\rangle[\/latex] be nonzero vectors. We want to find a vector [latex]\\mathbf{w} =\\langle w_1, w_2, w_3 \\rangle[\/latex] orthogonal to both [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex]\u2014that is, we want to find [latex]\\mathbf{w}[\/latex] such that [latex]\\mathbf{u} \\cdot \\mathbf{w} = 0[\/latex] and [latex]\\mathbf{v} \\cdot \\mathbf{w} =0[\/latex]. Therefore, [latex]w_1[\/latex], [latex]w_2[\/latex], and [latex]w_3[\/latex] must satisfy<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{aligned}  u_1w_1 + u_2w_2 + u_3w_3 &= 0 \\\\  v_1w_1 + v_2w_2 + v_3w_3 &= 0.\\\\  \\end{aligned}[\/latex]<\/div>\n<p>If we multiply the top equation by [latex]v_3[\/latex] and the bottom equation by [latex]u_3[\/latex] and subtract, we can eliminate the variable [latex]w_3[\/latex], which gives<\/p>\n<div style=\"text-align: center;\">[latex](u_1v_3 - v_1u_3)w_1 + (u_2v_3 - v_2u_3)w_2 = 0[\/latex].<\/div>\n<p id=\"fs-id1163723884318\">If we select<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  w_1 &= u_2v_3 - v_2u_3\\\\  w_2 &= -(u_1v_3 - u_3v_1),\\\\  \\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1163723528456\">we get a possible solution vector. Substituting these values back into the original equations gives<\/p>\n<div style=\"text-align: center;\">[latex]w_3 = u_1v_2 - u_2v_1[\/latex].<\/div>\n<p id=\"fs-id1163723670069\">That is, vector<\/p>\n<div style=\"text-align: center;\">[latex]\\mathbf{w} = \\langle u_2v_3 - u_3v_2, -(u_1v_3 - u_3v_1), u_1v_2 - u_2v_1 \\rangle[\/latex]<\/div>\n<p id=\"fs-id1163724043762\">is orthogonal to both [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex], which leads us to define the following operation, called the cross product.<\/p>\n<\/section>\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\" style=\"text-align: center;\" data-type=\"title\"><span id=\"2\" class=\"os-title-label\" data-type=\"\">DEFINITION<\/span><\/h3>\n<hr \/>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-id1163723775830\">Let [latex]\\mathbf{u} =\\langle u_1, u_2, u_3 \\rangle[\/latex] and [latex]\\mathbf{v} =\\langle v_1, v_2, v_3 \\rangle[\/latex]. Then, the\u00a0<strong>cross product<\/strong> [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex]\u00a0is vector<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{u} \\times \\mathbf{v} &= (u_2v_3 - u_3v_2)\\mathbf{i} -(u_1v_3 - u_3v_1)\\mathbf{j}+ (u_1v_2 - u_2v_1)\\mathbf{k}\\\\  &= \\langle u_2v_3 - u_3v_2, -(u_1v_3 - u_3v_1), u_1v_2 - u_2v_1 \\rangle.\\\\  \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/section>\n<\/div>\n<p id=\"fs-id1163724072465\">From the way we have developed [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex], it should be clear that the cross product is orthogonal to both [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex]. However, it never hurts to check. To show that [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] is orthogonal to [latex]\\textbf u[\/latex], we calculate the dot product of [latex]\\textbf u[\/latex] and [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex].<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{u} \\cdot (\\mathbf{u} \\times \\mathbf{v}) &= \\langle u_1,u_2,u_3 \\rangle \\cdot \\langle u_2v_3 - u_3v_2, -u_1v_3 - u_3v_1, u_1v_2 - u_2v_1 \\rangle\\\\  &= u_1(u_2v_3 - u_3v_2) + u_2 (-u_1v_3 + u_3v_1) +u_3( u_1v_2 - u_2v_1)\\\\  &= u_1u_2v_3 - u_1u_3v_2 -u_1u_2v_3+u_2u_3v_1 +u_1u_3v_2-u_2u_3v_1\\\\  &= (u_1u_2v_3 - u_1u_2v_3) +(-u_1u_3v_2+u_1u_3v_2) + (u_2u_3v_1-u_2u_3v_1)\\\\  &= 0\\\\  \\end{align*}[\/latex]<\/div>\n<p>In a similar manner, we can show that the cross product is also orthogonal to [latex]\\textbf v[\/latex].<\/p>\n<div id=\"fs-id1163724060614\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\n<div id=\"fs-id1163723497714\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: finding a cross product<\/h3>\n<p>Let [latex]\\mathbf{p} =\\langle -1,2,5 \\rangle[\/latex] and [latex]\\mathbf{q} =\\langle 4,0,-3 \\rangle[\/latex] (Figure 1). Find [latex]\\mathbf{p} \\times \\mathbf{q}[\/latex].<\/p>\n<div id=\"attachment_5136\" style=\"width: 511px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5136\" class=\"size-full wp-image-5136\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27155235\/Figure-2.53.jpg\" alt=\"This figure is the 3-dimensional coordinate system. It has two vectors in standard position. The first vector is labeled \u201cp = &lt;-1, 2, 5&gt;.\u201d The second vector is labeled \u201cq = &lt;4, 0, -3&gt;.\u201d\" width=\"501\" height=\"497\" \/><\/p>\n<p id=\"caption-attachment-5136\" class=\"wp-caption-text\">Figure 1. Finding a cross product to two given vectors.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q934723610\">Show Solution<\/span><\/p>\n<div id=\"q934723610\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723747703\">Substitute the components of the vectors into\u00a0the Cross Product Equation:<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{p} \\times \\mathbf{q} &= \\langle -1,2,5 \\rangle \\times \\langle 4,0,-3 \\rangle\\\\  &= \\langle p_2q_3-p_3q_2, p_3q_1 - p_1q_3,p_1q_2 -p_2q_1 \\rangle \\\\  &= \\langle 2(-3)-5(0), -(-1)(-3)+5(4), (-1)(0)-2(4) \\rangle\\\\  &= \\langle -6,17,-8 \\rangle.\\\\\\end{align*}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find [latex]\\mathbf{p} \\times \\mathbf{q}[\/latex] for [latex]\\mathbf{p} =\\langle 5,1,2 \\rangle[\/latex] and [latex]\\mathbf{q} =\\langle -2,0,1 \\rangle[\/latex]. Express the answer using standard unit vectors.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q125237156\">Show Solution<\/span><\/p>\n<div id=\"q125237156\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\mathbf{i} -9\\mathbf{j}+2\\mathbf{k}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try IT.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7753558&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=fEMzUqQTmoo&amp;video_target=tpm-plugin-aghj6e1q-fEMzUqQTmoo\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\"><span style=\"font-size: 1rem; orphans: 1; text-align: initial;\">You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.30_transcript.html\">\u201cCP 2.30\u201d here (opens in new window).<\/a><\/span><\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm5986\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=5986&theme=oea&iframe_resize_id=ohm5986&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<p>Although it may not be obvious from\u00a0the Cross Product Equation, the direction of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] is given by the right-hand rule. If we hold the right hand out with the fingers pointing in the direction of [latex]\\textbf u[\/latex], then curl the fingers toward vector [latex]\\textbf v[\/latex], the thumb points in the direction of the cross product, as shown.<\/p>\n<div id=\"attachment_5138\" style=\"width: 676px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5138\" class=\"size-full wp-image-5138\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27155328\/Figure-2.54.jpg\" alt=\"This figure has two images. The first image has three vectors with the same initial point. Two of the vectors are labeled \u201cu\u201d and \u201cv.\u201d The angle between u and v is theta. The third vector is perpendicular to u and v. It is labeled \u201cu cross v.\u201d The second image has three vectors. The vectors are labeled \u201cu, v, and u cross v.\u201d \u201cu cross v\u201d is perpendicular to u and v. Also, on the image of these three vectors is a right hand. The fingers are in the direction of u. As the hand is closing, the direction of the closing fingers is the direction of v. The thumb is up and in the direction of \u201cu cross v.\u201d\" width=\"666\" height=\"416\" \/><\/p>\n<p id=\"caption-attachment-5138\" class=\"wp-caption-text\">Figure 2. The direction of [latex]{\\bf{u}}\\times{\\bf{v}}[\/latex] is determined by the right-hand rule.<\/p>\n<\/div>\n<p>Notice what this means for the direction of [latex]\\mathbf{v} \\times \\mathbf{u}[\/latex]. If we apply the right-hand rule to [latex]\\mathbf{v} \\times \\mathbf{u}[\/latex], we start with our fingers pointed in the direction of [latex]\\textbf v[\/latex], then curl our fingers toward the vector [latex]\\textbf u[\/latex]. In this case, the thumb points in the opposite direction of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex]. (Try it!)<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: anticommutativity of the cross product<\/h3>\n<p>Let [latex]\\mathbf{u} =\\langle 0,2,1 \\rangle[\/latex] and [latex]\\mathbf{v} =\\langle 3,-1,0 \\rangle[\/latex]. Calculate [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] and [latex]\\mathbf{v} \\times \\mathbf{u}[\/latex] and graph them.<\/p>\n<div id=\"attachment_5141\" style=\"width: 534px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5141\" class=\"size-full wp-image-5141\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27155653\/Figure-2.55.jpg\" alt=\"This figure is the 3-dimensional coordinate system. It has two vectors in standard position. The first vector is labeled \u201cu = &lt;0, 2, 1&gt;.\u201d The second vector is labeled \u201cv = &lt;3, -1, 0&gt;.\u201d\" width=\"524\" height=\"497\" \/><\/p>\n<p id=\"caption-attachment-5141\" class=\"wp-caption-text\">Figure 3. Are the cross products [latex]{\\bf{u}}\\times{\\bf{v}}[\/latex] and [latex]{\\bf{v}}\\times{\\bf{u}}[\/latex] in the same direction?<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q856612738\">Show Solution<\/span><\/p>\n<div id=\"q856612738\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723474704\">We have<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{u} \\times \\mathbf{v} &= \\langle (0+1),-(0-3),(0-6) \\rangle = \\langle 1,3,-6\\rangle\\\\  \\mathbf{v} \\times \\mathbf{u}&= \\langle (-1-0),-(3-0),(6-0) \\rangle = \\langle -1,-3,6\\rangle.\\\\\\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1163724049208\">We see that, in this case, [latex]\\mathbf{u} \\times \\mathbf{v} = -(\\mathbf{v} \\times \\mathbf{u})[\/latex] (<a class=\"autogenerated-content\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/2-4-the-cross-product#CNX_Calc_Figure_12_04_013\">Figure 2.56<\/a>). We prove this in general later in this section.<\/p>\n<div id=\"attachment_5142\" style=\"width: 511px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5142\" class=\"size-full wp-image-5142\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27155808\/Figure-2.56.jpg\" alt=\"\" width=\"501\" height=\"672\" \/><\/p>\n<p id=\"caption-attachment-5142\" class=\"wp-caption-text\">Figure 4. The cross products [latex]{\\bf{u}}\\times{\\bf{v}}[\/latex] and [latex]{\\bf{v}}\\times{\\bf{u}}[\/latex] are both orthogonal to [latex]{\\bf{u}}[\/latex] and [latex]{\\bf{v}}[\/latex], but in opposite directions.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Suppose vectors [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] lie in the [latex]xy[\/latex]-plane (the [latex]z[\/latex]-component of each vector is zero). Now suppose the [latex]x[\/latex]&#8211; and [latex]y[\/latex]-components of [latex]\\textbf u[\/latex] and the [latex]y[\/latex]-component of [latex]\\textbf v[\/latex] are all positive, whereas the [latex]x[\/latex]-component of [latex]\\textbf v[\/latex] is negative. Assuming the coordinate axes are oriented in the usual positions, in which direction does [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] point?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q373093789\">Show Solution<\/span><\/p>\n<div id=\"q373093789\" class=\"hidden-answer\" style=\"display: none\">\n<p>Up (the positive [latex]z[\/latex]-direction)<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1163724097644\">The cross products of the standard unit vectors [latex]\\textbf i[\/latex], [latex]\\textbf j[\/latex], and [latex]\\textbf k[\/latex] can be useful for simplifying some calculations, so let\u2019s consider these cross products. A straightforward application of the definition shows that<\/p>\n<div style=\"text-align: center;\">[latex]\\mathbf{i} \\times \\mathbf{i} = \\mathbf{j} \\times \\mathbf{j}= \\mathbf{k} \\times \\mathbf{k} = 0[\/latex].<\/div>\n<p id=\"fs-id1163723478160\">(The cross product of two vectors is a vector, so each of these products results in the zero vector, not the scalar 0.) It\u2019s up to you to verify the calculations on your own.<\/p>\n<p id=\"fs-id1163723096820\">Furthermore, because the cross product of two vectors is orthogonal to each of these vectors, we know that the cross product of [latex]\\textbf i[\/latex] and [latex]\\textbf j[\/latex] is parallel to [latex]\\textbf k[\/latex]. Similarly, the vector product of [latex]\\textbf i[\/latex] and [latex]\\textbf k[\/latex] is parallel to [latex]\\textbf j[\/latex], and the vector product of [latex]\\textbf j[\/latex] and [latex]\\textbf k[\/latex] is parallel to [latex]\\textbf i[\/latex]. We can use the right-hand rule to determine the direction of each product. Then we have<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{aligned}  \\mathbf{i} \\times \\mathbf{j} &= \\mathbf{k} & \\mathbf{j} \\times \\mathbf{i} &= -\\mathbf{k}\\\\  \\mathbf{j} \\times \\mathbf{k} &= \\mathbf{i} & \\mathbf{k} \\times \\mathbf{j} &= -\\mathbf{i} \\\\  \\mathbf{k} \\times \\mathbf{i} &= \\mathbf{j} & \\mathbf{i} \\times \\mathbf{k} &= -\\mathbf{j}.\\\\\\end{aligned}[\/latex]<\/div>\n<p id=\"fs-id1163723472405\">These formulas come in handy later.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: cross product of standard unit vectors<\/h3>\n<p>Find [latex]\\mathbf{i} \\times (\\mathbf{j} \\times \\mathbf{k})[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q734661173\">Show Solution<\/span><\/p>\n<div id=\"q734661173\" class=\"hidden-answer\" style=\"display: none\">\n<p>We know that [latex]\\mathbf{j} \\times \\mathbf{k} = \\mathbf{i}[\/latex]. Therefore, [latex]\\mathbf{i} \\times (\\mathbf{j} \\times \\mathbf{k}) = \\mathbf{i} \\times \\mathbf{i} =0[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find [latex](\\mathbf{i} \\times \\mathbf{j}) \\times (\\mathbf{k} \\times \\mathbf{i})[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q092380347\">Show Solution<\/span><\/p>\n<div id=\"q092380347\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-\\mathbf{i}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>As we have seen, the dot product is often called the\u00a0<em data-effect=\"italics\">scalar product<\/em>\u00a0because it results in a scalar. The cross product results in a vector, so it is sometimes called the\u00a0<strong><span id=\"d53b2426-3a76-4d95-af4f-6532ad8e4e9a_term82\" data-type=\"term\">vector product<\/span><\/strong>. These operations are both versions of vector multiplication, but they have very different properties and applications. Let\u2019s explore some properties of the cross product. We prove only a few of them. Proofs of the other properties are left as exercises.<\/p>\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: properties of the cross product<\/span><\/h3>\n<hr \/>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-id1163723277869\">Let [latex]\\textbf u[\/latex], [latex]\\textbf v[\/latex], and [latex]\\textbf w[\/latex] be vectors in space, and let [latex]c[\/latex] be a scalar.<\/p>\n<p>\\[ \\begin{array}{lrl}<br \/>\n\\mbox{i.} &amp; \\mathbf{u \\times v} &amp;= -( \\mathbf{v \\times u}) &amp; \\mbox{Anticommutative property} \\\\<br \/>\n\\mbox{ii.} &amp; \\mathbf{u} \\times (\\mathbf{v} + \\textbf{w}) &amp;= \\mathbf{u \\times v} + \\mathbf{u \\times w} &amp; \\mbox{Distributive property} \\\\<br \/>\n\\mbox{iii.} &amp; c(\\mathbf{u \\times v}) &amp;= (c\\mathbf{u}) \\times \\mathbf{v} = \\mathbf{u} \\times (c\\mathbf{v}) &amp; \\mbox{ Multiplication by a constant} \\\\<br \/>\n\\mbox{iv.} &amp; \\mathbf{u} \\times \\mathbf{0} &amp;= \\mathbf{0} \\times \\mathbf{u} = \\mathbf{0} &amp; \\mbox{ Cross product of the zero vector} \\\\<br \/>\n\\mbox{iv.} &amp; \\mathbf{v} \\times \\mathbf{v} &amp;= \\mathbf{0} &amp; \\mbox{ \u00a0Cross product of a vector with itself} \\\\<br \/>\n\\mbox{iv.} &amp; \\mathbf{u} \\times (\\mathbf{v} \\times \\textbf{w}) &amp;= (\\mathbf{u} \\times \\textbf{v}) \\times \\mathbf{w} &amp; \\mbox{ Scalar triple product} \\\\<br \/>\n\\end{array}\\]<\/p>\n<\/div>\n<\/section>\n<\/div>\n<h3 data-type=\"title\">Proof<\/h3>\n<p id=\"fs-id1163723338502\">For property i., we want to show [latex]\\mathbf{u \\times v} = -( \\mathbf{v \\times u})[\/latex]. We have<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{u} \\times \\mathbf{v} &= \\langle u_1, u_2, u_3 \\rangle \\times \\langle v_1, v_2, v_3 \\rangle\\\\  &= \\langle u_2v_3 - u_3v_2, -u_1v_3 + u_3v_1, u_1v_2 - u_2v_1 \\rangle \\\\  &= -\\langle u_3v_2 - u_2v_3, -u_3v_1 + u_1v_3, u_2v_1 - u_1v_2 \\rangle\\\\  &= -\\langle v_1, v_2, v_3 \\rangle \\times \\langle u_1, u_2, u_3 \\rangle \\\\  &= -( \\mathbf{v \\times u}).\\\\  \\end{align*}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1163723523497\">Unlike most operations we\u2019ve seen, the cross product is not commutative. This makes sense if we think about the right-hand rule.<\/p>\n<p id=\"fs-id1163723523503\">For property iv., this follows directly from the definition of the cross product. We have<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{u} \\times \\mathbf{0} &= \\langle u_2(0) - u_3(0), -(u_2(0) - u_3(0)), u_1(0) - u_2(0) \\rangle\\\\  &= \\langle 0,0,0\\rangle = \\mathbf{0}.\\\\  \\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1163723200827\">Then, by property i., [latex]\\mathbf{0} \\times \\mathbf{u} = \\mathbf{0}[\/latex] as well. Remember that the dot product of a vector and the zero vector is the\u00a0<em data-effect=\"italics\">scalar<\/em> [latex]0[\/latex], whereas the cross product of a vector with the zero vector is the\u00a0<em data-effect=\"italics\">vector <\/em>[latex]\\textbf 0[\/latex].<\/p>\n<p id=\"fs-id1163723525985\">Property vi. looks like the associative property, but note the change in operations:<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{u} \\cdot (\\mathbf{v \\times w}) &= \\mathbf{u} \\cdot \\langle v_2w_3 - v_3w_2, -v_1w_3 + v_3w_1,v_1w_2 - v_2w_1 \\rangle\\\\  &= u_1(v_2w_3 - v_3w_2) + u_2(-v_1w_3 + v_3w_1) + u_3(v_1w_2 - v_2w_1) \\\\  &= u_1v_2w_3 - u_1v_3w_2 - u_2v_1w_3 + u_2v_3w_1 + u_3v_1w_2 - u_3v_2w_1 \\\\  &= (u_2v_3 - u_3v_2)w_1 + (u_3v_1 - u_1v_3)w_2 + (u_1v_2 - u_2v_1)w_3\\\\  &= \\langle u_2v_3 - u_3v_2,u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \\rangle \\times \\langle w_1, w_2, w_3 \\rangle \\\\  &= (\\mathbf{u \\times v}) \\cdot \\mathbf{w}.\\\\  \\end{align*}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<div id=\"fs-id1163723525994\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\n<div id=\"fs-id1163723500201\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: using the properties of the cross product<\/h3>\n<p>Use the cross product properties to calculate [latex](2\\mathbf{i} \\times 3\\mathbf{j}) \\times \\mathbf{j}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q859323711\">Show Solution<\/span><\/p>\n<div id=\"q859323711\" class=\"hidden-answer\" style=\"display: none\">\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  (2\\mathbf{i} \\times 3\\mathbf{j}) \\times \\mathbf{j} &= 2(\\mathbf{i} \\times 3\\mathbf{j}) \\times \\mathbf{j}\\\\  &= 2(3)(\\mathbf{i} \\times \\mathbf{j}) \\times \\mathbf{j} \\\\  &= (6\\mathbf{k}) \\times \\mathbf{j}\\\\  &= 6 (\\mathbf{k} \\times \\mathbf{j})\\\\  &= 6(-\\mathbf{i}) = -6\\mathbf{i}. \\\\  \\end{align*}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Use the cross product properties to calculate [latex](\\mathbf{i} \\times \\mathbf{k}) \\times (\\mathbf{k} \\times \\mathbf{j})[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q092367174\">Show Solution<\/span><\/p>\n<div id=\"q092367174\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]-\\mathbf{k}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>So far in this section, we have been concerned with the direction of the vector [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex], but we have not discussed its magnitude. It turns out there is a simple expression for the magnitude of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] involving the magnitudes of [latex]\\mathbf u[\/latex] and [latex]\\mathbf v[\/latex], and the sine of the angle between them.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: magnitude of the cross product<\/span><\/h3>\n<hr \/>\n<p>Let [latex]\\mathbf u[\/latex] and [latex]\\mathbf v[\/latex] be vectors, and let [latex]\\theta[\/latex] be the angle between them. Then, [latex]||\\mathbf{u} \\times \\mathbf{v}|| = ||\\mathbf{u}|| \\cdot ||\\mathbf{v}||\\cdot \\sin{\\theta}[\/latex].<\/p>\n<\/div>\n<h3 data-type=\"title\">Proof<\/h3>\n<p id=\"fs-id1163723236451\">Let [latex]\\mathbf{u} = \\langle u_1,u_2, u_3\\rangle[\/latex] and [latex]\\mathbf{v} = \\langle v_1,v_2, v_3\\rangle[\/latex] be vectors, and let [latex]\\theta[\/latex] denote the angle between them. Then<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  ||\\mathbf{u} \\times \\mathbf{v}||^2 &= (u_2v_3 - u_3v_2)^2 + (u_3v_1 - u_1v_3)^2 + (u_1v_2 - u_2v_1)^2 \\\\  &= u_2^2v_3^2 -2u_2u_3v_2v_3 + u_3^2v_2^2 + u_3^2v_1^2 -2u_1u_3v_1v_3 + u_1^2v_3^2 + u_1^2v_2^2 -2u_1u_2v_1v_2 + u_2^2v_1^2 \\\\  &= u_1^2v_1^2 + u_1^2v_2^2 + u_1^2v_3^2 +u_2^2v_1^2 + u_2^2v_2^2 +u_2^2v_3^2 +u_3^2v_1^2 +u_3^2v_2^2+u_3^2v_3^2 - (u_1^2v_1^2 + u_2^2v_2^2 +u_3^2v_3^2 +2u_1u_2v_1v_2 + 2u_1u_3v_1v_3 + 2u_2u_3v_2v_3) \\\\  &= (u_1^2 + u_2^2 + u_3^2)(v_1^2 +v_2^2+v_3^2) - (u_1v_1 + u_2v_2 +u_3v_3)^2 \\\\  &= ||\\mathbf{u}||^2||\\mathbf{v}||^2 - (\\mathbf{u} \\cdot \\mathbf{v})^2 \\\\  &= ||\\mathbf{u}||^2||\\mathbf{v}||^2 - ||\\mathbf{u}||^2||\\mathbf{v}||^2\\cos^2{\\theta} \\\\  &= ||\\mathbf{u}||^2||\\mathbf{v}||^2 - (1-\\cos^2{\\theta}) \\\\  &= ||\\mathbf{u}||^2||\\mathbf{v}||^2 - (\\sin^2{\\theta}). \\\\  \\end{align*}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1163724074652\">Taking square roots and noting that [latex]\\sqrt{\\sin^2{\\theta}} = \\sin{\\theta}[\/latex] for [latex]0\\geq \\theta \\geq 180[\/latex], we have the desired result:<\/p>\n<div style=\"text-align: center;\">[latex]||\\mathbf{u} \\times \\mathbf{v}|| = ||\\mathbf{u}||\\;||\\mathbf{v}||\\;\\sin{\\theta}[\/latex].<\/div>\n<div id=\"fs-id1163724074696\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">[latex]_\\blacksquare[\/latex]<\/div>\n<div data-type=\"equation\" data-label=\"\"><\/div>\n<p id=\"fs-id1163724074769\">This definition of the cross product allows us to visualize or interpret the product geometrically. It is clear, for example, that the cross product is defined only for vectors in three dimensions, not for vectors in two dimensions. In two dimensions, it is impossible to generate a vector simultaneously orthogonal to two nonparallel vectors.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: calculating the cross product<\/h3>\n<p>Use\u00a0<a class=\"autogenerated-content\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/2-4-the-cross-product#fs-id1163724130388\">Properties of the Cross Product<\/a>\u00a0to find the magnitude of the cross product of [latex]\\mathbf{u} = \\langle 0,4,0\\rangle[\/latex] and\u00a0[latex]\\mathbf{v} = \\langle 0,0,-3\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q466215391\">Show Solution<\/span><\/p>\n<div id=\"q466215391\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163724049532\">We have<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  ||\\mathbf{u} \\times \\mathbf{v}|| &= ||\\mathbf{u}||\\cdot||\\mathbf{v}||\\cdot\\sin{\\theta} \\\\  &= \\sqrt{0^2+4^2+0^2} \\cdot \\sqrt{0^2+0^2+(-3)^2} \\cdot \\sin{\\frac{\\pi}{2}}\\\\  &= 4(3)(1) = 12. \\\\  \\end{align*}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Use\u00a0<a class=\"autogenerated-content\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/2-4-the-cross-product#fs-id1163724130388\">Properties of the Cross Product<\/a>\u00a0to find the magnitude of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex], where [latex]\\mathbf{u} = \\langle -8,0,0\\rangle[\/latex] and\u00a0[latex]\\mathbf{v} = \\langle 0,2,0\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q635451352\">Show Solution<\/span><\/p>\n<div id=\"q635451352\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]16[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2 data-type=\"title\">Determinants and the Cross Product<\/h2>\n<p id=\"fs-id1163723093323\">Using\u00a0the Cross Product Equation\u00a0to find the cross product of two vectors is straightforward, and it presents the cross product in the useful component form. The formula, however, is complicated and difficult to remember. Fortunately, we have an alternative. We can calculate the cross product of two vectors using\u00a0<strong><span id=\"d53b2426-3a76-4d95-af4f-6532ad8e4e9a_term83\" data-type=\"term\">determinant<\/span><\/strong>\u00a0notation.<\/p>\n<p id=\"fs-id1163723093336\">A [latex]2 \\times 2[\/latex] determinant is defined by<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{vmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end{vmatrix} = a_1b_2 - b_1a_2[\/latex].<\/div>\n<p id=\"fs-id1163724065992\">For example,<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{vmatrix} 3 & -2 \\\\ 5 & 1 \\end{vmatrix} = 3(1) - 5(-2) = 3+10 = 13[\/latex].<\/div>\n<p id=\"fs-id1163724066065\">A [latex]3 \\times 3[\/latex] determinant is defined in terms of [latex]2 \\times 2[\/latex] determinants as follows:<\/p>\n<div style=\"text-align: center;\">[latex]\\begin{vmatrix} a_1 & a_2 & a_3 \\\\ b_1 & b_2 & b_3 \\\\ c_1 & c_2 & c_3\\end{vmatrix} = a_1\\begin{vmatrix} b_2 & b_3 \\\\ c_2 & c_3 \\end{vmatrix} - a_2 \\begin{vmatrix} b_1 & b_3 \\\\ c_1 & c_3 \\end{vmatrix} + a_3\\begin{vmatrix} b_1 & b_2 \\\\ c_1 & c_2 \\end{vmatrix}[\/latex].<\/div>\n<p id=\"fs-id1163724050342\"><a class=\"autogenerated-content\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/2-4-the-cross-product#fs-id1163724066100\">Equation 2.10<\/a>\u00a0is referred to as the\u00a0<em data-effect=\"italics\">expansion of the determinant along the first row<\/em>. Notice that the multipliers of each of the [latex]2 \\times 2[\/latex] determinants on the right side of this expression are the entries in the first row of the [latex]3 \\times 3[\/latex] determinant. Furthermore, each of the [latex]2 \\times 2[\/latex] determinants contains the entries from the [latex]3 \\times 3[\/latex] determinant that would remain if you crossed out the row and column containing the multiplier. Thus, for the first term on the right, [latex]a_1[\/latex] is the multiplier, and the [latex]2 \\times 2[\/latex] determinant contains the entries that remain if you cross out the first row and first column of the [latex]3 \\times 3[\/latex] determinant. Similarly, for the second term, the multiplier is [latex]a_2[\/latex], and the [latex]2 \\times 2[\/latex] determinant contains the entries that remain if you cross out the first row and second column of the [latex]3 \\times 3[\/latex] determinant. Notice, however, that the coefficient of the second term is negative. The third term can be calculated in similar fashion.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example: using Expansion along the first row to compute a\u00a0[latex]3 \\times 3[\/latex] determinant<\/h3>\n<p>Evaluate the determinant [latex]\\begin{vmatrix} 2 & 5 & -1 \\\\ -1 & 1 & 3 \\\\ -2 & 3 & 4\\end{vmatrix}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q157384568\">Show Solution<\/span><\/p>\n<div id=\"q157384568\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723183874\">We have<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\begin{vmatrix} 2 & 5 & -1 \\\\ -1 & 1 & 3 \\\\ -2 & 3 & 4\\end{vmatrix} &= 2\\begin{vmatrix} 1 & 3 \\\\ 3 & 4 \\end{vmatrix} -5 \\begin{vmatrix} -1 & 3 \\\\ -2 & 4 \\end{vmatrix} -1\\begin{vmatrix} -1 & 1 \\\\ -2 & 3 \\end{vmatrix} \\\\  &= 2(4-9) -5(-4+6) -1(-3+2)\\\\  &= 2(-5) -5(2) -1(-1) = -10 - 10 +1\\\\  &= -19. \\\\  \\end{align*}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Evaluate the determinant [latex]\\begin{vmatrix} 1 & -2 & -1 \\\\ 3 & 2 & -3 \\\\ 1 & 5 & 4\\end{vmatrix}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q178463041\">Show Solution<\/span><\/p>\n<div id=\"q178463041\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]40[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1163723298997\">Technically, determinants are defined only in terms of arrays of real numbers. However, the determinant notation provides a useful mnemonic device for the cross product formula.<\/p>\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\" style=\"text-align: center;\" data-type=\"title\"><span id=\"21\" class=\"os-title-label\" data-type=\"\">RULE: CROSS PRODUCT CALCULATED BY A DETERMINANT<\/span><\/h3>\n<hr \/>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-id1163723299007\">Let [latex]\\mathbf{u} = \\langle u_1,u_2, u_3\\rangle[\/latex] and [latex]\\mathbf{v} = \\langle v_1,v_2, v_3\\rangle[\/latex] be vectors. Then the cross product [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] is given by<\/p>\n<div style=\"text-align: center;\">[latex]\\mathbf{u} \\times \\mathbf{v} = \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ u_1 & u_2 & u_3 \\\\ v_1 & v_2 & v_3\\end{vmatrix} = \\begin{vmatrix} u_2 & u_3 \\\\ v_2 & v_3 \\end{vmatrix}\\mathbf{i}-\\begin{vmatrix} u_1 & u_3 \\\\ v_1 & v_3 \\end{vmatrix}\\mathbf{j} + \\begin{vmatrix} u_1 & u_2 \\\\ v_1 & v_2 \\end{vmatrix}\\mathbf{k}[\/latex].<\/div>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: using determinant notation to find\u00a0[latex]\\mathbf{p} \\times \\mathbf{q}[\/latex]<\/h3>\n<p>Let [latex]\\mathbf{p} = \\langle -1,2,5\\rangle[\/latex] and [latex]\\mathbf{q} = \\langle 4,0,-3\\rangle[\/latex]. Find [latex]\\mathbf{p} \\times \\mathbf{q}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q188409830\">Show Solution<\/span><\/p>\n<div id=\"q188409830\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163724057826\">We set up our determinant by putting the standard unit vectors across the first row, the components of [latex]\\textbf u[\/latex] in the second row, and the components of [latex]\\textbf v[\/latex] in the third row. Then, we have<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{p} \\times \\mathbf{q} &= \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ -1 & 2 & 5 \\\\ 4 & 0 & -3\\end{vmatrix} = \\begin{vmatrix} 2 & 5 \\\\ 0 & -3 \\end{vmatrix}\\mathbf{i} - \\begin{vmatrix} -1 & 5 \\\\ 4 & -3 \\end{vmatrix}\\mathbf{j} +\\begin{vmatrix} -1 & 2 \\\\ 4 & 0 \\end{vmatrix}\\mathbf{k} \\\\  &=(-6-0)\\mathbf{i} -(3-20)\\mathbf{j} +(0-8)\\mathbf{k}\\\\  &= -6\\mathbf{i} + 17\\mathbf{j} -8\\mathbf{k}.\\\\  \\end{align*}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1163723314873\">Notice that this answer confirms the calculation of the cross product in\u00a0<a class=\"autogenerated-content\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/2-4-the-cross-product#fs-id1163723497714\">Example 2.31<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Use determinant notation to find [latex]\\mathbf{a} \\times \\mathbf{b}[\/latex], where [latex]\\mathbf{a} = \\langle 8,2,3\\rangle[\/latex] and\u00a0[latex]\\mathbf{b} = \\langle -1,0,4\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q457345863\">Show Solution<\/span><\/p>\n<div id=\"q457345863\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]8\\mathbf{i} -35\\mathbf{j} +2\\mathbf{k}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to the above Try IT.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" src=\"\/\/plugin.3playmedia.com\/show?mf=7809775&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=umfZKb2xRMw&amp;video_target=tpm-plugin-ma5mr889-umfZKb2xRMw\" width=\"800px\" height=\"450px\" frameborder=\"0\" marginwidth=\"0px\" marginheight=\"0px\"><\/iframe><\/div>\n<div style=\"text-align: center;\">You can view the transcript for <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Calculus+3\/Calc+3+transcripts\/CP2.36_transcript.html\">\u201cCP 2.36\u201d here (opens in new window).<\/a><\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-5451\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>CP 2.30. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>CP 2.36. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 3. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":349141,"menu_order":17,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 3\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-3\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"CP 2.30\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 2.36\",\"author\":\"Ryan 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