{"id":520,"date":"2025-10-09T15:17:36","date_gmt":"2025-10-09T15:17:36","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=520"},"modified":"2026-05-26T14:46:41","modified_gmt":"2026-05-26T14:46:41","slug":"cross-product-and-applications-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/cross-product-and-applications-learn-it-1\/","title":{"raw":"Cross Product and Applications: Learn It 1","rendered":"Cross Product and Applications: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Calculate the cross product of two vectors and understand what it tells us<\/li>\r\n \t<li>Find a vector orthogonal to two given vectors<\/li>\r\n \t<li>Determine areas and volumes by using the cross product<\/li>\r\n \t<li>Calculate the torque of a given force and position vector<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>The Cross Product<\/h2>\r\n<p class=\"whitespace-normal break-words\">Imagine a mechanic turning a wrench to tighten a bolt. The mechanic applies force at the end of the wrench, creating rotation (torque) that tightens the bolt.<\/p>\r\n<p class=\"whitespace-normal break-words\">We can represent both the applied force and the radius from bolt to wrench end as vectors. The resulting torque is also a vector, but it points along the axis of rotation\u2014perpendicular to both the force vector and radius vector. How do we find a vector perpendicular to two given vectors? The <strong>cross product<\/strong> provides the answer.<\/p>\r\nThe dot product multiplies two vectors to get a scalar. The cross product multiplies two vectors to get a third vector that's orthogonal to both original vectors. Let's see how this works.\r\n\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p class=\"whitespace-normal break-words\">Given vectors [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], we want to find vector [latex]\\mathbf{w} = \\langle w_1, w_2, w_3 \\rangle[\/latex] such that:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\mathbf{u} \\cdot \\mathbf{w} = 0[\/latex] (orthogonal to [latex]\\mathbf{u}[\/latex])<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\mathbf{v} \\cdot \\mathbf{w} = 0[\/latex] (orthogonal to [latex]\\mathbf{v}[\/latex])<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-pre-wrap break-words\">This gives us the system of equations:<\/p>\r\n<p style=\"text-align: center;\">[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]<\/p>\r\nThrough algebraic manipulation (multiplying and eliminating variables), we find:\r\n<p style=\"text-align: center;\">[latex]w_1 = u_2v_3 - u_3v_2[\/latex]\r\n[latex]w_2 = -(u_1v_3 - u_3v_1) = u_3v_1 - u_1v_3[\/latex]\r\n[latex]w_3 = u_1v_2 - u_2v_1[\/latex]<\/p>\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]<\/p>\r\n\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">To eliminate [latex]w_3[\/latex], multiply the first equation by [latex]v_3[\/latex] and the second by [latex]u_3[\/latex], then subtract. This algebraic manipulation leads to the cross product formula<\/section><section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>cross product<\/h3>\r\n<p class=\"whitespace-normal break-words\">For vectors [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]:<\/p>\r\n&nbsp;\r\n<p class=\"whitespace-normal break-words\"><strong>Component Form<\/strong>:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\times \\mathbf{v} = \\langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \\rangle[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\"><strong>Standard Unit Vector Form<\/strong>:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\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}[\/latex]<\/p>\r\n\r\n<\/section>\r\n<h3>Orthogonality Check<\/h3>\r\nLet's verify that [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] is indeed orthogonal to [latex]\\mathbf{u}[\/latex] by computing their dot product:\r\n\r\n<center>[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>The terms cancel out perfectly! Similarly, we can show that [latex]\\mathbf{v} \\cdot (\\mathbf{u} \\times \\mathbf{v}) = 0[\/latex].\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\r\n<p class=\"whitespace-normal break-words\"><strong>Cross Product vs. Dot Product<\/strong>:<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Dot product<\/strong>: Two vectors \u2192 scalar<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Cross product<\/strong>: Two vectors \u2192 vector (orthogonal to both inputs)<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">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].<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=\"&quot;This\" \/>[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<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]325837[\/ohm_question]<\/section>\r\n<h3>Direction: The Right-Hand Rule<\/h3>\r\n<p class=\"whitespace-normal break-words\">The cross product formula tells us the components of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex], but how do we determine its direction? The <strong>right-hand rule<\/strong> provides the answer.<\/p>\r\n\r\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\r\n<p class=\"whitespace-normal break-words\"><strong>Right-Hand Rule Steps<\/strong>:<\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">Point your right hand fingers in the direction of the first vector [latex]\\mathbf{u}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Curl your fingers toward the second vector [latex]\\mathbf{v}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">Your thumb points in the direction of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex]<\/li>\r\n<\/ol>\r\n<\/section>[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]<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Order Matters<\/strong>: Try the right-hand rule for [latex]\\mathbf{v} \\times \\mathbf{u}[\/latex]. Start with fingers pointing toward [latex]\\mathbf{v}[\/latex], curl toward [latex]\\mathbf{u}[\/latex]\u2014your thumb points in the opposite direction! This shows that [latex]\\mathbf{u} \\times \\mathbf{v} = -(\\mathbf{v} \\times \\mathbf{u})[\/latex].<\/section><section class=\"textbox example\" aria-label=\"Example\">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.<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=\"&quot;This\" \/>[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]<\/section>\r\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Cross Products of Standard Unit Vectors<\/h3>\r\n<p class=\"whitespace-normal break-words\">Understanding how [latex]\\mathbf{i}[\/latex], [latex]\\mathbf{j}[\/latex], and [latex]\\mathbf{k}[\/latex] interact under cross multiplication simplifies many calculations.<\/p>\r\n<p class=\"whitespace-normal break-words\">When crossing a unit vector with itself:<\/p>\r\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{i} \\times \\mathbf{i} = \\mathbf{j} \\times \\mathbf{j} = \\mathbf{k} \\times \\mathbf{k} = \\mathbf{0}[\/latex]<\/p>\r\n<p class=\"whitespace-normal break-words\">This makes sense\u2014a vector crossed with itself has no meaningful perpendicular direction, so the result is the zero vector.<\/p>\r\n<p class=\"whitespace-normal break-words\">When crossing different unit vectors:<\/p>\r\n<p class=\"whitespace-normal break-words\">Since the cross product of two different unit vectors must be perpendicular to both, it must be parallel to the third unit vector. The right-hand rule determines whether it's positive or negative:<\/p>\r\n\r\n<center>[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><section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>unit vector cross products<\/h3>\r\n<p class=\"whitespace-normal break-words\"><strong>Positive Results<\/strong> (right-hand rule gives positive direction):<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\mathbf{i} \\times \\mathbf{j} = \\mathbf{k}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\mathbf{j} \\times \\mathbf{k} = \\mathbf{i}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\mathbf{k} \\times \\mathbf{i} = \\mathbf{j}[\/latex]<\/li>\r\n<\/ul>\r\n<p class=\"whitespace-normal break-words\"><strong>Negative Results<\/strong> (opposite order gives negative direction):<\/p>\r\n\r\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\mathbf{j} \\times \\mathbf{i} = -\\mathbf{k}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\mathbf{k} \\times \\mathbf{j} = -\\mathbf{i}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">[latex]\\mathbf{i} \\times \\mathbf{k} = -\\mathbf{j}[\/latex]<\/li>\r\n<\/ul>\r\n<\/section><section class=\"textbox proTip\" aria-label=\"Pro Tip\">Notice the cyclic pattern [latex]\\mathbf{i} \\to \\mathbf{j} \\to \\mathbf{k} \\to \\mathbf{i}[\/latex]. Going forward in the cycle gives positive results, backward gives negative results.<\/section><section class=\"textbox example\" aria-label=\"Example\">Find [latex]\\mathbf{i} \\times (\\mathbf{j} \\times \\mathbf{k})[\/latex].[reveal-answer q=\"734661173\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"734661173\"]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].[\/hidden-answer]\r\n\r\n<\/section>\r\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Properties of the Cross Product<\/h3>\r\nThe dot product and cross product are both forms of vector multiplication, but they have very different properties and applications. The dot product is called the <strong>scalar product<\/strong> because it produces a scalar, while the cross product is called the <strong>vector product<\/strong> because it produces a vector.\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>properties of the cross product:<\/h3>\r\n<p class=\"whitespace-normal break-words\">Let [latex]\\mathbf{u}[\/latex], [latex]\\mathbf{v}[\/latex], and [latex]\\mathbf{w}[\/latex] be vectors, and let [latex]c[\/latex] be a scalar.<\/p>\r\n\r\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\r\n \t<li class=\"whitespace-normal break-words\"><strong>Anticommutative<\/strong>: [latex]\\mathbf{u} \\times \\mathbf{v} = -(\\mathbf{v} \\times \\mathbf{u})[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Distributive<\/strong>: [latex]\\mathbf{u} \\times (\\mathbf{v} + \\mathbf{w}) = \\mathbf{u} \\times \\mathbf{v} + \\mathbf{u} \\times \\mathbf{w}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Scalar multiplication<\/strong>: [latex]c(\\mathbf{u} \\times \\mathbf{v}) = (c\\mathbf{u}) \\times \\mathbf{v} = \\mathbf{u} \\times (c\\mathbf{v})[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Zero vector<\/strong>: [latex]\\mathbf{u} \\times \\mathbf{0} = \\mathbf{0} \\times \\mathbf{u} = \\mathbf{0}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Self cross product<\/strong>: [latex]\\mathbf{v} \\times \\mathbf{v} = \\mathbf{0}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Scalar triple product<\/strong>: [latex]\\mathbf{u} \\times (\\mathbf{v} \\times \\mathbf{w}) \\neq (\\mathbf{u} \\times \\mathbf{v}) \\times \\mathbf{w}[\/latex] (in general)<\/li>\r\n<\/ol>\r\n<\/section><section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n\r\n\r\n<hr \/>\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\r\n<\/section><section class=\"textbox example\" aria-label=\"Example\">Use the cross product properties to calculate [latex](2\\mathbf{i} \\times 3\\mathbf{j}) \\times \\mathbf{j}[\/latex].[reveal-answer q=\"859323711\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"859323711\"]<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<\/section><section class=\"textbox example\" aria-label=\"Example\">Use\u00a0the properties of the cross product\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].[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<\/section>\r\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Magnitude of the Cross Product<\/h3>\r\nWe've focused on the direction of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex], but what about 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<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>magnitude of the cross product<\/h3>\r\nFor vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] with angle [latex]\\theta[\/latex] between them:\r\n<p style=\"text-align: center;\">[latex]||\\mathbf{u} \\times \\mathbf{v}|| = ||\\mathbf{u}|| \\cdot ||\\mathbf{v}|| \\cdot \\sin \\theta[\/latex]<\/p>\r\n\r\n<\/section><section class=\"textbox connectIt\">\r\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\r\n\r\n\r\n<hr \/>\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<\/section>\r\n<div data-type=\"equation\" data-label=\"\"><section class=\"textbox proTip\" aria-label=\"Pro Tip\">Compare this to the dot product formula [latex]\\mathbf{u} \\cdot \\mathbf{v} = ||\\mathbf{u}|| \\cdot ||\\mathbf{v}|| \\cdot \\cos \\theta[\/latex]. The cross product uses sine instead of cosine!<\/section>\r\n<p class=\"whitespace-normal break-words\">The cross product only works in three dimensions. In 2D, there's no direction simultaneously perpendicular to two nonparallel vectors\u2014you'd need to point \"out of the plane,\" which requires a third dimension. This geometric requirement explains why cross products are fundamental in 3D applications like torque, rotation, and magnetic fields, but don't exist in 2D vector algebra.<\/p>\r\n\r\n<\/div>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Calculate the cross product of two vectors and understand what it tells us<\/li>\n<li>Find a vector orthogonal to two given vectors<\/li>\n<li>Determine areas and volumes by using the cross product<\/li>\n<li>Calculate the torque of a given force and position vector<\/li>\n<\/ul>\n<\/section>\n<h2>The Cross Product<\/h2>\n<p class=\"whitespace-normal break-words\">Imagine a mechanic turning a wrench to tighten a bolt. The mechanic applies force at the end of the wrench, creating rotation (torque) that tightens the bolt.<\/p>\n<p class=\"whitespace-normal break-words\">We can represent both the applied force and the radius from bolt to wrench end as vectors. The resulting torque is also a vector, but it points along the axis of rotation\u2014perpendicular to both the force vector and radius vector. How do we find a vector perpendicular to two given vectors? The <strong>cross product<\/strong> provides the answer.<\/p>\n<p>The dot product multiplies two vectors to get a scalar. The cross product multiplies two vectors to get a third vector that&#8217;s orthogonal to both original vectors. Let&#8217;s see how this works.<\/p>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p class=\"whitespace-normal break-words\">Given vectors [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], we want to find vector [latex]\\mathbf{w} = \\langle w_1, w_2, w_3 \\rangle[\/latex] such that:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\mathbf{u} \\cdot \\mathbf{w} = 0[\/latex] (orthogonal to [latex]\\mathbf{u}[\/latex])<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\mathbf{v} \\cdot \\mathbf{w} = 0[\/latex] (orthogonal to [latex]\\mathbf{v}[\/latex])<\/li>\n<\/ul>\n<p class=\"whitespace-pre-wrap break-words\">This gives us the system of equations:<\/p>\n<p style=\"text-align: center;\">[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]<\/p>\n<p>Through algebraic manipulation (multiplying and eliminating variables), we find:<\/p>\n<p style=\"text-align: center;\">[latex]w_1 = u_2v_3 - u_3v_2[\/latex]<br \/>\n[latex]w_2 = -(u_1v_3 - u_3v_1) = u_3v_1 - u_1v_3[\/latex]<br \/>\n[latex]w_3 = u_1v_2 - u_2v_1[\/latex]<\/p>\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]<\/p>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">To eliminate [latex]w_3[\/latex], multiply the first equation by [latex]v_3[\/latex] and the second by [latex]u_3[\/latex], then subtract. This algebraic manipulation leads to the cross product formula<\/section>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>cross product<\/h3>\n<p class=\"whitespace-normal break-words\">For vectors [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]:<\/p>\n<p>&nbsp;<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Component Form<\/strong>:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\times \\mathbf{v} = \\langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \\rangle[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\"><strong>Standard Unit Vector Form<\/strong>:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\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}[\/latex]<\/p>\n<\/section>\n<h3>Orthogonality Check<\/h3>\n<p>Let&#8217;s verify that [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] is indeed orthogonal to [latex]\\mathbf{u}[\/latex] by computing their dot product:<\/p>\n<div style=\"text-align: center;\">[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>The terms cancel out perfectly! Similarly, we can show that [latex]\\mathbf{v} \\cdot (\\mathbf{u} \\times \\mathbf{v}) = 0[\/latex].<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">\n<p class=\"whitespace-normal break-words\"><strong>Cross Product vs. Dot Product<\/strong>:<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Dot product<\/strong>: Two vectors \u2192 scalar<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Cross product<\/strong>: Two vectors \u2192 vector (orthogonal to both inputs)<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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].<img decoding=\"async\" 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=\"&quot;This\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q934723610\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm325837\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=325837&theme=lumen&iframe_resize_id=ohm325837&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h3>Direction: The Right-Hand Rule<\/h3>\n<p class=\"whitespace-normal break-words\">The cross product formula tells us the components of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex], but how do we determine its direction? The <strong>right-hand rule<\/strong> provides the answer.<\/p>\n<section class=\"textbox questionHelp\" aria-label=\"Question Help\">\n<p class=\"whitespace-normal break-words\"><strong>Right-Hand Rule Steps<\/strong>:<\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">Point your right hand fingers in the direction of the first vector [latex]\\mathbf{u}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Curl your fingers toward the second vector [latex]\\mathbf{v}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Your thumb points in the direction of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex]<\/li>\n<\/ol>\n<\/section>\n<figure id=\"attachment_5138\" aria-describedby=\"caption-attachment-5138\" style=\"width: 666px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption 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.<\/figcaption><\/figure>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\"><strong>Order Matters<\/strong>: Try the right-hand rule for [latex]\\mathbf{v} \\times \\mathbf{u}[\/latex]. Start with fingers pointing toward [latex]\\mathbf{v}[\/latex], curl toward [latex]\\mathbf{u}[\/latex]\u2014your thumb points in the opposite direction! This shows that [latex]\\mathbf{u} \\times \\mathbf{v} = -(\\mathbf{v} \\times \\mathbf{u})[\/latex].<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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.<img decoding=\"async\" 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=\"&quot;This\" \/><\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q856612738\">Show Solution<\/button><\/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<figure id=\"attachment_5142\" aria-describedby=\"caption-attachment-5142\" style=\"width: 501px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" 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\" \/><figcaption 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.<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<\/section>\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Cross Products of Standard Unit Vectors<\/h3>\n<p class=\"whitespace-normal break-words\">Understanding how [latex]\\mathbf{i}[\/latex], [latex]\\mathbf{j}[\/latex], and [latex]\\mathbf{k}[\/latex] interact under cross multiplication simplifies many calculations.<\/p>\n<p class=\"whitespace-normal break-words\">When crossing a unit vector with itself:<\/p>\n<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{i} \\times \\mathbf{i} = \\mathbf{j} \\times \\mathbf{j} = \\mathbf{k} \\times \\mathbf{k} = \\mathbf{0}[\/latex]<\/p>\n<p class=\"whitespace-normal break-words\">This makes sense\u2014a vector crossed with itself has no meaningful perpendicular direction, so the result is the zero vector.<\/p>\n<p class=\"whitespace-normal break-words\">When crossing different unit vectors:<\/p>\n<p class=\"whitespace-normal break-words\">Since the cross product of two different unit vectors must be perpendicular to both, it must be parallel to the third unit vector. The right-hand rule determines whether it&#8217;s positive or negative:<\/p>\n<div style=\"text-align: center;\">[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<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>unit vector cross products<\/h3>\n<p class=\"whitespace-normal break-words\"><strong>Positive Results<\/strong> (right-hand rule gives positive direction):<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\mathbf{i} \\times \\mathbf{j} = \\mathbf{k}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\mathbf{j} \\times \\mathbf{k} = \\mathbf{i}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\mathbf{k} \\times \\mathbf{i} = \\mathbf{j}[\/latex]<\/li>\n<\/ul>\n<p class=\"whitespace-normal break-words\"><strong>Negative Results<\/strong> (opposite order gives negative direction):<\/p>\n<ul class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-disc space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\">[latex]\\mathbf{j} \\times \\mathbf{i} = -\\mathbf{k}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\mathbf{k} \\times \\mathbf{j} = -\\mathbf{i}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]\\mathbf{i} \\times \\mathbf{k} = -\\mathbf{j}[\/latex]<\/li>\n<\/ul>\n<\/section>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Notice the cyclic pattern [latex]\\mathbf{i} \\to \\mathbf{j} \\to \\mathbf{k} \\to \\mathbf{i}[\/latex]. Going forward in the cycle gives positive results, backward gives negative results.<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Find [latex]\\mathbf{i} \\times (\\mathbf{j} \\times \\mathbf{k})[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q734661173\">Show Solution<\/button><\/p>\n<div id=\"q734661173\" class=\"hidden-answer\" style=\"display: none\">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].<\/div>\n<\/div>\n<\/section>\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Properties of the Cross Product<\/h3>\n<p>The dot product and cross product are both forms of vector multiplication, but they have very different properties and applications. The dot product is called the <strong>scalar product<\/strong> because it produces a scalar, while the cross product is called the <strong>vector product<\/strong> because it produces a vector.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>properties of the cross product:<\/h3>\n<p class=\"whitespace-normal break-words\">Let [latex]\\mathbf{u}[\/latex], [latex]\\mathbf{v}[\/latex], and [latex]\\mathbf{w}[\/latex] be vectors, and let [latex]c[\/latex] be a scalar.<\/p>\n<ol class=\"[&amp;:not(:last-child)_ul]:pb-1 [&amp;:not(:last-child)_ol]:pb-1 list-decimal space-y-2.5 pl-7\">\n<li class=\"whitespace-normal break-words\"><strong>Anticommutative<\/strong>: [latex]\\mathbf{u} \\times \\mathbf{v} = -(\\mathbf{v} \\times \\mathbf{u})[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Distributive<\/strong>: [latex]\\mathbf{u} \\times (\\mathbf{v} + \\mathbf{w}) = \\mathbf{u} \\times \\mathbf{v} + \\mathbf{u} \\times \\mathbf{w}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Scalar multiplication<\/strong>: [latex]c(\\mathbf{u} \\times \\mathbf{v}) = (c\\mathbf{u}) \\times \\mathbf{v} = \\mathbf{u} \\times (c\\mathbf{v})[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Zero vector<\/strong>: [latex]\\mathbf{u} \\times \\mathbf{0} = \\mathbf{0} \\times \\mathbf{u} = \\mathbf{0}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Self cross product<\/strong>: [latex]\\mathbf{v} \\times \\mathbf{v} = \\mathbf{0}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Scalar triple product<\/strong>: [latex]\\mathbf{u} \\times (\\mathbf{v} \\times \\mathbf{w}) \\neq (\\mathbf{u} \\times \\mathbf{v}) \\times \\mathbf{w}[\/latex] (in general)<\/li>\n<\/ol>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q859323711\">Show Solution<\/button><\/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<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">Use\u00a0the properties of the cross product\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\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q466215391\">Show Solution<\/button><\/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<\/section>\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Magnitude of the Cross Product<\/h3>\n<p>We&#8217;ve focused on the direction of [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex], but what about 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<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>magnitude of the cross product<\/h3>\n<p>For vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] with angle [latex]\\theta[\/latex] between them:<\/p>\n<p style=\"text-align: center;\">[latex]||\\mathbf{u} \\times \\mathbf{v}|| = ||\\mathbf{u}|| \\cdot ||\\mathbf{v}|| \\cdot \\sin \\theta[\/latex]<\/p>\n<\/section>\n<section class=\"textbox connectIt\">\n<p style=\"text-align: center;\"><strong>Proof<\/strong><\/p>\n<hr \/>\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<\/section>\n<div data-type=\"equation\" data-label=\"\">\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Compare this to the dot product formula [latex]\\mathbf{u} \\cdot \\mathbf{v} = ||\\mathbf{u}|| \\cdot ||\\mathbf{v}|| \\cdot \\cos \\theta[\/latex]. The cross product uses sine instead of cosine!<\/section>\n<p class=\"whitespace-normal break-words\">The cross product only works in three dimensions. In 2D, there&#8217;s no direction simultaneously perpendicular to two nonparallel vectors\u2014you&#8217;d need to point &#8220;out of the plane,&#8221; which requires a third dimension. This geometric requirement explains why cross products are fundamental in 3D applications like torque, rotation, and magnetic fields, but don&#8217;t exist in 2D vector algebra.<\/p>\n<\/div>\n","protected":false},"author":15,"menu_order":4,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-520","chapter","type-chapter","status-publish","hentry"],"part":422,"module-header":"learn_it","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/520","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/520\/revisions"}],"predecessor-version":[{"id":2882,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/520\/revisions\/2882"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/422"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/520\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=520"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=520"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=520"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=520"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}