{"id":810,"date":"2021-08-27T19:55:38","date_gmt":"2021-08-27T19:55:38","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=810"},"modified":"2022-10-21T00:18:00","modified_gmt":"2022-10-21T00:18:00","slug":"using-the-cross-product","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/using-the-cross-product\/","title":{"raw":"Using the Cross Product","rendered":"Using 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\">Find a vector orthogonal to two given vectors.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Determine areas and volumes by using the cross product.<\/span><\/li>\r\n \t<li><span class=\"os-abstract-content\">Calculate the torque of a given force and position vector.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2 data-type=\"title\">Using the Cross Product<\/h2>\r\n<p id=\"fs-id1163723339475\">The cross product is very useful for several types of calculations, including finding a vector orthogonal to two given vectors, computing areas of triangles and parallelograms, and even determining the volume of the three-dimensional geometric shape made of parallelograms known as a\u00a0<em data-effect=\"italics\">parallelepiped<\/em>. The following examples illustrate these calculations.<\/p>\r\n\r\n<div id=\"fs-id1163723339486\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding a unit vector orthogonal to two given vectors<\/h3>\r\nLet [latex]\\mathbf{a} = \\langle 5,2,-1\\rangle[\/latex] and\u00a0[latex]\\mathbf{b} = \\langle 0,-1,4\\rangle[\/latex]. Find a unit vector orthogonal to both [latex]\\textbf a[\/latex] and\u00a0[latex]\\textbf b[\/latex].\r\n\r\n[reveal-answer q=\"27965428\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"27965428\"]\r\n<p id=\"fs-id1163724138497\">The cross product [latex]\\mathbf{a} \\times \\mathbf{b}[\/latex] is orthogonal to both vectors [latex]\\textbf a[\/latex] and [latex]\\textbf b[\/latex]. We can calculate it with a determinant:<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{a} \\times \\mathbf{b} &amp;= \\begin{vmatrix} \\mathbf{i} &amp; \\mathbf{j} &amp; \\mathbf{k} \\\\ 5 &amp; 2 &amp; -1 \\\\ 0 &amp; -1 &amp; 4\\end{vmatrix} = \\begin{vmatrix} 2 &amp; -1 \\\\ -1 &amp; 4 \\end{vmatrix}\\mathbf{i} - \\begin{vmatrix} 5 &amp; -1 \\\\ 0 &amp; 4 \\end{vmatrix}\\mathbf{j} +\\begin{vmatrix} 5 &amp; 2 \\\\ 0 &amp; -1 \\end{vmatrix}\\mathbf{k} \\\\\r\n&amp;=(8-1)\\mathbf{i} -(20-0)\\mathbf{j} +(-5-0)\\mathbf{k}\\\\\r\n&amp;= 7\\mathbf{i} - 20\\mathbf{j} -5\\mathbf{k}.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>&nbsp;\r\n<p id=\"fs-id1163723341785\">Normalize this vector to find a unit vector in the same direction:<\/p>\r\n\r\n<center>[latex]||\\mathbf{a} \\times \\mathbf{b}|| = \\sqrt{(7)^2 +(-20)^2 +(-5)^2} = \\sqrt{474}[\/latex].<\/center>\r\n<p id=\"fs-id1163723171302\">Thus, [latex]\\langle \\frac{7}{\\sqrt{474}},\\frac{-20}{\\sqrt{474}},\\frac{-5}{\\sqrt{474}}\\rangle[\/latex] is a unit vector orthogonal to [latex]\\textbf a[\/latex] and\u00a0[latex]\\textbf b[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind a unit vector orthogonal to both [latex]\\textbf a[\/latex] and [latex]\\textbf b[\/latex], where [latex]\\mathbf{a} = \\langle 4,0,3\\rangle[\/latex] and\u00a0[latex]\\mathbf{b} = \\langle 1,1,4\\rangle[\/latex].\r\n\r\n[reveal-answer q=\"748528349\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"748528349\"]\r\n\r\n[latex]\\langle \\frac{-3}{\\sqrt{194}},\\frac{-13}{\\sqrt{194}},\\frac{4}{\\sqrt{194}}\\rangle[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\nTo use the cross product for calculating areas, we state and prove the following theorem.\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: area of a parallelogram<\/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-id1163723995111\">If we locate vectors [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] such that they form adjacent sides of a parallelogram, then the area of the parallelogram is given by [latex]||\\mathbf{u} \\times \\mathbf{v}||[\/latex] (Figure 1).<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n[caption id=\"attachment_5144\" align=\"aligncenter\" width=\"422\"]<img class=\"size-full wp-image-5144\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27210440\/Figure-2.57.jpg\" alt=\"This figure is a parallelogram. One side is represented with a vector labeled \u201cv.\u201d The second side, the base, has the same initial point as vector v and is labeled \u201cu.\u201d The angle between u and v is theta. Also, a perpendicular line segment is drawn from the terminal point of v to vector u. It is labeled \u201c|v|sin(theta).\u201d\" width=\"422\" height=\"259\" \/> Figure 1. The parallelogram with adjacent sides [latex]{\\bf{u}}[\/latex] and [latex]{\\bf{v}}[\/latex] has base [latex]\u2225{\\bf{u}}\u2225[\/latex] and height [latex]\u2225{\\bf{v}}\u2225\\sin\\theta[\/latex].[\/caption]\r\n<h3 data-type=\"title\">Proof<\/h3>\r\n<p id=\"fs-id1163723995254\">We show that the magnitude of the cross product is equal to the base times height of the parallelogram.<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\text{Area of parallelogram} &amp;= \\text{base} \\times \\text{height} \\\\\r\n&amp;=||\\mathbf{u}||(||\\mathbf{v}||\\sin{\\theta})\\\\\r\n&amp;= ||\\mathbf{u} \\times \\mathbf{v}||\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>&nbsp;\r\n\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding the area of a triangle<\/h3>\r\nLet [latex]P=(1, 0, 0)[\/latex], [latex]Q(0, 1, 0)[\/latex], and [latex]R=(0, 0, 1)[\/latex] be the vertices of a triangle (Figure 2). Find its area.\r\n\r\n[caption id=\"attachment_5147\" align=\"aligncenter\" width=\"501\"]<img class=\"size-full wp-image-5147\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27210841\/Figure-2.58.jpg\" alt=\"This figure is the 3-dimensional coordinate system. It has a triangle drawn in the first octant. The vertices of the triangle are points P(1, 0, 0); Q(0, 1, 0); and R(0, 0, 1).\" width=\"501\" height=\"497\" \/> Figure 2. Finding the area of a triangle by using the cross product.[\/caption]\r\n\r\n[reveal-answer q=\"452836520\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"452836520\"]\r\nWe have [latex]\\overrightarrow{PQ} = \\langle 0-1,1-0,0-0\\rangle = \\langle -1,1,0\\rangle[\/latex] and [latex]\\overrightarrow{PR} = \\langle 0-1,0-0,1-0\\rangle = \\langle -1,0,1\\rangle[\/latex]. The area of the parallelogram with adjacent sides [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{PR}[\/latex] is given by\u00a0[latex]||\\overrightarrow{PQ} \\times \\overrightarrow{PR}||[\/latex]:\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\overrightarrow{PQ} \\times \\overrightarrow{PR} &amp;= \\begin{vmatrix} \\mathbf{i} &amp; \\mathbf{j} &amp; \\mathbf{k} \\\\ -1 &amp; 1 &amp; 0 \\\\ -1 &amp; 0 &amp; 1\\end{vmatrix} = (1-0)\\mathbf{i} -(-1-0)\\mathbf{j} +(0-(-1))\\mathbf{k} = \\mathbf{i} +\\mathbf{j} +\\mathbf{k} \\\\\r\n||\\overrightarrow{PQ} \\times \\overrightarrow{PR}||&amp;= ||\\langle 1,1,1\\rangle|| = \\sqrt{1^2 +1^2 +1^2} = \\sqrt{3}.\r\n\\end{align*}\r\n[\/latex]<\/center>\r\n<p id=\"fs-id1163723499884\">The area of [latex]\\Delta PQR[\/latex] is half the area of the parallelogram, or [latex]\\frac{\\sqrt{3}}{2}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the area of the parallelogram [latex]PQRS[\/latex] with vertices\u00a0[latex]P(1, 1, 0)[\/latex],\u00a0[latex]Q(7, 1, 0)[\/latex],\u00a0[latex]R(9, 4, 2)[\/latex], and\u00a0[latex]S(3, 4, 2)[\/latex].\r\n\r\n[reveal-answer q=\"387493274\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"387493274\"]\r\n\r\n[latex]6\\sqrt{13}[\/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=7753559&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=L2XRUTcifoM&amp;video_target=tpm-plugin-mi2dpg43-L2XRUTcifoM\" 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\/CP2.38_transcript.html\">\u201cCP 2.38\u201d here (opens in new window).<\/a><\/center>\r\n<h2 data-type=\"title\">The Triple Scalar Product<\/h2>\r\n<p id=\"fs-id1163723326759\">Because the cross product of two vectors is a vector, it is possible to combine the dot product and the cross product. The dot product of a vector with the cross product of two other vectors is called the triple scalar product because the result is a scalar.<\/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=\"31\" 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-id1163723326769\">The\u00a0<strong><span id=\"d53b2426-3a76-4d95-af4f-6532ad8e4e9a_term84\" data-type=\"term\">triple scalar product<\/span>\u00a0<\/strong>of vectors [latex]\\textbf u[\/latex],\u00a0[latex]\\textbf v[\/latex], and\u00a0[latex]\\textbf w[\/latex] is\u00a0[latex]{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: calculating a triple scalar product<\/span><\/h3>\r\n<\/header><section>\r\n<div class=\"os-note-body\">\r\n\r\n<hr \/>\r\n<p id=\"fs-id1163723326861\">The triple scalar product of vectors [latex]{\\bf{u}}=u_1{\\bf{i}}+u_2{\\bf{j}}+u_3{\\bf{k}}[\/latex], [latex]{\\bf{v}}=v_1{\\bf{i}}+v_2{\\bf{j}}+v_3{\\bf{k}}[\/latex], and [latex]{\\bf{w}}=w_1{\\bf{i}}+w_2{\\bf{j}}+w_3{\\bf{k}}[\/latex] is the determinant of the [latex]3\\times3[\/latex] matrix formed by the components of the vectors:<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})=\\begin{vmatrix}u_1&amp;u_2&amp;u_3 \\\\ v_1&amp;v_2&amp;v_3 \\\\w_1&amp;w_2&amp;w_3 \\end{vmatrix}[\/latex].<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n<h3 data-type=\"title\">Proof<\/h3>\r\n<p id=\"fs-id1163723477915\">The calculation is straightforward.<\/p>\r\n\r\n<div id=\"fs-id1163723477918\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\r\n<div style=\"text-align: center;\" data-type=\"example\">[latex]\\large{\\begin{aligned}\r\n{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&amp;=\\langle u_1,u_2,u_3\\rangle\\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_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;=\\begin{vmatrix}u_1&amp;u_2&amp;u_3 \\\\ v_1&amp;v_2&amp;v_3 \\\\w_1&amp;w_2&amp;w_3 \\end{vmatrix}.\r\n\\end{aligned}}[\/latex]<\/div>\r\n<div data-type=\"example\">[latex]_\\blacksquare[\/latex]<\/div>\r\n<div data-type=\"example\"><\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: calculating the triple scalar product<\/h3>\r\nLet [latex]{\\bf{u}}=\\langle1,3,5\\rangle[\/latex], [latex]{\\bf{v}}=\\langle2,-1,0\\rangle[\/latex], and [latex]{\\bf{w}}=\\langle-3,0,-1\\rangle[\/latex]. Calculate the triple scalar product [latex]{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})[\/latex].\r\n\r\n[reveal-answer q=\"348759826\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"348759826\"]\r\n\r\nApply\u00a0Theorem: Calculating a Triple Scalar Product directly:\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&amp;=\\begin{vmatrix}1&amp;3&amp;5 \\\\ 2&amp;-1&amp;0 \\\\ -3&amp;0&amp;-1 \\end{vmatrix} \\\\\r\n&amp;=1\\begin{vmatrix}-1&amp;0 \\\\ 0&amp;-1 \\end{vmatrix}-3\\begin{vmatrix} 2&amp;0 \\\\-3&amp;-1 \\end{vmatrix}+5\\begin{vmatrix} 2&amp;-1 \\\\-3&amp;0 \\end{vmatrix} \\\\\r\n&amp;=(1-0)-3(-2-0)+5(0-3) \\\\\r\n&amp;=1+6-15=-8.\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nCalculate the triple scalar product [latex]{\\bf{a}}\\cdot({\\bf{b}}\\times{\\bf{c}})[\/latex], where [latex]{\\bf{a}}=\\langle2,-4,1\\rangle[\/latex],\u00a0[latex]{\\bf{b}}=\\langle0,\r\n\r\n3,-1\\rangle[\/latex], and\u00a0[latex]{\\bf{c}}=\\langle5,-3,3\\rangle[\/latex].\r\n\r\n[reveal-answer q=\"849527436\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"849527436\"]\r\n\r\n[latex]17[\/latex]\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1163724060988\">When we create a matrix from three vectors, we must be careful about the order in which we list the vectors. If we list them in a matrix in one order and then rearrange the rows, the absolute value of the determinant remains unchanged. However, each time two rows switch places, the determinant changes sign:<\/p>\r\n<p style=\"text-align: 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}=d \\quad\\begin{vmatrix}b_1&amp;b_2&amp;b_3 \\\\a_1&amp;a_2&amp;a_3 \\\\ c_1&amp;c_2&amp;c_3\\end{vmatrix}=-d \\quad \\begin{vmatrix}b_1&amp;b_2&amp;b_3 \\\\c_1&amp;c_2&amp;c_3 \\\\ a_1&amp;a_2&amp;a_3\\end{vmatrix}=d \\quad\\begin{vmatrix}c_1&amp;c_2&amp;c_3 \\\\b_1&amp;b_2&amp;b_3 \\\\ a_1&amp;a_2&amp;a_3\\end{vmatrix}=-d[\/latex].<\/p>\r\n<p id=\"fs-id1163724153222\">Verifying this fact is straightforward, but rather messy. Let\u2019s take a look at this with an example:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\begin{vmatrix}1&amp;2&amp;1 \\\\ -2&amp;0&amp;3 \\\\4&amp;1&amp;-1\\end{vmatrix} &amp;=\\begin{vmatrix}0&amp;3 \\\\1&amp;-1\\end{vmatrix}-2\\begin{vmatrix}-2&amp;3 \\\\4&amp;-1\\end{vmatrix}+\\begin{vmatrix}-2&amp;0 \\\\4&amp;1\\end{vmatrix} \\\\\r\n&amp;=(0-3)-2(2-12)+(-2-0)=-3+20-2=15.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1163723979960\">Switching the top two rows we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\begin{vmatrix}-2&amp;0&amp;3 \\\\1&amp;2&amp;1\\\\ 4&amp;1&amp;-1\\end{vmatrix} &amp;=-2\\begin{vmatrix}2&amp;1 \\\\1&amp;-1\\end{vmatrix}+3\\begin{vmatrix}1&amp;2 \\\\4&amp;1\\end{vmatrix} \\\\\r\n&amp;=-2(-2-1)+3(1-8)=6-21=-15.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1163723124321\">Rearranging vectors in the triple products is equivalent to reordering the rows in the matrix of the determinant. Let [latex]\r\n{\\bf{u}}=u_1{\\bf{i}}+u_2{\\bf{j}}+u_3{\\bf{k}}[\/latex], [latex]{\\bf{v}}=v_1{\\bf{i}}+v_2{\\bf{j}}+v_3{\\bf{k}}[\/latex], and [latex]{\\bf{w}}=w_1{\\bf{i}}+w_2{\\bf{j}}+w_3{\\bf{k}}[\/latex]. Applying\u00a0Theorem: Calculating a Triple Scalar Product, we have<\/p>\r\n<p style=\"text-align: center;\">[latex]{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})=\\begin{vmatrix}u_1&amp;u_2&amp;u_3 \\\\ v_1&amp;v_2&amp;v_3 \\\\w_1&amp;w_2&amp;w_3\\end{vmatrix}\\text{ and }{\\bf{u}}\\cdot({\\bf{w}}\\times{\\bf{v}})=\\begin{vmatrix}u_1&amp;u_2&amp;u_3 \\\\w_1&amp;w_2&amp;w_3 \\\\v_1&amp;v_2&amp;v_3\\end{vmatrix}[\/latex].<\/p>\r\n<p id=\"fs-id1163724055509\">We can obtain the determinant for calculating [latex]{\\bf{u}}\\cdot({\\bf{w}}\\times{\\bf{v}})[\/latex] by switching the bottom two rows of [latex]{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})[\/latex]. Therefore, [latex]{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})=-{\\bf{u}}\\cdot({\\bf{w}}\\times{\\bf{v}})[\/latex].<\/p>\r\n<p id=\"fs-id1163724055682\">Following this reasoning and exploring the different ways we can interchange variables in the triple scalar product lead to the following identities:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&amp;=-{\\bf{u}}\\cdot({\\bf{w}}\\times{\\bf{v}}) \\\\\r\n{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&amp;={\\bf{v}}\\cdot({\\bf{w}}\\times{\\bf{u}})={\\bf{w}}\\cdot({\\bf{u}}\\times{\\bf{v}}).\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1163724042230\">Let [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] be two vectors in standard position. If [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] are not scalar multiples of each other, then these vectors form adjacent sides of a parallelogram. We saw in\u00a0Theorem: Area of a Parallelogram\u00a0that the area of this parallelogram is [latex]||{\\bf{u}}\\times{\\bf{v}}||[\/latex]. Now suppose we add a third vector [latex]\\textbf w[\/latex] that does not lie in the same plane as [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] but still shares the same initial point. Then these vectors form three edges of a\u00a0<span id=\"d53b2426-3a76-4d95-af4f-6532ad8e4e9a_term85\" data-type=\"term\">parallelepiped<\/span>, a three-dimensional prism with six faces that are each parallelograms, as shown in\u00a0Figure 3. The volume of this prism is the product of the figure\u2019s height and the area of its base. The triple scalar product of [latex]\\textbf u[\/latex], [latex]\\textbf v[\/latex], and [latex]\\textbf w[\/latex] provides a simple method for calculating the volume of the parallelepiped defined by these vectors.<\/p>\r\n\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: volume of a parallelepiped<\/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-id1163724035012\">The volume of a parallelepiped with adjacent edges given by the vectors [latex]\\textbf u[\/latex], [latex]\\textbf v[\/latex], and [latex]\\textbf w[\/latex] is the absolute value of the triple scalar product:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\large{V=|{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})|}[\/latex].<\/p>\r\n<p id=\"fs-id1163723289962\">See\u00a0Figure 3.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\nNote that, as the name indicates, the triple scalar product produces a scalar. The volume formula just presented uses the absolute value of a scalar quantity.\r\n\r\n[caption id=\"attachment_5149\" align=\"aligncenter\" width=\"415\"]<img class=\"size-full wp-image-5149\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27210947\/Figure-2.59.jpg\" alt=\"This figure is a parallelepided, a three dimensional parallelogram. Three of the sides are represented with vectors. The base has vectors v and w. The vertical side has vector u. All three vectors have the same initial point. A perpendicular vector is drawn from this common point. It is labeled \u201cproj sub (v x w) u.\u201d\" width=\"415\" height=\"201\" \/> Figure 3. The height of the parallelepiped is given by [latex]||\\text{proj}_{{\\bf{v}}\\times{\\bf{w}}}{\\bf{u}}||[\/latex].[\/caption]\r\n<h3 data-type=\"title\">Proof<\/h3>\r\n<p id=\"fs-id1163723290028\">The area of the base of the parallelepiped is given by [latex]||{\\bf{v}}\\times{\\bf{w}}||[\/latex]. The height of the figure is given by [latex]||\\text{proj}_{{\\bf{v}}\\times{\\bf{w}}}{\\bf{u}}||[\/latex]. The volume of the parallelepiped is the product of the height and the area of the base, so we have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\nv&amp;=||\\text{proj}_{{\\bf{v}}\\times{\\bf{w}}}{\\bf{u}}|| \\ ||{\\bf{v}}\\times{\\bf{w}}|| \\\\\r\n&amp;=\\left|\\frac{{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})}{||{\\bf{v}}\\times{\\bf{w}}||}\\right| \\||{\\bf{v}}\\times{\\bf{w}}|| \\\\\r\n&amp;=|{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})|.\r\n\\end{aligned}[\/latex]<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: calculating the volume of a parallelepiped<\/h3>\r\nLet [latex]{\\bf{u}}=\\langle-1,-2,1\\rangle[\/latex],\u00a0[latex]{\\bf{v}}=\\langle4,3,2\\rangle[\/latex], and\u00a0[latex]{\\bf{w}}=\\langle0,-5,-2\\rangle[\/latex]. Find the volume of the parallelepiped with adjacent edges [latex]\\textbf u[\/latex], [latex]\\textbf v[\/latex], and [latex]\\textbf w[\/latex] (Figure 4).\r\n\r\n[caption id=\"attachment_5151\" align=\"aligncenter\" width=\"417\"]<img class=\"size-full wp-image-5151\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27211252\/Figure-2.60.jpg\" alt=\"This figure is the 3-dimensional coordinate system. It has three vectors in standard position. The vectors are u = &lt;-1, -2, 1&gt;; v = &lt;4, 3, 2&gt;; and w = &lt;0, -5, -2&gt;.\" width=\"417\" height=\"497\" \/> Figure 4. A parallelepiped with adjacent edges [latex]\\textbf u[\/latex], [latex]\\textbf v[\/latex], and [latex]\\textbf w[\/latex][\/caption][reveal-answer q=\"384527634\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"384527634\"]\r\n<p id=\"fs-id1163724159278\">We have<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&amp;=\\begin{vmatrix}-1&amp;-2&amp;1 \\\\4&amp;3&amp;2 \\\\0&amp;-5&amp;-2\\end{vmatrix}=(-1)\\begin{vmatrix}3&amp;2 \\\\-5&amp;-2\\end{vmatrix}+2\\begin{vmatrix}4&amp;2 \\\\ 0&amp;-2\\end{vmatrix}+\\begin{vmatrix} 4&amp;3 \\\\ 0&amp;-5\\end{vmatrix} \\\\\r\n&amp;=(-1)(-6+10)+2(-8-0)+(-20-0) \\\\\r\n&amp;=-4-16-20 \\\\\r\n&amp;=-40.\r\n\\end{aligned}[\/latex]<\/p>\r\nThus, the volume of the parallelepiped is [latex]|-40|=40[\/latex] units<sup>3<\/sup>.\r\n<div id=\"fs-id1163724159281\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nFind the volume of the parallelepiped formed by the vectors\u00a0[latex]{\\bf{a}}=3{\\bf{i}}+4{\\bf{j}}-{\\bf{k}}[\/latex],\u00a0[latex]{\\bf{b}}=2{\\bf{i}}-{\\bf{j}}-{\\bf{k}}[\/latex], and\u00a0[latex]{\\bf{c}}=3{\\bf{j}}+{\\bf{k}}[\/latex].\r\n\r\n[reveal-answer q=\"983459723\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"983459723\"]\r\n\r\n[latex]8[\/latex] units<sup>3<\/sup>.\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=7753560&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=4Jv4U5G3qxA&amp;video_target=tpm-plugin-wayy3i76-4Jv4U5G3qxA\" 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\/CP2.40_transcript.html\">\u201cCP 2.40\u201d here (opens in new window).<\/a><\/center>\r\n<h2 data-type=\"title\">Applications of the Cross Product<\/h2>\r\n<p id=\"fs-id1163724163689\">The cross product appears in many practical applications in mathematics, physics, and engineering. Let\u2019s examine some of these applications here, including the idea of torque, with which we began this section. Other applications show up in later chapters, particularly in our study of vector fields such as gravitational and electromagnetic fields (<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/why-it-matters-vector-calculus\/\" data-page-uuid=\"41358830-7753-4f50-af9d-ec46020d7b0c\" data-page-fragment=\"page_41358830-7753-4f50-af9d-ec46020d7b0c\">Introduction to Vector Calculus<\/a>).<\/p>\r\n\r\n<\/div>\r\n<div data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: using the triple scalar product<\/h3>\r\nUse the triple scalar product to show that vectors [latex]{\\bf{u}}=\\langle2,0,5\\rangle[\/latex], [latex]{\\bf{v}}=\\langle2,2,4\\rangle[\/latex], and [latex]{\\bf{w}}=\\langle1,-1,3\\rangle[\/latex] are coplanar\u2014that is, show that these vectors lie in the same plane.\r\n\r\n[reveal-answer q=\"094352634\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"094352634\"]\r\n<p id=\"fs-id1163724182515\">Start by calculating the triple scalar product to find the volume of the parallelepiped defined by [latex]\\textbf u[\/latex],\u00a0[latex]\\textbf v[\/latex], and\u00a0[latex]\\textbf w[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&amp;=\\begin{vmatrix}2&amp;0&amp;5\\\\ 2&amp;2&amp;4 \\\\ 1&amp;-1&amp;3\\end{vmatrix} \\\\\r\n&amp;=[2(2)(3)+(0)(4)(1)+5(2)(-1)]-[5(2)(1)+(2)(4)(-1)+(0)(2)(3)] \\\\&amp;=2-2 \\\\\r\n&amp;=0.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1163724130658\">The volume of the parallelepiped is [latex]0[\/latex] units<sup>3<\/sup>, so one of the dimensions must be zero. Therefore, the three vectors all lie in the same plane.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nAre the vectors [latex]{\\bf{a}}={\\bf{i}}+{\\bf{j}}-{\\bf{k}}[\/latex], [latex]{\\bf{b}}={\\bf{i}}-{\\bf{j}}+{\\bf{k}}[\/latex], and [latex]{\\bf{c}}={\\bf{i}}+{\\bf{j}}+{\\bf{k}}[\/latex] coplanar?\r\n\r\n[reveal-answer q=\"308475728\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"308475728\"]\r\n\r\nNo, the triple scalar product is [latex]-4\\ne0[\/latex], so the three vectors form the adjacent edges of a parallelepiped. They are not coplanar.\r\n\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example: finding an orthogonal vector<\/h3>\r\nOnly a single plane can pass through any set of three noncolinear points. Find a vector orthogonal to the plane containing points [latex]P=(9, -3, -2)[\/latex],\u00a0[latex]Q=(1, 3, 0)[\/latex], and\u00a0[latex]R=(-2,5,0)[\/latex].\r\n\r\n[reveal-answer q=\"314652076\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"314652076\"]\r\n<p id=\"fs-id1163723475226\">The plane must contain vectors[latex]\\overrightarrow{PQ}[\/latex] and\u00a0[latex]\\overrightarrow{QR}[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\overrightarrow{PQ}&amp;=\\langle1-9,3-(-3),0-(-2)\\rangle =\\langle-8,6,2\\rangle \\\\\r\n\\overrightarrow{QR}&amp;=\\langle-2-1,5-3,0-0\\rangle=\\langle-3,2,0\\rangle.\r\n\\end{aligned}[\/latex]<\/p>\r\n<p id=\"fs-id1163724062686\">The cross product [latex]\\overrightarrow{PQ}\\times\\overrightarrow{QR}[\/latex] produces a vector orthogonal to both [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{QR}[\/latex]. Therefore, the cross product is orthogonal to the plane that contains these two vectors:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\begin{aligned}\r\n\\overrightarrow{PQ}\\times\\overrightarrow{QR}&amp;=\\begin{vmatrix} {\\bf{i}}&amp;{\\bf{j}}&amp;{\\bf{k}} \\\\ -8&amp;6&amp;2 \\\\ -3&amp;2&amp;0\\end{vmatrix} \\\\\r\n&amp;=0{\\bf{i}}-6{\\bf{j}}-16{\\bf{k}}-(-18{\\bf{k}}+4{\\bf{i}}+0{\\bf{j}}) \\\\\r\n&amp;=-4{\\bf{i}}-6{\\bf{j}}+2{\\bf{k}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<p id=\"fs-id1163723336017\">We have seen how to use the triple scalar product and how to find a vector orthogonal to a plane. Now we apply the cross product to real-world situations.<\/p>\r\n<p id=\"fs-id1163723336021\">Sometimes a force causes an object to rotate. For example, turning a screwdriver or a wrench creates this kind of rotational effect, called torque.<\/p>\r\n\r\n<div id=\"fs-id1163723336026\" class=\"ui-has-child-title\" data-type=\"note\">\r\n<div class=\"textbox shaded\"><header>\r\n<h3 class=\"os-title\" style=\"text-align: center;\" data-type=\"title\"><span id=\"43\" 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-id1163723336031\"><strong><span id=\"d53b2426-3a76-4d95-af4f-6532ad8e4e9a_term86\" data-type=\"term\">Torque<\/span><\/strong>, [latex]\\tau[\/latex] (the Greek letter\u00a0<em data-effect=\"italics\">tau<\/em>), measures the tendency of a force to produce rotation about an axis of rotation. Let [latex]\\textbf r[\/latex] be a vector with an initial point located on the axis of rotation and with a terminal point located at the point where the force is applied, and let vector [latex]\\textbf F[\/latex] represent the force. Then torque is equal to the cross product of [latex]\\textbf r[\/latex] and\u00a0[latex]\\textbf F[\/latex]:<\/p>\r\n<p style=\"text-align: center;\">[latex]\\tau={\\bf{r}}\\times{\\bf{F}}[\/latex].<\/p>\r\n<p id=\"fs-id1163723149206\">See\u00a0Figure 5.<\/p>\r\n\r\n<\/div>\r\n<\/section><\/div>\r\n\r\n[caption id=\"attachment_5153\" align=\"aligncenter\" width=\"469\"]<img class=\"size-full wp-image-5153\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27211439\/Figure-2.61.jpg\" alt=\"This figure has a vector r from an \u201caxis of rotation\u201d. At the terminal point of r there is a vector labeled \u201cF\u201d. The angle between r and F is theta.\" width=\"469\" height=\"176\" \/> Figure 5. Torque measures how a force causes an object to rotate.[\/caption]\r\n<p id=\"fs-id1163723149234\">Think about using a wrench to tighten a bolt. The torque [latex]\\tau[\/latex] applied to the bolt depends on how hard we push the wrench (force) and how far up the handle we apply the force (distance). The torque increases with a greater force on the wrench at a greater distance from the bolt. Common units of torque are the newton-meter or foot-pound. Although torque is dimensionally equivalent to work (it has the same units), the two concepts are distinct. Torque is used specifically in the context of rotation, whereas work typically involves motion along a line.<\/p>\r\n\r\n<div id=\"fs-id1163723149246\" class=\"ui-has-child-title\" data-type=\"example\">\r\n<div class=\"textbox exercises\">\r\n<h3>Example: evaluating torque<\/h3>\r\nA bolt is tightened by applying a force of [latex]6[\/latex] N to a [latex]0.15[\/latex]-m wrench (Figure 6). The angle between the wrench and the force vector is [latex]40^{\\small\\circ}[\/latex]. Find the magnitude of the torque about the center of the bolt. Round the answer to two decimal places.\r\n\r\n[caption id=\"attachment_5155\" align=\"aligncenter\" width=\"281\"]<img class=\"size-full wp-image-5155\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27211539\/Figure-2.62.jpg\" alt=\"This figure is the image of an open-end wrench. The length of the wrench is labeled \u201c0.15 m.\u201d The angle the wrench makes with a vertical vector is 40 degrees. The vector is labeled with \u201c6 N.\u201d\" width=\"281\" height=\"322\" \/> Figure 6. Torque describes the twisting action of the wrench.[\/caption]\r\n\r\n[reveal-answer q=\"853047651\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"853047651\"]\r\n\r\nSubstitute the given information into the equation defining torque:\r\n<p style=\"text-align: center;\">[latex]||\\tau||=||{\\bf{r}}\\times{\\bf{F}}||=||{\\bf{r}}|| \\ ||{\\bf{F}}||\\sin\\theta=(0.15\\text{ m})(6\\text{ N})\\sin{40^{\\small\\circ}}\\approx0.58\\text{N}\\cdot\\text{m}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>try it<\/h3>\r\nCalculate the force required to produce [latex]15\\text{N}\\cdot\\text{m}[\/latex] torque at an angle of [latex]30^{\\small\\circ}[\/latex] from a [latex]150[\/latex]-cm rod.\r\n\r\n[reveal-answer q=\"473650281\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"473650281\"]\r\n\r\n[latex]20[\/latex] N.\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=7753561&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=VgSpno5eD9k&amp;video_target=tpm-plugin-ydgl3n50-VgSpno5eD9k\" 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\/CP2.42_transcript.html\">\u201cCP 2.42\u201d here (opens in new window).<\/a><\/center><\/div>\r\n<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]168788[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<ul class=\"os-abstract\">\n<li><span class=\"os-abstract-content\">Find a vector orthogonal to two given vectors.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Determine areas and volumes by using the cross product.<\/span><\/li>\n<li><span class=\"os-abstract-content\">Calculate the torque of a given force and position vector.<\/span><\/li>\n<\/ul>\n<\/div>\n<h2 data-type=\"title\">Using the Cross Product<\/h2>\n<p id=\"fs-id1163723339475\">The cross product is very useful for several types of calculations, including finding a vector orthogonal to two given vectors, computing areas of triangles and parallelograms, and even determining the volume of the three-dimensional geometric shape made of parallelograms known as a\u00a0<em data-effect=\"italics\">parallelepiped<\/em>. The following examples illustrate these calculations.<\/p>\n<div id=\"fs-id1163723339486\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: finding a unit vector orthogonal to two given vectors<\/h3>\n<p>Let [latex]\\mathbf{a} = \\langle 5,2,-1\\rangle[\/latex] and\u00a0[latex]\\mathbf{b} = \\langle 0,-1,4\\rangle[\/latex]. Find a unit vector orthogonal to both [latex]\\textbf a[\/latex] and\u00a0[latex]\\textbf b[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q27965428\">Show Solution<\/span><\/p>\n<div id=\"q27965428\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163724138497\">The cross product [latex]\\mathbf{a} \\times \\mathbf{b}[\/latex] is orthogonal to both vectors [latex]\\textbf a[\/latex] and [latex]\\textbf b[\/latex]. We can calculate it with a determinant:<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{a} \\times \\mathbf{b} &= \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ 5 & 2 & -1 \\\\ 0 & -1 & 4\\end{vmatrix} = \\begin{vmatrix} 2 & -1 \\\\ -1 & 4 \\end{vmatrix}\\mathbf{i} - \\begin{vmatrix} 5 & -1 \\\\ 0 & 4 \\end{vmatrix}\\mathbf{j} +\\begin{vmatrix} 5 & 2 \\\\ 0 & -1 \\end{vmatrix}\\mathbf{k} \\\\  &=(8-1)\\mathbf{i} -(20-0)\\mathbf{j} +(-5-0)\\mathbf{k}\\\\  &= 7\\mathbf{i} - 20\\mathbf{j} -5\\mathbf{k}.\\\\  \\end{align*}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-id1163723341785\">Normalize this vector to find a unit vector in the same direction:<\/p>\n<div style=\"text-align: center;\">[latex]||\\mathbf{a} \\times \\mathbf{b}|| = \\sqrt{(7)^2 +(-20)^2 +(-5)^2} = \\sqrt{474}[\/latex].<\/div>\n<p id=\"fs-id1163723171302\">Thus, [latex]\\langle \\frac{7}{\\sqrt{474}},\\frac{-20}{\\sqrt{474}},\\frac{-5}{\\sqrt{474}}\\rangle[\/latex] is a unit vector orthogonal to [latex]\\textbf a[\/latex] and\u00a0[latex]\\textbf b[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find a unit vector orthogonal to both [latex]\\textbf a[\/latex] and [latex]\\textbf b[\/latex], where [latex]\\mathbf{a} = \\langle 4,0,3\\rangle[\/latex] and\u00a0[latex]\\mathbf{b} = \\langle 1,1,4\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q748528349\">Show Solution<\/span><\/p>\n<div id=\"q748528349\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]\\langle \\frac{-3}{\\sqrt{194}},\\frac{-13}{\\sqrt{194}},\\frac{4}{\\sqrt{194}}\\rangle[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p>To use the cross product for calculating areas, we state and prove the following theorem.<\/p>\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: area of a parallelogram<\/span><\/h3>\n<hr \/>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-id1163723995111\">If we locate vectors [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] such that they form adjacent sides of a parallelogram, then the area of the parallelogram is given by [latex]||\\mathbf{u} \\times \\mathbf{v}||[\/latex] (Figure 1).<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"attachment_5144\" style=\"width: 432px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5144\" class=\"size-full wp-image-5144\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27210440\/Figure-2.57.jpg\" alt=\"This figure is a parallelogram. One side is represented with a vector labeled \u201cv.\u201d The second side, the base, has the same initial point as vector v and is labeled \u201cu.\u201d The angle between u and v is theta. Also, a perpendicular line segment is drawn from the terminal point of v to vector u. It is labeled \u201c|v|sin(theta).\u201d\" width=\"422\" height=\"259\" \/><\/p>\n<p id=\"caption-attachment-5144\" class=\"wp-caption-text\">Figure 1. The parallelogram with adjacent sides [latex]{\\bf{u}}[\/latex] and [latex]{\\bf{v}}[\/latex] has base [latex]\u2225{\\bf{u}}\u2225[\/latex] and height [latex]\u2225{\\bf{v}}\u2225\\sin\\theta[\/latex].<\/p>\n<\/div>\n<h3 data-type=\"title\">Proof<\/h3>\n<p id=\"fs-id1163723995254\">We show that the magnitude of the cross product is equal to the base times height of the parallelogram.<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\text{Area of parallelogram} &= \\text{base} \\times \\text{height} \\\\  &=||\\mathbf{u}||(||\\mathbf{v}||\\sin{\\theta})\\\\  &= ||\\mathbf{u} \\times \\mathbf{v}||\\\\  \\end{align*}[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: finding the area of a triangle<\/h3>\n<p>Let [latex]P=(1, 0, 0)[\/latex], [latex]Q(0, 1, 0)[\/latex], and [latex]R=(0, 0, 1)[\/latex] be the vertices of a triangle (Figure 2). Find its area.<\/p>\n<div id=\"attachment_5147\" style=\"width: 511px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5147\" class=\"size-full wp-image-5147\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27210841\/Figure-2.58.jpg\" alt=\"This figure is the 3-dimensional coordinate system. It has a triangle drawn in the first octant. The vertices of the triangle are points P(1, 0, 0); Q(0, 1, 0); and R(0, 0, 1).\" width=\"501\" height=\"497\" \/><\/p>\n<p id=\"caption-attachment-5147\" class=\"wp-caption-text\">Figure 2. Finding the area of a triangle by using the cross product.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q452836520\">Show Solution<\/span><\/p>\n<div id=\"q452836520\" class=\"hidden-answer\" style=\"display: none\">\nWe have [latex]\\overrightarrow{PQ} = \\langle 0-1,1-0,0-0\\rangle = \\langle -1,1,0\\rangle[\/latex] and [latex]\\overrightarrow{PR} = \\langle 0-1,0-0,1-0\\rangle = \\langle -1,0,1\\rangle[\/latex]. The area of the parallelogram with adjacent sides [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{PR}[\/latex] is given by\u00a0[latex]||\\overrightarrow{PQ} \\times \\overrightarrow{PR}||[\/latex]:<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\overrightarrow{PQ} \\times \\overrightarrow{PR} &= \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ -1 & 1 & 0 \\\\ -1 & 0 & 1\\end{vmatrix} = (1-0)\\mathbf{i} -(-1-0)\\mathbf{j} +(0-(-1))\\mathbf{k} = \\mathbf{i} +\\mathbf{j} +\\mathbf{k} \\\\  ||\\overrightarrow{PQ} \\times \\overrightarrow{PR}||&= ||\\langle 1,1,1\\rangle|| = \\sqrt{1^2 +1^2 +1^2} = \\sqrt{3}.  \\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1163723499884\">The area of [latex]\\Delta PQR[\/latex] is half the area of the parallelogram, or [latex]\\frac{\\sqrt{3}}{2}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the area of the parallelogram [latex]PQRS[\/latex] with vertices\u00a0[latex]P(1, 1, 0)[\/latex],\u00a0[latex]Q(7, 1, 0)[\/latex],\u00a0[latex]R(9, 4, 2)[\/latex], and\u00a0[latex]S(3, 4, 2)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q387493274\">Show Solution<\/span><\/p>\n<div id=\"q387493274\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]6\\sqrt{13}[\/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=7753559&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=L2XRUTcifoM&amp;video_target=tpm-plugin-mi2dpg43-L2XRUTcifoM\" 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\/CP2.38_transcript.html\">\u201cCP 2.38\u201d here (opens in new window).<\/a><\/div>\n<h2 data-type=\"title\">The Triple Scalar Product<\/h2>\n<p id=\"fs-id1163723326759\">Because the cross product of two vectors is a vector, it is possible to combine the dot product and the cross product. The dot product of a vector with the cross product of two other vectors is called the triple scalar product because the result is a scalar.<\/p>\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\" style=\"text-align: center;\" data-type=\"title\"><span id=\"31\" 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-id1163723326769\">The\u00a0<strong><span id=\"d53b2426-3a76-4d95-af4f-6532ad8e4e9a_term84\" data-type=\"term\">triple scalar product<\/span>\u00a0<\/strong>of vectors [latex]\\textbf u[\/latex],\u00a0[latex]\\textbf v[\/latex], and\u00a0[latex]\\textbf w[\/latex] is\u00a0[latex]{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})[\/latex].<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: calculating a triple scalar product<\/span><\/h3>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<hr \/>\n<p id=\"fs-id1163723326861\">The triple scalar product of vectors [latex]{\\bf{u}}=u_1{\\bf{i}}+u_2{\\bf{j}}+u_3{\\bf{k}}[\/latex], [latex]{\\bf{v}}=v_1{\\bf{i}}+v_2{\\bf{j}}+v_3{\\bf{k}}[\/latex], and [latex]{\\bf{w}}=w_1{\\bf{i}}+w_2{\\bf{j}}+w_3{\\bf{k}}[\/latex] is the determinant of the [latex]3\\times3[\/latex] matrix formed by the components of the vectors:<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})=\\begin{vmatrix}u_1&u_2&u_3 \\\\ v_1&v_2&v_3 \\\\w_1&w_2&w_3 \\end{vmatrix}[\/latex].<\/p>\n<\/div>\n<\/section>\n<\/div>\n<h3 data-type=\"title\">Proof<\/h3>\n<p id=\"fs-id1163723477915\">The calculation is straightforward.<\/p>\n<div id=\"fs-id1163723477918\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\n<div style=\"text-align: center;\" data-type=\"example\">[latex]\\large{\\begin{aligned}  {\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&=\\langle u_1,u_2,u_3\\rangle\\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_1(v_2w_3-v_3w_2)-u_2(v_1w_3-v_3w_1)+u_3(v_1w_2-v_2w_1) \\\\  &=\\begin{vmatrix}u_1&u_2&u_3 \\\\ v_1&v_2&v_3 \\\\w_1&w_2&w_3 \\end{vmatrix}.  \\end{aligned}}[\/latex]<\/div>\n<div data-type=\"example\">[latex]_\\blacksquare[\/latex]<\/div>\n<div data-type=\"example\"><\/div>\n<div data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: calculating the triple scalar product<\/h3>\n<p>Let [latex]{\\bf{u}}=\\langle1,3,5\\rangle[\/latex], [latex]{\\bf{v}}=\\langle2,-1,0\\rangle[\/latex], and [latex]{\\bf{w}}=\\langle-3,0,-1\\rangle[\/latex]. Calculate the triple scalar product [latex]{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q348759826\">Show Solution<\/span><\/p>\n<div id=\"q348759826\" class=\"hidden-answer\" style=\"display: none\">\n<p>Apply\u00a0Theorem: Calculating a Triple Scalar Product directly:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&=\\begin{vmatrix}1&3&5 \\\\ 2&-1&0 \\\\ -3&0&-1 \\end{vmatrix} \\\\  &=1\\begin{vmatrix}-1&0 \\\\ 0&-1 \\end{vmatrix}-3\\begin{vmatrix} 2&0 \\\\-3&-1 \\end{vmatrix}+5\\begin{vmatrix} 2&-1 \\\\-3&0 \\end{vmatrix} \\\\  &=(1-0)-3(-2-0)+5(0-3) \\\\  &=1+6-15=-8.  \\end{aligned}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Calculate the triple scalar product [latex]{\\bf{a}}\\cdot({\\bf{b}}\\times{\\bf{c}})[\/latex], where [latex]{\\bf{a}}=\\langle2,-4,1\\rangle[\/latex],\u00a0[latex]{\\bf{b}}=\\langle0,    3,-1\\rangle[\/latex], and\u00a0[latex]{\\bf{c}}=\\langle5,-3,3\\rangle[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q849527436\">Show Solution<\/span><\/p>\n<div id=\"q849527436\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]17[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1163724060988\">When we create a matrix from three vectors, we must be careful about the order in which we list the vectors. If we list them in a matrix in one order and then rearrange the rows, the absolute value of the determinant remains unchanged. However, each time two rows switch places, the determinant changes sign:<\/p>\n<p 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}=d \\quad\\begin{vmatrix}b_1&b_2&b_3 \\\\a_1&a_2&a_3 \\\\ c_1&c_2&c_3\\end{vmatrix}=-d \\quad \\begin{vmatrix}b_1&b_2&b_3 \\\\c_1&c_2&c_3 \\\\ a_1&a_2&a_3\\end{vmatrix}=d \\quad\\begin{vmatrix}c_1&c_2&c_3 \\\\b_1&b_2&b_3 \\\\ a_1&a_2&a_3\\end{vmatrix}=-d[\/latex].<\/p>\n<p id=\"fs-id1163724153222\">Verifying this fact is straightforward, but rather messy. Let\u2019s take a look at this with an example:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\begin{vmatrix}1&2&1 \\\\ -2&0&3 \\\\4&1&-1\\end{vmatrix} &=\\begin{vmatrix}0&3 \\\\1&-1\\end{vmatrix}-2\\begin{vmatrix}-2&3 \\\\4&-1\\end{vmatrix}+\\begin{vmatrix}-2&0 \\\\4&1\\end{vmatrix} \\\\  &=(0-3)-2(2-12)+(-2-0)=-3+20-2=15.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1163723979960\">Switching the top two rows we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\begin{vmatrix}-2&0&3 \\\\1&2&1\\\\ 4&1&-1\\end{vmatrix} &=-2\\begin{vmatrix}2&1 \\\\1&-1\\end{vmatrix}+3\\begin{vmatrix}1&2 \\\\4&1\\end{vmatrix} \\\\  &=-2(-2-1)+3(1-8)=6-21=-15.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1163723124321\">Rearranging vectors in the triple products is equivalent to reordering the rows in the matrix of the determinant. Let [latex]{\\bf{u}}=u_1{\\bf{i}}+u_2{\\bf{j}}+u_3{\\bf{k}}[\/latex], [latex]{\\bf{v}}=v_1{\\bf{i}}+v_2{\\bf{j}}+v_3{\\bf{k}}[\/latex], and [latex]{\\bf{w}}=w_1{\\bf{i}}+w_2{\\bf{j}}+w_3{\\bf{k}}[\/latex]. Applying\u00a0Theorem: Calculating a Triple Scalar Product, we have<\/p>\n<p style=\"text-align: center;\">[latex]{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})=\\begin{vmatrix}u_1&u_2&u_3 \\\\ v_1&v_2&v_3 \\\\w_1&w_2&w_3\\end{vmatrix}\\text{ and }{\\bf{u}}\\cdot({\\bf{w}}\\times{\\bf{v}})=\\begin{vmatrix}u_1&u_2&u_3 \\\\w_1&w_2&w_3 \\\\v_1&v_2&v_3\\end{vmatrix}[\/latex].<\/p>\n<p id=\"fs-id1163724055509\">We can obtain the determinant for calculating [latex]{\\bf{u}}\\cdot({\\bf{w}}\\times{\\bf{v}})[\/latex] by switching the bottom two rows of [latex]{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})[\/latex]. Therefore, [latex]{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})=-{\\bf{u}}\\cdot({\\bf{w}}\\times{\\bf{v}})[\/latex].<\/p>\n<p id=\"fs-id1163724055682\">Following this reasoning and exploring the different ways we can interchange variables in the triple scalar product lead to the following identities:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&=-{\\bf{u}}\\cdot({\\bf{w}}\\times{\\bf{v}}) \\\\  {\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&={\\bf{v}}\\cdot({\\bf{w}}\\times{\\bf{u}})={\\bf{w}}\\cdot({\\bf{u}}\\times{\\bf{v}}).  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1163724042230\">Let [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] be two vectors in standard position. If [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] are not scalar multiples of each other, then these vectors form adjacent sides of a parallelogram. We saw in\u00a0Theorem: Area of a Parallelogram\u00a0that the area of this parallelogram is [latex]||{\\bf{u}}\\times{\\bf{v}}||[\/latex]. Now suppose we add a third vector [latex]\\textbf w[\/latex] that does not lie in the same plane as [latex]\\textbf u[\/latex] and [latex]\\textbf v[\/latex] but still shares the same initial point. Then these vectors form three edges of a\u00a0<span id=\"d53b2426-3a76-4d95-af4f-6532ad8e4e9a_term85\" data-type=\"term\">parallelepiped<\/span>, a three-dimensional prism with six faces that are each parallelograms, as shown in\u00a0Figure 3. The volume of this prism is the product of the figure\u2019s height and the area of its base. The triple scalar product of [latex]\\textbf u[\/latex], [latex]\\textbf v[\/latex], and [latex]\\textbf w[\/latex] provides a simple method for calculating the volume of the parallelepiped defined by these vectors.<\/p>\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\" style=\"text-align: center;\"><span class=\"os-title-label\">THEOREM: volume of a parallelepiped<\/span><\/h3>\n<hr \/>\n<\/header>\n<section>\n<div class=\"os-note-body\">\n<p id=\"fs-id1163724035012\">The volume of a parallelepiped with adjacent edges given by the vectors [latex]\\textbf u[\/latex], [latex]\\textbf v[\/latex], and [latex]\\textbf w[\/latex] is the absolute value of the triple scalar product:<\/p>\n<p style=\"text-align: center;\">[latex]\\large{V=|{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})|}[\/latex].<\/p>\n<p id=\"fs-id1163723289962\">See\u00a0Figure 3.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<p>Note that, as the name indicates, the triple scalar product produces a scalar. The volume formula just presented uses the absolute value of a scalar quantity.<\/p>\n<div id=\"attachment_5149\" style=\"width: 425px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5149\" class=\"size-full wp-image-5149\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27210947\/Figure-2.59.jpg\" alt=\"This figure is a parallelepided, a three dimensional parallelogram. Three of the sides are represented with vectors. The base has vectors v and w. The vertical side has vector u. All three vectors have the same initial point. A perpendicular vector is drawn from this common point. It is labeled \u201cproj sub (v x w) u.\u201d\" width=\"415\" height=\"201\" \/><\/p>\n<p id=\"caption-attachment-5149\" class=\"wp-caption-text\">Figure 3. The height of the parallelepiped is given by [latex]||\\text{proj}_{{\\bf{v}}\\times{\\bf{w}}}{\\bf{u}}||[\/latex].<\/p>\n<\/div>\n<h3 data-type=\"title\">Proof<\/h3>\n<p id=\"fs-id1163723290028\">The area of the base of the parallelepiped is given by [latex]||{\\bf{v}}\\times{\\bf{w}}||[\/latex]. The height of the figure is given by [latex]||\\text{proj}_{{\\bf{v}}\\times{\\bf{w}}}{\\bf{u}}||[\/latex]. The volume of the parallelepiped is the product of the height and the area of the base, so we have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  v&=||\\text{proj}_{{\\bf{v}}\\times{\\bf{w}}}{\\bf{u}}|| \\ ||{\\bf{v}}\\times{\\bf{w}}|| \\\\  &=\\left|\\frac{{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})}{||{\\bf{v}}\\times{\\bf{w}}||}\\right| \\||{\\bf{v}}\\times{\\bf{w}}|| \\\\  &=|{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})|.  \\end{aligned}[\/latex]<\/p>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: calculating the volume of a parallelepiped<\/h3>\n<p>Let [latex]{\\bf{u}}=\\langle-1,-2,1\\rangle[\/latex],\u00a0[latex]{\\bf{v}}=\\langle4,3,2\\rangle[\/latex], and\u00a0[latex]{\\bf{w}}=\\langle0,-5,-2\\rangle[\/latex]. Find the volume of the parallelepiped with adjacent edges [latex]\\textbf u[\/latex], [latex]\\textbf v[\/latex], and [latex]\\textbf w[\/latex] (Figure 4).<\/p>\n<div id=\"attachment_5151\" style=\"width: 427px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5151\" class=\"size-full wp-image-5151\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27211252\/Figure-2.60.jpg\" alt=\"This figure is the 3-dimensional coordinate system. It has three vectors in standard position. The vectors are u = &lt;-1, -2, 1&gt;; v = &lt;4, 3, 2&gt;; and w = &lt;0, -5, -2&gt;.\" width=\"417\" height=\"497\" \/><\/p>\n<p id=\"caption-attachment-5151\" class=\"wp-caption-text\">Figure 4. A parallelepiped with adjacent edges [latex]\\textbf u[\/latex], [latex]\\textbf v[\/latex], and [latex]\\textbf w[\/latex]<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q384527634\">Show Solution<\/span><\/p>\n<div id=\"q384527634\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163724159278\">We have<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&=\\begin{vmatrix}-1&-2&1 \\\\4&3&2 \\\\0&-5&-2\\end{vmatrix}=(-1)\\begin{vmatrix}3&2 \\\\-5&-2\\end{vmatrix}+2\\begin{vmatrix}4&2 \\\\ 0&-2\\end{vmatrix}+\\begin{vmatrix} 4&3 \\\\ 0&-5\\end{vmatrix} \\\\  &=(-1)(-6+10)+2(-8-0)+(-20-0) \\\\  &=-4-16-20 \\\\  &=-40.  \\end{aligned}[\/latex]<\/p>\n<p>Thus, the volume of the parallelepiped is [latex]|-40|=40[\/latex] units<sup>3<\/sup>.<\/p>\n<div id=\"fs-id1163724159281\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Find the volume of the parallelepiped formed by the vectors\u00a0[latex]{\\bf{a}}=3{\\bf{i}}+4{\\bf{j}}-{\\bf{k}}[\/latex],\u00a0[latex]{\\bf{b}}=2{\\bf{i}}-{\\bf{j}}-{\\bf{k}}[\/latex], and\u00a0[latex]{\\bf{c}}=3{\\bf{j}}+{\\bf{k}}[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q983459723\">Show Solution<\/span><\/p>\n<div id=\"q983459723\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]8[\/latex] units<sup>3<\/sup>.<\/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=7753560&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=4Jv4U5G3qxA&amp;video_target=tpm-plugin-wayy3i76-4Jv4U5G3qxA\" 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\/CP2.40_transcript.html\">\u201cCP 2.40\u201d here (opens in new window).<\/a><\/div>\n<h2 data-type=\"title\">Applications of the Cross Product<\/h2>\n<p id=\"fs-id1163724163689\">The cross product appears in many practical applications in mathematics, physics, and engineering. Let\u2019s examine some of these applications here, including the idea of torque, with which we began this section. Other applications show up in later chapters, particularly in our study of vector fields such as gravitational and electromagnetic fields (<a href=\"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/why-it-matters-vector-calculus\/\" data-page-uuid=\"41358830-7753-4f50-af9d-ec46020d7b0c\" data-page-fragment=\"page_41358830-7753-4f50-af9d-ec46020d7b0c\">Introduction to Vector Calculus<\/a>).<\/p>\n<\/div>\n<div data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: using the triple scalar product<\/h3>\n<p>Use the triple scalar product to show that vectors [latex]{\\bf{u}}=\\langle2,0,5\\rangle[\/latex], [latex]{\\bf{v}}=\\langle2,2,4\\rangle[\/latex], and [latex]{\\bf{w}}=\\langle1,-1,3\\rangle[\/latex] are coplanar\u2014that is, show that these vectors lie in the same plane.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q094352634\">Show Solution<\/span><\/p>\n<div id=\"q094352634\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163724182515\">Start by calculating the triple scalar product to find the volume of the parallelepiped defined by [latex]\\textbf u[\/latex],\u00a0[latex]\\textbf v[\/latex], and\u00a0[latex]\\textbf w[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  {\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&=\\begin{vmatrix}2&0&5\\\\ 2&2&4 \\\\ 1&-1&3\\end{vmatrix} \\\\  &=[2(2)(3)+(0)(4)(1)+5(2)(-1)]-[5(2)(1)+(2)(4)(-1)+(0)(2)(3)] \\\\&=2-2 \\\\  &=0.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1163724130658\">The volume of the parallelepiped is [latex]0[\/latex] units<sup>3<\/sup>, so one of the dimensions must be zero. Therefore, the three vectors all lie in the same plane.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Are the vectors [latex]{\\bf{a}}={\\bf{i}}+{\\bf{j}}-{\\bf{k}}[\/latex], [latex]{\\bf{b}}={\\bf{i}}-{\\bf{j}}+{\\bf{k}}[\/latex], and [latex]{\\bf{c}}={\\bf{i}}+{\\bf{j}}+{\\bf{k}}[\/latex] coplanar?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q308475728\">Show Solution<\/span><\/p>\n<div id=\"q308475728\" class=\"hidden-answer\" style=\"display: none\">\n<p>No, the triple scalar product is [latex]-4\\ne0[\/latex], so the three vectors form the adjacent edges of a parallelepiped. They are not coplanar.<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example: finding an orthogonal vector<\/h3>\n<p>Only a single plane can pass through any set of three noncolinear points. Find a vector orthogonal to the plane containing points [latex]P=(9, -3, -2)[\/latex],\u00a0[latex]Q=(1, 3, 0)[\/latex], and\u00a0[latex]R=(-2,5,0)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q314652076\">Show Solution<\/span><\/p>\n<div id=\"q314652076\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1163723475226\">The plane must contain vectors[latex]\\overrightarrow{PQ}[\/latex] and\u00a0[latex]\\overrightarrow{QR}[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\overrightarrow{PQ}&=\\langle1-9,3-(-3),0-(-2)\\rangle =\\langle-8,6,2\\rangle \\\\  \\overrightarrow{QR}&=\\langle-2-1,5-3,0-0\\rangle=\\langle-3,2,0\\rangle.  \\end{aligned}[\/latex]<\/p>\n<p id=\"fs-id1163724062686\">The cross product [latex]\\overrightarrow{PQ}\\times\\overrightarrow{QR}[\/latex] produces a vector orthogonal to both [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{QR}[\/latex]. Therefore, the cross product is orthogonal to the plane that contains these two vectors:<\/p>\n<p style=\"text-align: center;\">[latex]\\begin{aligned}  \\overrightarrow{PQ}\\times\\overrightarrow{QR}&=\\begin{vmatrix} {\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\ -8&6&2 \\\\ -3&2&0\\end{vmatrix} \\\\  &=0{\\bf{i}}-6{\\bf{j}}-16{\\bf{k}}-(-18{\\bf{k}}+4{\\bf{i}}+0{\\bf{j}}) \\\\  &=-4{\\bf{i}}-6{\\bf{j}}+2{\\bf{k}}.  \\end{aligned}[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<p id=\"fs-id1163723336017\">We have seen how to use the triple scalar product and how to find a vector orthogonal to a plane. Now we apply the cross product to real-world situations.<\/p>\n<p id=\"fs-id1163723336021\">Sometimes a force causes an object to rotate. For example, turning a screwdriver or a wrench creates this kind of rotational effect, called torque.<\/p>\n<div id=\"fs-id1163723336026\" class=\"ui-has-child-title\" data-type=\"note\">\n<div class=\"textbox shaded\">\n<header>\n<h3 class=\"os-title\" style=\"text-align: center;\" data-type=\"title\"><span id=\"43\" 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-id1163723336031\"><strong><span id=\"d53b2426-3a76-4d95-af4f-6532ad8e4e9a_term86\" data-type=\"term\">Torque<\/span><\/strong>, [latex]\\tau[\/latex] (the Greek letter\u00a0<em data-effect=\"italics\">tau<\/em>), measures the tendency of a force to produce rotation about an axis of rotation. Let [latex]\\textbf r[\/latex] be a vector with an initial point located on the axis of rotation and with a terminal point located at the point where the force is applied, and let vector [latex]\\textbf F[\/latex] represent the force. Then torque is equal to the cross product of [latex]\\textbf r[\/latex] and\u00a0[latex]\\textbf F[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]\\tau={\\bf{r}}\\times{\\bf{F}}[\/latex].<\/p>\n<p id=\"fs-id1163723149206\">See\u00a0Figure 5.<\/p>\n<\/div>\n<\/section>\n<\/div>\n<div id=\"attachment_5153\" style=\"width: 479px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5153\" class=\"size-full wp-image-5153\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27211439\/Figure-2.61.jpg\" alt=\"This figure has a vector r from an \u201caxis of rotation\u201d. At the terminal point of r there is a vector labeled \u201cF\u201d. The angle between r and F is theta.\" width=\"469\" height=\"176\" \/><\/p>\n<p id=\"caption-attachment-5153\" class=\"wp-caption-text\">Figure 5. Torque measures how a force causes an object to rotate.<\/p>\n<\/div>\n<p id=\"fs-id1163723149234\">Think about using a wrench to tighten a bolt. The torque [latex]\\tau[\/latex] applied to the bolt depends on how hard we push the wrench (force) and how far up the handle we apply the force (distance). The torque increases with a greater force on the wrench at a greater distance from the bolt. Common units of torque are the newton-meter or foot-pound. Although torque is dimensionally equivalent to work (it has the same units), the two concepts are distinct. Torque is used specifically in the context of rotation, whereas work typically involves motion along a line.<\/p>\n<div id=\"fs-id1163723149246\" class=\"ui-has-child-title\" data-type=\"example\">\n<div class=\"textbox exercises\">\n<h3>Example: evaluating torque<\/h3>\n<p>A bolt is tightened by applying a force of [latex]6[\/latex] N to a [latex]0.15[\/latex]-m wrench (Figure 6). The angle between the wrench and the force vector is [latex]40^{\\small\\circ}[\/latex]. Find the magnitude of the torque about the center of the bolt. Round the answer to two decimal places.<\/p>\n<div id=\"attachment_5155\" style=\"width: 291px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-5155\" class=\"size-full wp-image-5155\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27211539\/Figure-2.62.jpg\" alt=\"This figure is the image of an open-end wrench. The length of the wrench is labeled \u201c0.15 m.\u201d The angle the wrench makes with a vertical vector is 40 degrees. The vector is labeled with \u201c6 N.\u201d\" width=\"281\" height=\"322\" \/><\/p>\n<p id=\"caption-attachment-5155\" class=\"wp-caption-text\">Figure 6. Torque describes the twisting action of the wrench.<\/p>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q853047651\">Show Solution<\/span><\/p>\n<div id=\"q853047651\" class=\"hidden-answer\" style=\"display: none\">\n<p>Substitute the given information into the equation defining torque:<\/p>\n<p style=\"text-align: center;\">[latex]||\\tau||=||{\\bf{r}}\\times{\\bf{F}}||=||{\\bf{r}}|| \\ ||{\\bf{F}}||\\sin\\theta=(0.15\\text{ m})(6\\text{ N})\\sin{40^{\\small\\circ}}\\approx0.58\\text{N}\\cdot\\text{m}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>try it<\/h3>\n<p>Calculate the force required to produce [latex]15\\text{N}\\cdot\\text{m}[\/latex] torque at an angle of [latex]30^{\\small\\circ}[\/latex] from a [latex]150[\/latex]-cm rod.<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q473650281\">Show Solution<\/span><\/p>\n<div id=\"q473650281\" class=\"hidden-answer\" style=\"display: none\">\n<p>[latex]20[\/latex] N.<\/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=7753561&amp;p3sdk_version=1.10.1&amp;p=20361&amp;pt=375&amp;video_id=VgSpno5eD9k&amp;video_target=tpm-plugin-ydgl3n50-VgSpno5eD9k\" 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\/CP2.42_transcript.html\">\u201cCP 2.42\u201d here (opens in new window).<\/a><\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm168788\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=168788&theme=oea&iframe_resize_id=ohm168788&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/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-810\">\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 2.42. <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 2.38. <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 2.40. <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":18,"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 2.42\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 2.38\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"CP 2.40\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-810","chapter","type-chapter","status-publish","hentry"],"part":20,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/810","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/users\/349141"}],"version-history":[{"count":68,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/810\/revisions"}],"predecessor-version":[{"id":6437,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/810\/revisions\/6437"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/20"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/810\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=810"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=810"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=810"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=810"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}