{"id":503,"date":"2025-10-09T15:15:56","date_gmt":"2025-10-09T15:15:56","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&#038;p=503"},"modified":"2026-05-26T14:44:18","modified_gmt":"2026-05-26T14:44:18","slug":"vector-operations-in-space-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/vector-operations-in-space-learn-it-1\/","title":{"raw":"Vector Operations in Space: Learn It 1","rendered":"Vector Operations in Space: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\r\n<ul>\r\n \t<li>Calculate the dot product of two given vectors<\/li>\r\n \t<li>Find angles between vectors and determine when vectors are perpendicular using dot products<\/li>\r\n \t<li>Explain what is meant by the vector projection of one vector onto another vector, and describe how to compute it<\/li>\r\n \t<li>Calculate the work done by a given force<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>The Dot Product<\/h2>\r\n<p class=\"whitespace-normal break-words\">When you apply a force to move an object, you do work on that object. In earlier calculus courses, you calculated work as force times distance\u2014but only when the force acted in the same direction as the motion.<\/p>\r\n<p class=\"whitespace-normal break-words\">In reality, force and motion are vectors that often point in different directions. The <strong>dot product<\/strong> allows us to calculate work when the force vector and motion vector have different directions. It tells us how much of the force vector acts in the direction of the motion vector.<\/p>\r\n<p class=\"whitespace-normal break-words\">Beyond calculating work, the dot product helps us measure angles between vectors, determine if vectors are perpendicular, and analyze how vectors relate to coordinate axes.<\/p>\r\n\r\n<h3>What is the Dot Product?<\/h3>\r\n<p class=\"whitespace-normal break-words\">The dot product is a new way to \"multiply\" vectors, but unlike vector addition and scalar multiplication, the result is a scalar (a number), not a vector.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>dot 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<p class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot \\mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3[\/latex]<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\">For 2D vectors [latex]\\mathbf{u} = \\langle u_1, u_2 \\rangle[\/latex] and [latex]\\mathbf{v} = \\langle v_1, v_2 \\rangle[\/latex]:<\/p>\r\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot \\mathbf{v} = u_1v_1 + u_2v_2[\/latex]<\/p>\r\n\r\n<\/section>\r\n<p id=\"fs-id1163724101618\">The dot product is also called the <strong>scalar product<\/strong> (because the result is a scalar) or the <strong>inner product<\/strong>.<\/p>\r\n\r\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Multiply corresponding components and add the results. For [latex]\\langle 2,3,1 \\rangle \\cdot \\langle 4,-1,5 \\rangle[\/latex]:\r\n<center>[latex](2)(4) + (3)(-1) + (1)(5) = 8 - 3 + 5 = 10[\/latex]<\/center><\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<ol id=\"fs-id1163724076122\" type=\"a\">\r\n \t<li>Find the dot product of [latex]\\mathbf{u} = \\langle 3,5,2 \\rangle[\/latex] and\u00a0[latex]\\mathbf{v} =\\langle -1,3,0 \\rangle[\/latex].<\/li>\r\n \t<li>Find the scalar product of [latex]\\mathbf{p} = 10\\mathbf{i} - 4\\mathbf{j} +7 \\mathbf{k}[\/latex] and\u00a0[latex]\\mathbf{q} = -2\\mathbf{i} + \\mathbf{j} + 6\\mathbf{k}[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"471238733\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"471238733\"]\r\n<ol id=\"fs-id1163723802236\" type=\"a\">\r\n \t<li>Substitute the vector components into the formula for the dot product:<span data-type=\"newline\">\r\n<\/span>\r\n<div id=\"fs-id1163723138109\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\"><center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{u \\cdot v} &amp;= u_1v_1 + u_2v_2 + u_3v_3 \\\\\r\n&amp;= 3(-1)+5(3)+2(0)= -3+15+0 = 12.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/div><\/li>\r\n \t<li>The calculation is the same if the vectors are written using standard unit vectors. We still have three components for each vector to substitute into the formula for the dot product:\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{p \\cdot q} &amp;= u_1v_1 + u_2v_2 + u_3v_3 \\\\\r\n&amp;= 10(-2)+(-4)(1)+7(6)= -20-4+42 = 18.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section><section class=\"textbox tryIt\" aria-label=\"Try It\">[ohm_question hide_question_numbers=1]325836[\/ohm_question]<\/section>\r\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Properties of the Dot Product<\/h3>\r\n<p class=\"whitespace-normal break-words\">The dot product follows familiar algebraic rules, making it easy to work with in calculations.<\/p>\r\n\r\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\r\n<h3>properties of the dot 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>Commutative<\/strong>: [latex]\\mathbf{u} \\cdot \\mathbf{v} = \\mathbf{v} \\cdot \\mathbf{u}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Distributive<\/strong>: [latex]\\mathbf{u} \\cdot (\\mathbf{v} + \\mathbf{w}) = \\mathbf{u} \\cdot \\mathbf{v} + \\mathbf{u} \\cdot \\mathbf{w}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Associative with scalars<\/strong>: [latex]c(\\mathbf{u} \\cdot \\mathbf{v}) = (c\\mathbf{u}) \\cdot \\mathbf{v} = \\mathbf{u} \\cdot (c\\mathbf{v})[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\"><strong>Magnitude relationship<\/strong>: [latex]\\mathbf{v} \\cdot \\mathbf{v} = ||\\mathbf{v}||^2[\/latex]<\/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-id1163723709365\">Let [latex]\\mathbf{u} = \\langle u_1, u_2, u_3 \\rangle[\/latex] and [latex]\\mathbf{v} =\\langle v_1, v_2, v_3 \\rangle[\/latex]. Then<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{u \\cdot v} &amp;= \\langle u_1, u_2, u_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\\r\n&amp;= u_1v_1 + u_2v_2 + u_3v_3\\\\\r\n&amp;= v_1u_1 + v_2u_2 + v_3u_3\\\\\r\n&amp;= \\langle v_1, v_2, v_3 \\rangle \\cdot \\langle u_1, u_2, u_3 \\rangle\\\\\r\n&amp;= \\mathbf{v \\cdot u}.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>\r\n<p id=\"fs-id1163723953410\">The associative property looks like the associative property for real-number multiplication, but pay close attention to the difference between scalar and vector objects:<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\nc (\\mathbf{u \\cdot v}) &amp;= c(u_1v_1 + u_2v_2 + u_3v_3)\\\\\r\n&amp;= c(u_1v_1) + c(u_2v_2) + c(u_3v_3)\\\\\r\n&amp;= (cu_1)v_1 + (cu_2)v_2 + (cu_3)v_3\\\\\r\n&amp;= \\langle cu_1, cu_2, cu_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\\r\n&amp;= c\\langle u_1, u_2,u_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\\r\n&amp;= (c\\mathbf{u}) \\cdot \\mathbf{v}.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>\r\n<p id=\"fs-id1163723796994\">The proof that [latex]c (\\mathbf{u \\cdot v}) = \\mathbf{u}\\cdot (c\\mathbf{v})[\/latex] is similar.<\/p>\r\n<p id=\"fs-id1163723299158\">The fourth property shows the relationship between the magnitude of a vector and its dot product with itself:<\/p>\r\n\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{v} \\cdot \\mathbf{v} &amp;= \\langle v_1, v_2, v_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\\r\n&amp;= (v_1)^2 + (v_2)^2 + (v_3)^2\\\\\r\n&amp;= [\\sqrt{(v_1)^2 + (v_2)^2 + (v_3)^2}]^2\\\\\r\n&amp;= ||\\mathbf{v}||^2.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>\r\n<p class=\"whitespace-normal break-words\">From the definition and properties, we can make several key observations:<\/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{0} \\cdot \\mathbf{v} = 0[\/latex] for any vector [latex]\\mathbf{v}[\/latex]<\/li>\r\n \t<li class=\"whitespace-normal break-words\">If [latex]\\mathbf{v} \\cdot \\mathbf{v} = 0[\/latex], then [latex]\\mathbf{v} = \\mathbf{0}[\/latex] (only the zero vector has dot product zero with itself)<\/li>\r\n \t<li class=\"whitespace-normal break-words\">The dot product of a vector with itself gives the square of its magnitude<\/li>\r\n<\/ul>\r\n<section class=\"textbox recall\" aria-label=\"Recall\">Remember that [latex]||\\mathbf{v}|| = \\sqrt{v_1^2 + v_2^2 + v_3^2}[\/latex], so [latex]||\\mathbf{v}||^2 = v_1^2 + v_2^2 + v_3^2 = \\mathbf{v} \\cdot \\mathbf{v}[\/latex].<\/section><section class=\"textbox example\" aria-label=\"Example\">\r\n<p id=\"fs-id1163724075617\">Let [latex]\\mathbf{a} = \\langle 1,2,-3 \\rangle[\/latex], [latex]\\mathbf{b} = \\langle 0,2,4 \\rangle[\/latex], and [latex]\\mathbf{c} = \\langle 5,-1,3 \\rangle[\/latex]. Find each of the following products.<\/p>\r\n\r\n<ol id=\"fs-id1163724078047\" type=\"a\">\r\n \t<li>[latex](\\mathbf{a} \\cdot \\mathbf{b})\\mathbf{c} [\/latex]<\/li>\r\n \t<li>[latex]\\mathbf{a}\\cdot (2\\mathbf{c} )[\/latex]<\/li>\r\n \t<li>[latex]||\\mathbf{b}||^2[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"235187363\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"235187363\"]\r\n<ol id=\"fs-id1163723139665\" type=\"a\">\r\n \t<li>Note that this expression asks for the scalar multiple of\u00a0 [latex]\\mathbf{c}[\/latex]\u00a0by [latex]\\mathbf{a} \\cdot \\mathbf{b}[\/latex]:<center>\r\n[latex]\r\n\\begin{align*}\r\n(\\mathbf{a} \\cdot \\mathbf{b})\\mathbf{c} &amp;= (\\langle 1,2,-3 \\rangle \\cdot \\langle 0,2,4 \\rangle) \\langle 5,-1,3 \\rangle\\\\\r\n&amp;= (1(0)+2(2)+(-3)(4))\\langle 5,-1,3 \\rangle\\\\\r\n&amp;= -8\\langle 5,-1,3 \\rangle\\\\\r\n&amp;= \\langle -40,8,-24 \\rangle.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/li>\r\n \t<li>This expression is a dot product of vector\u00a0[latex]\\mathbf{a}[\/latex]\u00a0and scalar multiple 2[latex]\\mathbf{c}[\/latex]:\r\n<center>\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{a}\\cdot (2\\mathbf{c} ) &amp;= 2(\\mathbf{a}\\cdot \\mathbf{c})\\\\\r\n&amp;= 2(\\langle 1,2,-3 \\rangle \\cdot\\langle 5,-1,3 \\rangle)\\\\\r\n&amp;= 2(1(5) +2(-1)+(-3)(3))\\\\\r\n&amp;= 2(-6) = -12.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center><\/li>\r\n \t<li>Simplifying this expression is a straightforward application of the dot product:\r\n<center>[latex]||\\mathbf{b}||^2 = \\mathbf{b} \\cdot \\mathbf{b} = \\langle 0,2,4 \\rangle\\cdot\\langle 0,2,4 \\rangle = 0^2+2^2 +4^2 = 0+4+16 = 20 [\/latex].<\/center><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\" aria-label=\"Learning Goals\">\n<ul>\n<li>Calculate the dot product of two given vectors<\/li>\n<li>Find angles between vectors and determine when vectors are perpendicular using dot products<\/li>\n<li>Explain what is meant by the vector projection of one vector onto another vector, and describe how to compute it<\/li>\n<li>Calculate the work done by a given force<\/li>\n<\/ul>\n<\/section>\n<h2>The Dot Product<\/h2>\n<p class=\"whitespace-normal break-words\">When you apply a force to move an object, you do work on that object. In earlier calculus courses, you calculated work as force times distance\u2014but only when the force acted in the same direction as the motion.<\/p>\n<p class=\"whitespace-normal break-words\">In reality, force and motion are vectors that often point in different directions. The <strong>dot product<\/strong> allows us to calculate work when the force vector and motion vector have different directions. It tells us how much of the force vector acts in the direction of the motion vector.<\/p>\n<p class=\"whitespace-normal break-words\">Beyond calculating work, the dot product helps us measure angles between vectors, determine if vectors are perpendicular, and analyze how vectors relate to coordinate axes.<\/p>\n<h3>What is the Dot Product?<\/h3>\n<p class=\"whitespace-normal break-words\">The dot product is a new way to &#8220;multiply&#8221; vectors, but unlike vector addition and scalar multiplication, the result is a scalar (a number), not a vector.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>dot 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 class=\"whitespace-normal break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot \\mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3[\/latex]<\/p>\n<p class=\"whitespace-pre-wrap break-words\">For 2D vectors [latex]\\mathbf{u} = \\langle u_1, u_2 \\rangle[\/latex] and [latex]\\mathbf{v} = \\langle v_1, v_2 \\rangle[\/latex]:<\/p>\n<p class=\"whitespace-pre-wrap break-words\" style=\"text-align: center;\">[latex]\\mathbf{u} \\cdot \\mathbf{v} = u_1v_1 + u_2v_2[\/latex]<\/p>\n<\/section>\n<p id=\"fs-id1163724101618\">The dot product is also called the <strong>scalar product<\/strong> (because the result is a scalar) or the <strong>inner product<\/strong>.<\/p>\n<section class=\"textbox proTip\" aria-label=\"Pro Tip\">Multiply corresponding components and add the results. For [latex]\\langle 2,3,1 \\rangle \\cdot \\langle 4,-1,5 \\rangle[\/latex]:<\/p>\n<div style=\"text-align: center;\">[latex](2)(4) + (3)(-1) + (1)(5) = 8 - 3 + 5 = 10[\/latex]<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<ol id=\"fs-id1163724076122\" type=\"a\">\n<li>Find the dot product of [latex]\\mathbf{u} = \\langle 3,5,2 \\rangle[\/latex] and\u00a0[latex]\\mathbf{v} =\\langle -1,3,0 \\rangle[\/latex].<\/li>\n<li>Find the scalar product of [latex]\\mathbf{p} = 10\\mathbf{i} - 4\\mathbf{j} +7 \\mathbf{k}[\/latex] and\u00a0[latex]\\mathbf{q} = -2\\mathbf{i} + \\mathbf{j} + 6\\mathbf{k}[\/latex].<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q471238733\">Show Solution<\/button><\/p>\n<div id=\"q471238733\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1163723802236\" type=\"a\">\n<li>Substitute the vector components into the formula for the dot product:<span data-type=\"newline\"><br \/>\n<\/span><\/p>\n<div id=\"fs-id1163723138109\" class=\"unnumbered\" data-type=\"equation\" data-label=\"\">\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{u \\cdot v} &= u_1v_1 + u_2v_2 + u_3v_3 \\\\  &= 3(-1)+5(3)+2(0)= -3+15+0 = 12.\\\\  \\end{align*}[\/latex]<\/div>\n<\/div>\n<\/li>\n<li>The calculation is the same if the vectors are written using standard unit vectors. We still have three components for each vector to substitute into the formula for the dot product:\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{p \\cdot q} &= u_1v_1 + u_2v_2 + u_3v_3 \\\\  &= 10(-2)+(-4)(1)+7(6)= -20-4+42 = 18.\\\\  \\end{align*}[\/latex]<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\" aria-label=\"Try It\"><iframe loading=\"lazy\" id=\"ohm325836\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=325836&theme=lumen&iframe_resize_id=ohm325836&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><\/section>\n<h3 class=\"text-xl font-bold text-text-100 mt-1 -mb-0.5\">Properties of the Dot Product<\/h3>\n<p class=\"whitespace-normal break-words\">The dot product follows familiar algebraic rules, making it easy to work with in calculations.<\/p>\n<section class=\"textbox keyTakeaway\" aria-label=\"Key Takeaway\">\n<h3>properties of the dot 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>Commutative<\/strong>: [latex]\\mathbf{u} \\cdot \\mathbf{v} = \\mathbf{v} \\cdot \\mathbf{u}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Distributive<\/strong>: [latex]\\mathbf{u} \\cdot (\\mathbf{v} + \\mathbf{w}) = \\mathbf{u} \\cdot \\mathbf{v} + \\mathbf{u} \\cdot \\mathbf{w}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Associative with scalars<\/strong>: [latex]c(\\mathbf{u} \\cdot \\mathbf{v}) = (c\\mathbf{u}) \\cdot \\mathbf{v} = \\mathbf{u} \\cdot (c\\mathbf{v})[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\"><strong>Magnitude relationship<\/strong>: [latex]\\mathbf{v} \\cdot \\mathbf{v} = ||\\mathbf{v}||^2[\/latex]<\/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-id1163723709365\">Let [latex]\\mathbf{u} = \\langle u_1, u_2, u_3 \\rangle[\/latex] and [latex]\\mathbf{v} =\\langle v_1, v_2, v_3 \\rangle[\/latex]. Then<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{u \\cdot v} &= \\langle u_1, u_2, u_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\  &= u_1v_1 + u_2v_2 + u_3v_3\\\\  &= v_1u_1 + v_2u_2 + v_3u_3\\\\  &= \\langle v_1, v_2, v_3 \\rangle \\cdot \\langle u_1, u_2, u_3 \\rangle\\\\  &= \\mathbf{v \\cdot u}.\\\\  \\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1163723953410\">The associative property looks like the associative property for real-number multiplication, but pay close attention to the difference between scalar and vector objects:<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  c (\\mathbf{u \\cdot v}) &= c(u_1v_1 + u_2v_2 + u_3v_3)\\\\  &= c(u_1v_1) + c(u_2v_2) + c(u_3v_3)\\\\  &= (cu_1)v_1 + (cu_2)v_2 + (cu_3)v_3\\\\  &= \\langle cu_1, cu_2, cu_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\  &= c\\langle u_1, u_2,u_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\  &= (c\\mathbf{u}) \\cdot \\mathbf{v}.\\\\  \\end{align*}[\/latex]<\/div>\n<p id=\"fs-id1163723796994\">The proof that [latex]c (\\mathbf{u \\cdot v}) = \\mathbf{u}\\cdot (c\\mathbf{v})[\/latex] is similar.<\/p>\n<p id=\"fs-id1163723299158\">The fourth property shows the relationship between the magnitude of a vector and its dot product with itself:<\/p>\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{v} \\cdot \\mathbf{v} &= \\langle v_1, v_2, v_3 \\rangle \\cdot \\langle v_1, v_2, v_3 \\rangle\\\\  &= (v_1)^2 + (v_2)^2 + (v_3)^2\\\\  &= [\\sqrt{(v_1)^2 + (v_2)^2 + (v_3)^2}]^2\\\\  &= ||\\mathbf{v}||^2.\\\\  \\end{align*}[\/latex]<\/div>\n<p>[latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n<p class=\"whitespace-normal break-words\">From the definition and properties, we can make several key observations:<\/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{0} \\cdot \\mathbf{v} = 0[\/latex] for any vector [latex]\\mathbf{v}[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">If [latex]\\mathbf{v} \\cdot \\mathbf{v} = 0[\/latex], then [latex]\\mathbf{v} = \\mathbf{0}[\/latex] (only the zero vector has dot product zero with itself)<\/li>\n<li class=\"whitespace-normal break-words\">The dot product of a vector with itself gives the square of its magnitude<\/li>\n<\/ul>\n<section class=\"textbox recall\" aria-label=\"Recall\">Remember that [latex]||\\mathbf{v}|| = \\sqrt{v_1^2 + v_2^2 + v_3^2}[\/latex], so [latex]||\\mathbf{v}||^2 = v_1^2 + v_2^2 + v_3^2 = \\mathbf{v} \\cdot \\mathbf{v}[\/latex].<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p id=\"fs-id1163724075617\">Let [latex]\\mathbf{a} = \\langle 1,2,-3 \\rangle[\/latex], [latex]\\mathbf{b} = \\langle 0,2,4 \\rangle[\/latex], and [latex]\\mathbf{c} = \\langle 5,-1,3 \\rangle[\/latex]. Find each of the following products.<\/p>\n<ol id=\"fs-id1163724078047\" type=\"a\">\n<li>[latex](\\mathbf{a} \\cdot \\mathbf{b})\\mathbf{c}[\/latex]<\/li>\n<li>[latex]\\mathbf{a}\\cdot (2\\mathbf{c} )[\/latex]<\/li>\n<li>[latex]||\\mathbf{b}||^2[\/latex]<\/li>\n<\/ol>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q235187363\">Show Solution<\/button><\/p>\n<div id=\"q235187363\" class=\"hidden-answer\" style=\"display: none\">\n<ol id=\"fs-id1163723139665\" type=\"a\">\n<li>Note that this expression asks for the scalar multiple of\u00a0 [latex]\\mathbf{c}[\/latex]\u00a0by [latex]\\mathbf{a} \\cdot \\mathbf{b}[\/latex]:\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  (\\mathbf{a} \\cdot \\mathbf{b})\\mathbf{c} &= (\\langle 1,2,-3 \\rangle \\cdot \\langle 0,2,4 \\rangle) \\langle 5,-1,3 \\rangle\\\\  &= (1(0)+2(2)+(-3)(4))\\langle 5,-1,3 \\rangle\\\\  &= -8\\langle 5,-1,3 \\rangle\\\\  &= \\langle -40,8,-24 \\rangle.\\\\  \\end{align*}[\/latex]<\/div>\n<\/li>\n<li>This expression is a dot product of vector\u00a0[latex]\\mathbf{a}[\/latex]\u00a0and scalar multiple 2[latex]\\mathbf{c}[\/latex]:\n<div style=\"text-align: center;\">\n[latex]\\begin{align*}  \\mathbf{a}\\cdot (2\\mathbf{c} ) &= 2(\\mathbf{a}\\cdot \\mathbf{c})\\\\  &= 2(\\langle 1,2,-3 \\rangle \\cdot\\langle 5,-1,3 \\rangle)\\\\  &= 2(1(5) +2(-1)+(-3)(3))\\\\  &= 2(-6) = -12.\\\\  \\end{align*}[\/latex]<\/div>\n<\/li>\n<li>Simplifying this expression is a straightforward application of the dot product:\n<div style=\"text-align: center;\">[latex]||\\mathbf{b}||^2 = \\mathbf{b} \\cdot \\mathbf{b} = \\langle 0,2,4 \\rangle\\cdot\\langle 0,2,4 \\rangle = 0^2+2^2 +4^2 = 0+4+16 = 20[\/latex].<\/div>\n<\/li>\n<\/ol>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":16,"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-503","chapter","type-chapter","status-publish","hentry"],"part":420,"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\/503","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":6,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/503\/revisions"}],"predecessor-version":[{"id":2881,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/503\/revisions\/2881"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/parts\/420"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapters\/503\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/media?parent=503"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/pressbooks\/v2\/chapter-type?post=503"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/contributor?post=503"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus3\/wp-json\/wp\/v2\/license?post=503"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}