{"id":808,"date":"2021-08-27T19:55:14","date_gmt":"2021-08-27T19:55:14","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus3\/?post_type=chapter&p=808"},"modified":"2022-10-21T00:14:47","modified_gmt":"2022-10-21T00:14:47","slug":"projections-and-work","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus3\/chapter\/projections-and-work\/","title":{"raw":"Projections and Work","rendered":"Projections and Work"},"content":{"raw":"
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Learning Objectives<\/h3>\r\n
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  • Explain what is meant by the vector projection of one vector onto another vector, and describe how to compute it.<\/span><\/li>\r\n \t
  • Calculate the work done by a given force.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n

    Projections<\/h2>\r\n

    As we have seen, addition combines two vectors to create a resultant vector. But what if we are given a vector and we need to find its component parts? We use vector projections to perform the opposite process; they can break down a vector into its components. The magnitude of a vector projection is a scalar projection. For example, if a child is pulling the handle of a wagon at a [latex]55^\\circ[\/latex] angle, we can use projections to determine how much of the force on the handle is actually moving the wagon forward (Figure 1). We return to this example and learn how to solve it after we see how to calculate projections.<\/p>\r\n\r\n<\/div>\r\n

    \r\n\r\n[caption id=\"attachment_5122\" align=\"aligncenter\" width=\"292\"]\"This Figure 1. When a child pulls a wagon, only the horizontal component of the force propels the wagon forward.[\/caption]\r\n\r\n<\/div>\r\n
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    DEFINITION<\/span><\/h3>\r\n\r\n
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    The\u00a0vector projection<\/strong>\u00a0of [latex]\\mathbf{v}[\/latex] onto [latex]\\mathbf{u}[\/latex] is the vector labeled [latex]\\text{proj}_\\text{u}\\mathbf{v}[\/latex] in\u00a0Figure 2. It has the same initial point as [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] and the same direction as [latex]\\mathbf{u}[\/latex], and represents the component of [latex]\\mathbf{v}[\/latex] that acts in the direction of [latex]\\mathbf{u}[\/latex]. If [latex]\\theta[\/latex] represents the angle between [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex], then, by properties of triangles, we know the length of [latex]\\text{proj}_\\text{u}\\mathbf{v}[\/latex] is [latex]||\\text{proj}_\\text{u}\\mathbf{v}|| = ||\\mathbf{v}||\\cos{\\theta}[\/latex]. When expressing [latex]\\cos{\\theta}[\/latex] in terms of the dot product, this becomes<\/p>\r\n\r\n

    \r\n[latex]\r\n\\begin{align*}\r\n||\\text{proj}_\\text{u}\\mathbf{v}|| &= ||\\mathbf{v}||\\cos{\\theta}\\\\\r\n&= ||\\mathbf{v}||(\\frac{|\\mathbf{u}\\cdot\\mathbf{v}|}{||\\mathbf{u}||\\,||\\mathbf{v}||})\\\\\r\n&= \\frac{|\\mathbf{u}\\cdot\\mathbf{v}|}{||\\mathbf{u}||}.\\\\\\end{align*}\r\n[\/latex]<\/center>\r\nWe now multiply by a unit vector in the direction of [latex]\\mathbf{u}[\/latex] to get [latex]\\text{proj}_\\text{u}\\mathbf{v}[\/latex]:\r\n\r\n
    [latex]\\text{proj}_\\text{u}\\mathbf{v} = \\frac{|\\mathbf{u}\\cdot\\mathbf{v}|}{||\\mathbf{u}||}(\\frac{1}{||\\mathbf{u}||}\\mathbf{u}) = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}||^2}\\mathbf{u}[\/latex].<\/center>The length of this vector is also known as the\u00a0scalar projection<\/strong>\u00a0of [latex]\\mathbf{v}[\/latex] onto [latex]\\mathbf{u}[\/latex] and is denoted by\r\n\r\n
    [latex]||\\text{proj}_\\text{u}\\mathbf{v}|| = ||\\text{comp}_\\text{u}\\mathbf{v}|| = \\frac{|\\mathbf{u}\\cdot\\mathbf{v}|}{||\\mathbf{u}||}[\/latex].<\/center><\/div>\r\n<\/section><\/div>\r\n[caption id=\"attachment_5124\" align=\"aligncenter\" width=\"269\"]\"This Figure 2. The projection of [latex]{\\bf{v}}[\/latex] onto [latex]{\\bf{u}}[\/latex] shows the component of vector [latex]{\\bf{v}}[\/latex] in the direction of [latex]{\\bf{u}}[\/latex].[\/caption]\r\n
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    Example: finding projections<\/h3>\r\n

    Find the projection of [latex]\\mathbf{v}[\/latex] onto\u00a0[latex]\\mathbf{u}[\/latex].<\/p>\r\n\r\n

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    1. [latex]\\mathbf{v} = \\langle 3,5,1 \\rangle[\/latex] and\u00a0[latex]\\mathbf{u} = \\langle -1,4,3 \\rangle[\/latex]<\/li>\r\n \t
    2. [latex]\\mathbf{v} = 3\\mathbf{i} -2\\mathbf{j}[\/latex] and\u00a0[latex]\\mathbf{u} = \\mathbf{i} +6\\mathbf{j}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"824461483\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"824461483\"]\r\n
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      1. Substitute the components of [latex]\\mathbf{v}[\/latex] and [latex]\\mathbf{u}[\/latex] into the formula for the projection:\r\n
        \r\n[latex]\r\n\\begin{align*}\r\n\\text{proj}_\\text{u}\\mathbf{v} &= \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}||^2}\\mathbf{u}\\\\\r\n&= \\frac{\\langle -1,4,3 \\rangle \\cdot \\langle 3,5,1 \\rangle}{||\\langle -1,4,3 \\rangle||^2}\\langle -1,4,3 \\rangle\\\\\r\n&= \\frac{-3+20+3}{(-1)^2+4^2+3^2}\\langle -1,4,3 \\rangle\\\\\r\n&= \\frac{20}{26}\\langle -1,4,3 \\rangle\\\\\r\n&= \\langle -\\frac{10}{13},\\frac{40}{13},\\frac{30}{13} \\rangle.\\\\\\end{align*}\r\n[\/latex] \r\n\r\n<\/center><\/li>\r\n \t
      2. To find the two-dimensional projection, simply adapt the formula to the two-dimensional case:\r\n
        \r\n[latex]\r\n\\begin{align*}\r\n\\text{proj}_\\text{u}\\mathbf{v} &= \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}||^2}\\mathbf{u}\\\\\r\n&= \\frac{(\\mathbf{i} +6\\mathbf{j}) \\cdot ( 3\\mathbf{i} -2\\mathbf{j})}{||\\mathbf{i} +6\\mathbf{j}||^2}(\\mathbf{i} +6\\mathbf{j})\\\\\r\n&= \\frac{1(3)+6(-2)}{1^2+6^2}(\\mathbf{i} +6\\mathbf{j})\\\\\r\n&= -\\frac{9}{37}(\\mathbf{i} +6\\mathbf{j})\\\\\r\n&= -\\frac{9}{37}\\mathbf{i}- \\frac{54}{37}\\mathbf{j}.\\\\\\end{align*}\r\n[\/latex] \r\n\r\n<\/center><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n
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        Try It<\/h3>\r\n[ohm_question]245402[\/ohm_question]\r\n\r\n<\/div>\r\n

        Sometimes it is useful to decompose vectors\u2014that is, to break a vector apart into a sum. This process is called the\u00a0resolution of a vector into components<\/em>. Projections allow us to identify two orthogonal vectors having a desired sum. For example, let [latex]\\mathbf{v} = \\langle 6,-4 \\rangle[\/latex] and let [latex]\\mathbf{u} = \\langle 3,1 \\rangle[\/latex]. We want to decompose the vector [latex]\\mathbf{v}[\/latex] into orthogonal components such that one of the component vectors has the same direction as [latex]\\mathbf{u}[\/latex].<\/p>\r\n

        We first find the component that has the same direction as [latex]\\mathbf{u}[\/latex] by projecting [latex]\\mathbf{v}[\/latex] onto\u00a0[latex]\\mathbf{u}[\/latex]. Let [latex]\\mathbf{p} = \\text{proj}_\\text{u}\\mathbf{v}[\/latex]. Then, we have<\/p>\r\n\r\n

        \r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{p} &= \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}||^2}\\mathbf{u}\\\\\r\n&= \\frac{18-4}{9+1}\\mathbf{u}\\\\\r\n&= \\frac{7}{5}\\mathbf{u} = \\frac{7}{5} \\langle 3,1 \\rangle = \\langle \\frac{21}{5},\\frac{7}{5} \\rangle.\\\\\\end{align*}\r\n[\/latex]<\/center>\r\n

        Now consider the vector [latex]\\mathbf{q} = \\mathbf{v} - \\mathbf{p}[\/latex]. We have<\/p>\r\n\r\n

        \r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{q} &= \\mathbf{v} - \\mathbf{p}\\\\\r\n&= \\langle 6,-4 \\rangle - \\langle \\frac{21}{5},\\frac{7}{5} \\rangle\\\\\r\n&= \\langle \\frac{9}{5},-\\frac{27}{5} \\rangle.\\\\\\end{align*}\r\n[\/latex]<\/center>\r\n

        Clearly, by the way we defined [latex]\\mathbf{q}[\/latex], we have [latex]\\mathbf{v} = \\mathbf{q} + \\mathbf{p}[\/latex], and<\/p>\r\n\r\n

        \r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{q} \\cdot \\mathbf{p} &= \\langle \\frac{9}{5},-\\frac{27}{5} \\rangle \\cdot \\langle \\frac{21}{5},\\frac{7}{5} \\rangle\\\\\r\n&= \\frac{9(21)}{25}+\\frac{-27(7)}{25} \\\\\r\n&= \\frac{189}{25}-\\frac{189}{25} = 0 .\\\\\\end{align*}\r\n[\/latex]<\/center>\r\n

        Therefore, [latex]\\mathbf{q}[\/latex] and [latex]\\mathbf{p}[\/latex] are orthogonal.<\/p>\r\n\r\n<\/div>\r\n

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        Example: resolving vectors into components<\/h3>\r\nExpress [latex]\\mathbf{v} = \\langle 8,-3,-3 \\rangle[\/latex] as a sum of orthogonal vectors such that one of the vectors has the same direction as [latex]\\mathbf{u} = \\langle 2,3,2 \\rangle[\/latex].\r\n\r\n[reveal-answer q=\"836153152\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"836153152\"]\r\n

        Let [latex]\\mathbf{p}[\/latex] represent the projection of [latex]\\mathbf{v}[\/latex] onto\u00a0[latex]\\mathbf{u}[\/latex]:<\/p>\r\n\r\n

        \r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{p} & = \\text{proj}_\\text{u}\\mathbf{v}\\\\\r\n&= \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}||^2}\\mathbf{u} \\\\\r\n&=\\frac{\\langle 2,3,2 \\rangle \\cdot \\langle 8,-3,-3 \\rangle}{||\\langle 2,3,2 \\rangle||^2}\\langle 2,3,2 \\rangle \\\\\r\n&= \\frac{16-9-6}{2^2+3^2+2^2}\\langle 2,3,2 \\rangle\\\\\r\n&= \\frac{1}{17}\\langle 2,3,2 \\rangle\\\\\r\n&= \\langle \\frac{2}{17},\\frac{3}{17},\\frac{2}{17} \\rangle.\\\\\\end{align*}\r\n[\/latex]<\/center>\r\n

        Then,<\/p>\r\n\r\n

        [latex]\\mathbf{q} =\\mathbf{v} - \\mathbf{p} = \\langle 8,-3,-3 \\rangle - \\langle \\frac{2}{17},\\frac{3}{17},\\frac{2}{17} \\rangle = \\langle \\frac{134}{17},-\\frac{54}{17},-\\frac{53}{17} \\rangle.[\/latex]<\/center>\r\n

        To check our work, we can use the dot product to verify that [latex]\\textbf p[\/latex] and [latex]\\textbf q[\/latex] are orthogonal vectors:<\/p>\r\n\r\n

        [latex]\\mathbf{p} \\cdot \\mathbf{q} = \\langle \\frac{2}{17},\\frac{3}{17},\\frac{2}{17} \\rangle \\cdot \\langle \\frac{134}{17},-\\frac{54}{17},-\\frac{53}{17} \\rangle = \\frac{268}{289} - \\frac{162}{289}-\\frac{106}{289} = 0[\/latex].<\/center>\r\n

        Then,<\/p>\r\n\r\n

        [latex]\\mathbf{v} = \\mathbf{p} + \\mathbf{q} = \\langle \\frac{2}{17},\\frac{3}{17},\\frac{2}{17} \\rangle + \\langle \\frac{134}{17},-\\frac{54}{17},-\\frac{53}{17} \\rangle[\/latex].<\/center>[\/hidden-answer]\r\n\r\n<\/div>\r\n
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        try it<\/h3>\r\nExpress [latex]\\mathbf{v} = 5\\mathbf{i} - \\mathbf{j}[\/latex] as a sum of orthogonal vectors such that one of the vectors has the same direction as [latex]\\mathbf{u} = 4\\mathbf{i} +2\\mathbf{j}[\/latex].\r\n\r\n[reveal-answer q=\"114833389\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"114833389\"]\r\n\r\n[latex]\\mathbf{v} = \\mathbf{p} + \\mathbf{q}[\/latex], where\u00a0[latex]\\mathbf{p} = \\frac{18}{5}\\mathbf{i} + \\frac{19}{5}\\mathbf{j}[\/latex] and\u00a0[latex]\\mathbf{q} = \\frac{7}{5}\\mathbf{i} - \\frac{14}{5}\\mathbf{j}[\/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