{"id":887,"date":"2025-10-14T15:28:24","date_gmt":"2025-10-14T15:28:24","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=887"},"modified":"2026-08-25T16:04:02","modified_gmt":"2026-08-25T16:04:02","slug":"vector-operations-in-space-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/vector-operations-in-space-learn-it-3\/","title":{"raw":"Vector Operations in Space: Learn It 3","rendered":"Vector Operations in Space: Learn It 3"},"content":{"raw":"

Projections<\/h2>\r\n
\r\n

Vector addition combines vectors to create a resultant vector. But what if you need to work in reverse\u2014breaking a vector into its component parts? Vector projections<\/strong> solve this problem by showing how much of one vector acts in the direction of another.<\/p>\r\n

Consider a child pulling a wagon handle at a 55\u00b0 angle. Only part of that pulling force actually moves the wagon forward\u2014the horizontal component. Projections help us find exactly how much force is working in the direction we care about.<\/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\nThe vector projection<\/strong> of [latex]\\mathbf{v}[\/latex] onto [latex]\\mathbf{u}[\/latex], written as [latex]\\text{proj}_{\\mathbf{u}}\\mathbf{v}[\/latex], represents the component of [latex]\\mathbf{v}[\/latex] that acts in the direction of [latex]\\mathbf{u}[\/latex].\r\n\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

From the geometry of the situation, if [latex]\\theta[\/latex] is the angle between [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex], then the length of the projection is:<\/p>\r\n

[latex]||\\text{proj}_{\\mathbf{u}}\\mathbf{v}|| = ||\\mathbf{v}|| \\cos \\theta[\/latex]<\/p>\r\n

Using the dot product formula [latex]\\cos \\theta = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| , ||\\mathbf{v}||}[\/latex], this becomes:<\/p>\r\n

[latex]||\\text{proj}_{\\mathbf{u}}\\mathbf{v}|| = ||\\mathbf{v}|| \\cdot \\frac{|\\mathbf{u} \\cdot \\mathbf{v}|}{||\\mathbf{u}|| , ||\\mathbf{v}||} = \\frac{|\\mathbf{u} \\cdot \\mathbf{v}|}{||\\mathbf{u}||}[\/latex]<\/p>\r\n

To get the actual projection vector (not just its length), we multiply by a unit vector in the direction of [latex]\\mathbf{u}[\/latex]:<\/p>\r\n

[latex]||\\text{proj}_{\\mathbf{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].<\/p>\r\nThe 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>
\r\n

vector projection<\/h3>\r\n

Vector Projection Formula<\/strong><\/p>\r\n

[latex]\\text{proj}_{\\mathbf{u}}\\mathbf{v} = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}||^2}\\mathbf{u}[\/latex]<\/p>\r\n

Scalar Projection<\/strong> (length only)<\/p>\r\n

[latex]||\\text{proj}_{\\mathbf{u}}\\mathbf{v}|| = \\text{comp}_{\\mathbf{u}}\\mathbf{v} = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}||}[\/latex]<\/p>\r\n\r\n<\/section>

Think of projection as the \"shadow\" that vector [latex]\\mathbf{v}[\/latex] casts onto vector [latex]\\mathbf{u}[\/latex] when light shines perpendicular to [latex]\\mathbf{u}[\/latex].<\/section>
\r\n

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

    \r\n \t
  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
      \r\n \t
    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<\/section>\r\n

      Decomposing Vectors into Orthogonal Components<\/h3>\r\n

      Projections allow us to decompose (break apart) vectors into orthogonal components. This process is called resolution of a vector into components<\/strong>.<\/p>\r\n\r\n

      \r\n

      Let [latex]\\mathbf{v} = \\langle 6,-4 \\rangle[\/latex] and [latex]\\mathbf{u} = \\langle 3,1 \\rangle[\/latex]. Decompose [latex]\\mathbf{v}[\/latex] into orthogonal components where one component has the same direction as [latex]\\mathbf{u}[\/latex].<\/p>\r\n

      Step 1<\/strong>: Find the projection of [latex]\\mathbf{v}[\/latex] onto [latex]\\mathbf{u}[\/latex]: [latex]\\mathbf{p} = \\text{proj}_{\\mathbf{u}}\\mathbf{v} = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}||^2}\\mathbf{u}[\/latex]<\/p>\r\n

      [latex]\\mathbf{u} \\cdot \\mathbf{v} = (3)(6) + (1)(-4) = 18 - 4 = 14[\/latex]<\/p>\r\n

      [latex]||\\mathbf{u}||^2 = 3^2 + 1^2 = 10[\/latex]<\/p>\r\n

      [latex]\\mathbf{p} = \\frac{14}{10}\\langle 3,1 \\rangle = \\frac{7}{5}\\langle 3,1 \\rangle = \\left\\langle \\frac{21}{5}, \\frac{7}{5} \\right\\rangle[\/latex]<\/p>\r\n

      Step 2<\/strong>: Find the orthogonal component:<\/p>\r\n

      [latex]\\mathbf{q} = \\mathbf{v} - \\mathbf{p} = \\langle 6,-4 \\rangle - \\left\\langle \\frac{21}{5}, \\frac{7}{5} \\right\\rangle = \\left\\langle \\frac{9}{5}, -\\frac{27}{5} \\right\\rangle[\/latex]<\/p>\r\n

      Step 3<\/strong>: Verify orthogonality:<\/p>\r\n

      [latex]\\mathbf{q} \\cdot \\mathbf{p} = \\left\\langle \\frac{9}{5}, -\\frac{27}{5} \\right\\rangle \\cdot \\left\\langle \\frac{21}{5}, \\frac{7}{5} \\right\\rangle = \\frac{189}{25} - \\frac{189}{25} = 0[\/latex] \u2713<\/p>\r\n

      Therefore: [latex]\\mathbf{v} = \\mathbf{p} + \\mathbf{q}[\/latex] where [latex]\\mathbf{p}[\/latex] and [latex]\\mathbf{q}[\/latex] are orthogonal.<\/p>\r\n\r\n<\/section>

      \r\n

      vector decomposition<\/h3>\r\n

      Any vector [latex]\\mathbf{v}[\/latex] can be written as [latex]\\mathbf{v} = \\mathbf{p} + \\mathbf{q}[\/latex] where:<\/p>\r\n\r\n

        \r\n \t
      • [latex]\\mathbf{p} = \\text{proj}_{\\mathbf{u}}\\mathbf{v}[\/latex] (component parallel to [latex]\\mathbf{u}[\/latex])<\/li>\r\n \t
      • [latex]\\mathbf{q} = \\mathbf{v} - \\mathbf{p}[\/latex] (component orthogonal to [latex]\\mathbf{u}[\/latex])<\/li>\r\n<\/ul>\r\n

        This decomposition is unique and [latex]\\mathbf{p} \\cdot \\mathbf{q} = 0[\/latex].<\/p>\r\n\r\n<\/section>

        Express [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].[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<\/section>
        [ohm_question hide_question_numbers=1]328434[\/ohm_question]<\/section>
        A container ship leaves port traveling [latex]15^\\circ[\/latex] north of east. Its engine generates a speed of [latex]20[\/latex] knots along that path (see the following figure). In addition, the ocean current moves the ship northeast at a speed of [latex]2[\/latex] knots. Considering both the engine and the current, how fast is the ship moving in the direction [latex]15^\\circ[\/latex] north of east? Round the answer to two decimal places.[caption id=\"attachment_5127\" align=\"aligncenter\" width=\"604\"]\"This Figure 3. Find how fast the ship is moving in the direction of [latex]15^{\\circ}[\/latex] north of east.[\/caption][reveal-answer q=\"178946002\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"178946002\"]\r\n

        Let [latex]\\mathbf{v}[\/latex] be the velocity vector generated by the engine, and let [latex]\\mathbf{w}[\/latex] be the velocity vector of the current. We already know [latex]||\\mathbf{v}|| = 20[\/latex] along the desired route. We just need to add in the scalar projection of [latex]\\mathbf{w}[\/latex] onto [latex]\\mathbf{v}[\/latex]. We get<\/p>\r\n\r\n

        \r\n[latex]\r\n\\begin{align*}\r\n\\text{comp}_\\text{v}\\mathbf{w} &= \\frac{\\mathbf{v} \\cdot \\mathbf{w}}{||\\mathbf{v}||}\\\\\r\n&= \\frac{||\\mathbf{v}|| \\, ||\\mathbf{w}||\\cos{(30^\\circ)}}{||\\mathbf{v}||}\\\\\r\n&=||\\mathbf{w}||\\cos{(30^\\circ)} \\\\\r\n&= 2\\frac{\\sqrt{3}}{2} = \\sqrt{3} \\approx 1.73 \\text{ knots.}\\\\\\end{align*}\r\n[\/latex]<\/center>\r\n

        The ship is moving at [latex]21.73[\/latex] knots in the direction [latex]15^\\circ[\/latex] north of east.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section><\/div>","rendered":"

        Projections<\/h2>\n
        \n

        Vector addition combines vectors to create a resultant vector. But what if you need to work in reverse\u2014breaking a vector into its component parts? Vector projections<\/strong> solve this problem by showing how much of one vector acts in the direction of another.<\/p>\n

        Consider a child pulling a wagon handle at a 55\u00b0 angle. Only part of that pulling force actually moves the wagon forward\u2014the horizontal component. Projections help us find exactly how much force is working in the direction we care about.<\/p>\n<\/div>\n

        \n
        \"This
        Figure 1. When a child pulls a wagon, only the horizontal component of the force propels the wagon forward.<\/figcaption><\/figure>\n

        The vector projection<\/strong> of [latex]\\mathbf{v}[\/latex] onto [latex]\\mathbf{u}[\/latex], written as [latex]\\text{proj}_{\\mathbf{u}}\\mathbf{v}[\/latex], represents the component of [latex]\\mathbf{v}[\/latex] that acts in the direction of [latex]\\mathbf{u}[\/latex].<\/p>\n

        \"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].<\/figcaption><\/figure>\n

        From the geometry of the situation, if [latex]\\theta[\/latex] is the angle between [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex], then the length of the projection is:<\/p>\n

        [latex]||\\text{proj}_{\\mathbf{u}}\\mathbf{v}|| = ||\\mathbf{v}|| \\cos \\theta[\/latex]<\/p>\n

        Using the dot product formula [latex]\\cos \\theta = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| , ||\\mathbf{v}||}[\/latex], this becomes:<\/p>\n

        [latex]||\\text{proj}_{\\mathbf{u}}\\mathbf{v}|| = ||\\mathbf{v}|| \\cdot \\frac{|\\mathbf{u} \\cdot \\mathbf{v}|}{||\\mathbf{u}|| , ||\\mathbf{v}||} = \\frac{|\\mathbf{u} \\cdot \\mathbf{v}|}{||\\mathbf{u}||}[\/latex]<\/p>\n

        To get the actual projection vector (not just its length), we multiply by a unit vector in the direction of [latex]\\mathbf{u}[\/latex]:<\/p>\n

        [latex]||\\text{proj}_{\\mathbf{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].<\/p>\n

        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<\/p>\n

        [latex]||\\text{proj}_\\text{u}\\mathbf{v}|| = \\text{comp}_\\text{u}\\mathbf{v} = \\frac{\\mathbf{u}\\cdot\\mathbf{v}}{||\\mathbf{u}||}[\/latex].<\/div>\n
        \n

        vector projection<\/h3>\n

        Vector Projection Formula<\/strong><\/p>\n

        [latex]\\text{proj}_{\\mathbf{u}}\\mathbf{v} = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}||^2}\\mathbf{u}[\/latex]<\/p>\n

        Scalar Projection<\/strong> (length only)<\/p>\n

        [latex]||\\text{proj}_{\\mathbf{u}}\\mathbf{v}|| = \\text{comp}_{\\mathbf{u}}\\mathbf{v} = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}||}[\/latex]<\/p>\n<\/section>\n

        Think of projection as the “shadow” that vector [latex]\\mathbf{v}[\/latex] casts onto vector [latex]\\mathbf{u}[\/latex] when light shines perpendicular to [latex]\\mathbf{u}[\/latex].<\/section>\n
        \n

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

          \n
        1. [latex]\\mathbf{v} = \\langle 3,5,1 \\rangle[\/latex] and\u00a0[latex]\\mathbf{u} = \\langle -1,4,3 \\rangle[\/latex]<\/li>\n
        2. [latex]\\mathbf{v} = 3\\mathbf{i} -2\\mathbf{j}[\/latex] and\u00a0[latex]\\mathbf{u} = \\mathbf{i} +6\\mathbf{j}[\/latex]<\/li>\n<\/ol>\n