{"id":862,"date":"2025-10-14T14:57:46","date_gmt":"2025-10-14T14:57:46","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=862"},"modified":"2026-05-26T14:37:21","modified_gmt":"2026-05-26T14:37:21","slug":"vector-basics-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/vector-basics-learn-it-4\/","title":{"raw":"Vector Basics: Learn It 4","rendered":"Vector Basics: Learn It 4"},"content":{"raw":"

Unit Vectors<\/h2>\r\n

A unit vector<\/strong> is a vector with magnitude [latex]1[\/latex]. Unit vectors are useful because they indicate direction without being tied to a specific magnitude.<\/p>\r\n

For any nonzero vector [latex]\\mathbf{v}[\/latex], you can find a unit vector [latex]\\mathbf{u}[\/latex] that points in the same direction by multiplying the vector by the reciprocal of its magnitude:<\/p>\r\n

[latex]\\mathbf{u} = \\frac{1}{||\\mathbf{v}||}\\mathbf{v}[\/latex]<\/p>\r\n

This process is called normalization<\/strong>.<\/p>\r\n\r\n

\r\n

unit vector<\/h3>\r\nA unit vector has magnitude [latex]1[\/latex]. To find the unit vector in the direction of [latex]\\mathbf{v}[\/latex], use:\r\n

[latex]\\mathbf{u} = \\frac{1}{||\\mathbf{v}||}\\mathbf{v}[\/latex]<\/p>\r\n\r\n<\/section>Let's verify this works:\r\n\r\n

[latex]||\\mathbf{u}|| = \\left|\\left|\\frac{1}{||\\mathbf{v}||}\\mathbf{v}\\right|\\right| = \\frac{1}{||\\mathbf{v}||} \\cdot ||\\mathbf{v}|| = 1 \u2713[\/latex]<\/center> \r\n\r\n
[caption id=\"attachment_700\" align=\"alignnone\" width=\"285\"]\"This Figure 1.\u00a0The vector [latex]{\\bf{v}}[\/latex] and associated unit vector\u00a0[latex]{\\bf{u}}=\\frac{1}{||{\\bf{v}}||}{\\bf{v}}[\/latex]. In this case, [latex]||{\\bf{v}}||>1[\/latex].[\/caption]<\/center>
Let [latex]{\\bf{v}} = \\langle 1,2 \\rangle[\/latex].\r\n
    \r\n \t
  1. Find a unit vector with the same direction as [latex]{\\bf{v}}[\/latex].<\/li>\r\n \t
  2. Find a vector [latex]{\\bf{w}}[\/latex] with the same direction as [latex]{\\bf{v}}[\/latex] such that [latex]||{\\bf{w}}|| = 7[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1167794934754\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167794934754\"]\r\n
      \r\n \t
    1. First, find the magnitude of [latex]{\\bf{v}}[\/latex], then divide the components of [latex]{\\bf{v}}[\/latex] by the magnitude:\r\n
      [latex]||{\\bf{v}}|| = \\sqrt{1^2 + 2^2} = \\sqrt{1 + 4} = \\sqrt{5}[\/latex]<\/center>\r\n
      [latex]{\\bf{u}} = \\frac{1}{||{\\bf{v}}||} {\\bf{v}} = \\frac{1}{\\sqrt{5}} \\langle 1,2 \\rangle = \\langle \\frac{1}{\\sqrt{5}}, \\frac{2}{\\sqrt{5}} \\rangle[\/latex].<\/center><\/li>\r\n \t
    2. The vector [latex]{\\bf{u}}[\/latex] is in the same direction as [latex]{\\bf{v}}[\/latex] and [latex]||{\\bf{u}}|| = 1[\/latex]. Use scalar multiplication to increase the length of [latex]{\\bf{u}}[\/latex] without changing direction:\r\n
      [latex]{\\bf{w}} = 7{\\bf{u}} = 7 \\langle \\frac{1}{\\sqrt{5}}, \\frac{2}{\\sqrt{5}} \\rangle = \\langle \\frac{7}{\\sqrt{5}}, \\frac{14}{\\sqrt{5}} \\rangle.[\/latex]<\/center><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

      Standard Unit Vectors<\/h3>\r\n

      Sometimes it's convenient to express vectors in terms of horizontal and vertical components using standard unit vectors<\/strong>:<\/p>\r\n\r\n

        \r\n \t
      • [latex]\\mathbf{i} = \\langle 1,0 \\rangle[\/latex] (points along positive [latex]x[\/latex]-axis)<\/li>\r\n \t
      • [latex]\\mathbf{j} = \\langle 0,1 \\rangle[\/latex] (points along positive [latex]y[\/latex]-axis)<\/li>\r\n<\/ul>\r\n[caption id=\"attachment_733\" align=\"aligncenter\" width=\"365\"]\"This Figure 2. The standard unit vectors [latex]i[\/latex] and [latex]j[\/latex].[\/caption]\r\n

        Any vector [latex]\\mathbf{v} = \\langle x,y \\rangle[\/latex] can be written as a linear combination<\/strong> of [latex]\\mathbf{i}[\/latex] and [latex]\\mathbf{j}[\/latex]:<\/p>\r\n

        [latex]\\mathbf{v} = \\langle x,y \\rangle = x\\langle 1,0 \\rangle + y\\langle 0,1 \\rangle = x\\mathbf{i} + y\\mathbf{j}[\/latex]<\/p>\r\n\r\n

        <\/center>Thus, [latex]{\\bf{v}}[\/latex] is the sum of a horizontal vector with magnitude [latex]x[\/latex], and a vertical vector with magnitude [latex]y[\/latex], as in the following figure.\r\n\r\n[caption id=\"attachment_735\" align=\"aligncenter\" width=\"174\"]\"This Figure 3. The vector [latex]{\\bf{v}}[\/latex] is the sum of [latex]x{\\bf{i}}[\/latex] and [latex]y{\\bf{j}}[\/latex].[\/caption]
        \r\n
          \r\n \t
        1. Express the vector [latex]{\\bf{w}} = \\langle 3,-4 \\rangle [\/latex] in terms of standard unit vectors.<\/li>\r\n \t
        2. Vector [latex]{\\bf{u}}[\/latex] is a unit vector that forms an angle of [latex]60\u00b0[\/latex] with the positive [latex]x[\/latex]-axis. Use standard unit vectors to describe [latex]{\\bf{u}}[\/latex].<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"fs-id1167798634754\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167798634754\"]\r\n
            \r\n \t
          1. Resolve vector [latex]{\\bf{w}}[\/latex] into a vector with a zero [latex]y[\/latex]-component and a vector with a zero [latex]x[\/latex]-component:\r\n
            [latex]||{\\bf{w}}|| = \\langle 3,-4 \\rangle = 3{\\bf{i}} - 4{\\bf{j}}[\/latex]<\/center><\/li>\r\n \t
          2. Because [latex]{\\bf{u}}[\/latex] is a unit vector, the terminal point lies on the unit circle when the vector is placed in standard position (Figure 2.20).\r\n
            [latex]\\begin{array}{ccc}\\hfill {\\bf{u}} & =\\hfill & \\langle \\cos{60\u00b0}, \\sin{60\u00b0} \\rangle \\hfill \\\\ \\hfill & =\\hfill & \\langle \\frac{1}{2}, \\frac{\\sqrt{3}}{2} \\rangle \\hfill \\\\ \\hfill & =\\hfill & \\frac{1}{2}{\\bf{i}} - \\frac{\\sqrt{3}}{2}{\\bf{j}}\\end{array}[\/latex].<\/div>\r\n[caption id=\"attachment_736\" align=\"aligncenter\" width=\"342\"]\"This Figure 4.\u00a0The terminal point of [latex]{\\bf{u}}[\/latex] lies on the unit circle [latex](\\cos{\\theta},\\sin{\\theta})[\/latex].[\/caption]<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>
            [ohm_question hide_question_numbers=1]325834[\/ohm_question]<\/section>","rendered":"

            Unit Vectors<\/h2>\n

            A unit vector<\/strong> is a vector with magnitude [latex]1[\/latex]. Unit vectors are useful because they indicate direction without being tied to a specific magnitude.<\/p>\n

            For any nonzero vector [latex]\\mathbf{v}[\/latex], you can find a unit vector [latex]\\mathbf{u}[\/latex] that points in the same direction by multiplying the vector by the reciprocal of its magnitude:<\/p>\n

            [latex]\\mathbf{u} = \\frac{1}{||\\mathbf{v}||}\\mathbf{v}[\/latex]<\/p>\n

            This process is called normalization<\/strong>.<\/p>\n

            \n

            unit vector<\/h3>\n

            A unit vector has magnitude [latex]1[\/latex]. To find the unit vector in the direction of [latex]\\mathbf{v}[\/latex], use:<\/p>\n

            [latex]\\mathbf{u} = \\frac{1}{||\\mathbf{v}||}\\mathbf{v}[\/latex]<\/p>\n<\/section>\n

            Let’s verify this works:<\/p>\n

            [latex]||\\mathbf{u}|| = \\left|\\left|\\frac{1}{||\\mathbf{v}||}\\mathbf{v}\\right|\\right| = \\frac{1}{||\\mathbf{v}||} \\cdot ||\\mathbf{v}|| = 1 \u2713[\/latex]<\/div>\n

             <\/p>\n

            \n
            \"This
            Figure 1.\u00a0The vector [latex]{\\bf{v}}[\/latex] and associated unit vector\u00a0[latex]{\\bf{u}}=\\frac{1}{||{\\bf{v}}||}{\\bf{v}}[\/latex]. In this case, [latex]||{\\bf{v}}||>1[\/latex].<\/figcaption><\/figure>\n<\/div>\n
            Let [latex]{\\bf{v}} = \\langle 1,2 \\rangle[\/latex].<\/p>\n
              \n
            1. Find a unit vector with the same direction as [latex]{\\bf{v}}[\/latex].<\/li>\n
            2. Find a vector [latex]{\\bf{w}}[\/latex] with the same direction as [latex]{\\bf{v}}[\/latex] such that [latex]||{\\bf{w}}|| = 7[\/latex].<\/li>\n<\/ol>\n