{"id":877,"date":"2025-10-14T15:27:07","date_gmt":"2025-10-14T15:27:07","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=877"},"modified":"2026-08-25T15:53:57","modified_gmt":"2026-08-25T15:53:57","slug":"vector-operations-in-space-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/vector-operations-in-space-learn-it-2\/","title":{"raw":"Vector Operations in Space: Learn It 2","rendered":"Vector Operations in Space: Learn It 2"},"content":{"raw":"

Using the Dot Product to Find the Angle between Two Vectors<\/h2>\r\n

When you place two nonzero vectors in standard position (starting from the origin), they form an angle between them. The dot product provides a powerful way to find this angle.<\/p>\r\n\r\n[caption id=\"attachment_5071\" align=\"aligncenter\" width=\"277\"]\"This Figure 1. Let [latex]\u03b8[\/latex] be the angle between two nonzero vectors [latex]{\\bf{u}}[\/latex] and [latex]{\\bf{v}}[\/latex] such that [latex]0\u2264\u03b8\u2264\u03c0[\/latex].[\/caption]The dot product has both algebraic and geometric interpretations. We've seen the algebraic version (multiply components and add). The geometric version connects the dot product to angles.\r\n\r\n

\r\n

geometric dot product formula<\/h3>\r\n

The dot product of two vectors is the product of the magnitude of each vector and the cosine of the angle between them:<\/p>\r\n\r\n

[latex]\\mathbf{u} \\cdot \\mathbf{v} = ||\\mathbf{u} ||\\;||\\mathbf{v} ||\\cos{\\theta}[\/latex].<\/center><\/section>
\r\n

Proof<\/strong><\/p>\r\n\r\n\r\n

\r\n\r\nPlace vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] in standard position and consider the vector [latex]\\mathbf{v} - \\mathbf{u}[\/latex] (Figure 2). These three vectors form a triangle with side lengths [latex]||\\mathbf{u}||[\/latex],\u00a0[latex]||\\mathbf{v}||[\/latex], and\u00a0[latex]||\\mathbf{v} - \\mathbf{u}||[\/latex].\r\n\r\n[caption id=\"attachment_5073\" align=\"aligncenter\" width=\"277\"]\"This Figure 2. The lengths of the sides of the triangle are given by the magnitudes of the vectors that form the triangle.[\/caption]\r\n

The law of cosines describes the relationship among the side lengths of the triangle and the angle [latex]\\theta[\/latex]. Applying the law of cosines here gives<\/p>\r\n\r\n

[latex]||\\mathbf{v} - \\mathbf{u}||^2 = ||\\mathbf{u}||^2 + ||\\mathbf{v}||^2 - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}[\/latex].<\/center>\r\n

The dot product provides a way to rewrite the left side of this equation:<\/p>\r\n\r\n

\r\n[latex]\r\n\\begin{align*}\r\n||\\mathbf{v} - \\mathbf{u}||^2 &= (\\mathbf{v} - \\mathbf{u}) \\cdot (\\mathbf{v} - \\mathbf{u})\\\\\r\n&= (\\mathbf{v} - \\mathbf{u}) \\cdot \\mathbf{v}-(\\mathbf{v} - \\mathbf{u}) \\cdot \\mathbf{u}\\\\\r\n&= \\mathbf{v} \\cdot \\mathbf{v} - \\mathbf{u} \\cdot \\mathbf{v} - \\mathbf{v} \\cdot \\mathbf{u}+\\mathbf{u} \\cdot \\mathbf{u}\\\\\r\n&= \\mathbf{v} \\cdot \\mathbf{v} - \\mathbf{u} \\cdot \\mathbf{v} - \\mathbf{u} \\cdot \\mathbf{v}+\\mathbf{u} \\cdot \\mathbf{u}\\\\\r\n&= ||\\mathbf{v}||^2 - 2 \\mathbf{u} \\cdot \\mathbf{v} + ||\\mathbf{u}||^2.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>\r\n

Substituting into the law of cosines yields<\/p>\r\n\r\n

[latex]\r\n\\begin{align*}\r\n||\\mathbf{v} - \\mathbf{u}||^2 &= ||\\mathbf{u}||^2 + ||\\mathbf{v}||^2 - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}\\\\\r\n||\\mathbf{v}||^2 - 2 \\mathbf{u} \\cdot \\mathbf{v} + ||\\mathbf{u}||^2 &= ||\\mathbf{u}||^2 + ||\\mathbf{v}||^2 - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}\\\\\r\n- 2 \\mathbf{u} \\cdot \\mathbf{v} &= - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}\\\\\r\n\\mathbf{u} \\cdot \\mathbf{v} &= ||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center>[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>\r\n

Finding Angles Between Vectors<\/h3>\r\n

We can rearrange the geometric formula to solve for the angle:<\/p>\r\n

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

Since we measure the smallest angle between vectors, we have [latex]0\u00b0 \\leq \\theta \\leq 180\u00b0[\/latex] (or [latex]0 \\leq \\theta \\leq \\pi[\/latex] radians). The inverse cosine function is unique over this range, so we can determine [latex]\\theta[\/latex] precisely.<\/p>\r\n\r\n

\r\n

Find the measure of the angle between each pair of vectors.<\/p>\r\n\r\n

    \r\n \t
  1. [latex]\\mathbf{i} + \\mathbf{j} + \\mathbf{k}[\/latex] and\u00a0[latex]2\\mathbf{i} - \\mathbf{j} - 3\\mathbf{k}[\/latex]<\/li>\r\n \t
  2. [latex]\\langle 2,5,6 \\rangle [\/latex] and\u00a0[latex]\\langle -2,-4,4 \\rangle [\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"723893215\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"723893215\"]\r\n
      \r\n \t
    1. To find the cosine of the angle formed by the two vectors, substitute the components of the vectors into\u00a0the rearranged Dot Product:
      \r\n[latex]\r\n\\begin{align*}\r\n\\cos{\\theta} &= \\frac{(\\mathbf{i} + \\mathbf{j} + \\mathbf{k}) \\cdot (2\\mathbf{i} - \\mathbf{j} - 3\\mathbf{k})}{||\\mathbf{i} + \\mathbf{j} + \\mathbf{k}|| \\cdot ||2\\mathbf{i} - \\mathbf{j} - 3\\mathbf{k}||}\\\\\r\n&= \\frac{1(2)+(1)(-1)+(1)(-3)}{\\sqrt{1^2+1^2+1^2}\\sqrt{2^2+(-1)^2+(-3)^2}}\\\\\r\n&= \\frac{-2}{\\sqrt{3}\\sqrt{14}} = \\frac{-2}{\\sqrt{42}}.\\\\\\end{align*}\r\n[\/latex]<\/center>\r\nTherefore, [latex]{\\theta} = \\arccos{\\frac{-2}{\\sqrt{42}}}[\/latex] rad.<\/li>\r\n \t
    2. Start by finding the value of the cosine of the angle between the vectors:\r\n<\/span>
      \r\n[latex]\r\n\\begin{align*}\r\n\\cos{\\theta} &= \\frac{\\langle 2,5,6 \\rangle \\cdot \\langle -2,-4,4 \\rangle}{||\\langle 2,5,6 \\rangle|| \\cdot ||\\langle -2,-4,4 \\rangle||}\\\\\r\n&= \\frac{2(-2)+(5)(-4)+(6)(4)}{\\sqrt{2^2+5^2+6^2}\\sqrt{(-2)^2+(-4)^2+4^2}}\\\\\r\n&= \\frac{0}{\\sqrt{65}\\sqrt{36}} = 0.\\\\\\end{align*}\r\n[\/latex]<\/center>\r\nNow,[latex]\\cos{\\theta} = 0[\/latex] and\u00a0[latex]0 \\leq \\theta \\leq \\pi[\/latex], so\u00a0[latex]\\theta = \\frac{\\pi}{2}[\/latex].<\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

      Interpreting Different Angles<\/h3>\r\n

      The value of [latex]\\cos \\theta[\/latex] tells us about the relationship between the vectors:<\/p>\r\n\r\n

        \r\n \t
      • Acute angle<\/strong> [latex](0 < \\theta < 90\u00b0)[\/latex]: [latex]\\cos \\theta > 0[\/latex] (vectors point in \"similar\" directions)<\/li>\r\n \t
      • Right angle<\/strong> [latex](\\theta = 90\u00b0)[\/latex]: [latex]\\cos \\theta = 0[\/latex] (vectors are perpendicular)<\/li>\r\n \t
      • Obtuse angle<\/strong> [latex](90\u00b0 < \\theta < 180\u00b0)[\/latex]: [latex]\\cos \\theta < 0[\/latex] (vectors point in \"opposite\" directions)<\/li>\r\n \t
      • Same direction<\/strong> [latex](\\theta = 0\u00b0)[\/latex]: [latex]\\cos \\theta = 1[\/latex]<\/li>\r\n \t
      • Opposite directions<\/strong> [latex](\\theta = 180\u00b0)[\/latex]: [latex]\\cos \\theta = -1[\/latex]<\/li>\r\n<\/ul>\r\nIf [latex]\\cos{\\theta} = 0[\/latex], then the vectors, when placed in standard position, form a right angle.\r\n\r\n[caption id=\"attachment_5075\" align=\"aligncenter\" width=\"671\"]\"This Figure 3. (a) An acute angle has [latex]0< \\cos{\\theta}<1[\/latex]. (b) An obtuse angle has [latex]\u22121<\\cos\\theta<0[\/latex]. (c) A straight line has [latex]\\cos\\theta=\u22121[\/latex]. (d) If the vectors have the same direction, [latex]\\cos\\theta=1[\/latex]. (e) If the vectors are orthogonal (perpendicular), [latex]\\cos\\theta=0[\/latex][\/caption]\r\n

        Orthogonal Vectors<\/h3>\r\nThe right angle case is so important that it has special terminology and a simple test. Nonzero vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] are orthogonal<\/strong> (perpendicular) if and only if [latex]\\mathbf{u} \\cdot \\mathbf{v} = 0[\/latex].\r\n\r\n
        \r\n

        orthogonal vectors<\/h3>\r\nThe nonzero vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] are\u00a0orthogonal vectors<\/strong> if and only if [latex]\\mathbf{u} \\cdot \\mathbf{v} = 0[\/latex].\r\n\r\n<\/section>
        \r\n

        Proof<\/strong><\/p>\r\n\r\n\r\n

        \r\n

        Let [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] be nonzero vectors, and let [latex]\\theta[\/latex] denote the angle between them. First, assume [latex]\\mathbf{u} \\cdot \\mathbf{v} = 0[\/latex]. Then<\/p>\r\n\r\n

        [latex]||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta} = 0[\/latex].<\/center>\r\n

        However, [latex]||\\mathbf{u}|| \\neq 0[\/latex] and [latex]||\\mathbf{v}|| \\neq 0[\/latex], so we must have [latex]\\cos{\\theta} = 0[\/latex]. Hence, [latex]\\theta = 90^\\circ[\/latex], and the vectors are orthogonal.<\/p>\r\n

        Now assume [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] are orthogonal. Then [latex]\\theta = 90^\\circ[\/latex] and we have<\/p>\r\n\r\n

        [latex]\\mathbf{u} \\cdot \\mathbf{v} = ||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta} = ||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{90^\\circ} = ||\\mathbf{u}||\\;||\\mathbf{v}||(0) = 0[\/latex].<\/center>[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>
        Terminology:<\/strong>\u00a0We say vectors are orthogonal<\/strong>, lines are perpendicular<\/strong>, and surfaces have normal<\/strong> vectors. All three terms indicate 90\u00b0 angles, but context determines which to use.<\/section>
        Determine whether [latex]\\mathbf{p} = \\langle 1,0,5 \\rangle[\/latex] and [latex]\\mathbf{q} = \\langle 10,3,-2 \\rangle[\/latex] are orthogonal vectors.[reveal-answer q=\"934625143\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"934625143\"]\r\n

        Using the definition, we need only check the dot product of the vectors:<\/p>\r\n\r\n

        [latex]\\mathbf{p} \\cdot \\mathbf{q} = 1(10) +(0)(3) + (5)(-2) = 10+0-10 = 0[\/latex].<\/center>\r\n

        Because [latex]\\mathbf{p} \\cdot \\mathbf{q} = 0[\/latex], the vectors are orthogonal (Figure 4).<\/p>\r\n\r\n[caption id=\"attachment_5076\" align=\"aligncenter\" width=\"500\"]\"This Figure 4. Vectors [latex]{\\bf{p}}[\/latex] and [latex]{\\bf{q}}[\/latex] form a right angle when their initial points are aligned.[\/caption][\/hidden-answer]<\/section>

        [ohm_question hide_question_numbers=1]328433[\/ohm_question]<\/section><\/section>
        \r\n

        Let [latex]\\mathbf{v} = \\langle 2,3,3 \\rangle[\/latex]. Find the measures of the angles formed by the following vectors.<\/p>\r\n\r\n

          \r\n \t
        1. [latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{i}[\/latex]<\/li>\r\n \t
        2. [latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{j}[\/latex]<\/li>\r\n \t
        3. [latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{k}[\/latex]<\/li>\r\n<\/ol>\r\n[reveal-answer q=\"834690919\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"834690919\"]\r\n
            \r\n \t
          1. Let [latex]\\alpha[\/latex] be the angle formed by [latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{i}[\/latex]:\r\n
            \r\n[latex]\r\n\\begin{align*}\r\n\\cos{\\alpha} &= \\frac{\\mathbf{v} \\cdot \\mathbf{i} }{||\\mathbf{v}|| \\cdot ||\\mathbf{i}||}\\\\\r\n&= \\frac{\\langle 2,3,3 \\rangle \\cdot \\langle 1,0,0 \\rangle}{\\sqrt{2^2+3^2+3^2}\\sqrt{1}}\\\\\r\n&= \\frac{2}{\\sqrt{22}} \\end{align*}\r\n[\/latex] \r\n\r\n<\/center>
            \r\n[latex]\\alpha = \\arccos{\\frac{2}{\\sqrt{22}}} \\approx 1.130[\/latex] rad.<\/center><\/li>\r\n \t
          2. Let [latex]\\beta[\/latex] represent the angle formed by [latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{j}[\/latex]:\r\n
            \r\n[latex]\r\n\\begin{align*}\r\n\\cos{\\beta} &= \\frac{\\mathbf{v} \\cdot \\mathbf{j} }{||\\mathbf{v}|| \\cdot ||\\mathbf{j}||}\\\\\r\n&= \\frac{\\langle 2,3,3 \\rangle \\cdot \\langle 0,1,0 \\rangle}{\\sqrt{2^2+3^2+3^2}\\sqrt{1}}\\\\\r\n&= \\frac{3}{\\sqrt{22}} =.\\\\\\end{align*}\r\n[\/latex] \r\n\r\n<\/center>
            \r\n[latex]\\beta = \\arccos{\\frac{3}{\\sqrt{22}}} \\approx 0.877[\/latex] rad.<\/center><\/li>\r\n \t
          3. Let [latex]\\gamma[\/latex] represent the angle formed by [latex]\\mathbf{v}[\/latex] and\u00a0[latex]\\mathbf{k}[\/latex]:
            \r\n[latex]\r\n\\begin{align*}\r\n\\cos{\\gamma} &= \\frac{\\mathbf{v} \\cdot \\mathbf{k} }{||\\mathbf{v}|| \\cdot ||\\mathbf{k}||}\\\\\r\n&= \\frac{\\langle 2,3,3 \\rangle \\cdot \\langle 0,0,1\\rangle}{\\sqrt{2^2+3^2+3^2}\\sqrt{1}}\\\\\r\n&= \\frac{3}{\\sqrt{22}} .\\\\\\end{align*}\r\n[\/latex] \r\n\r\n<\/center>
            \r\n[latex]\\gamma = \\arccos{\\frac{3}{\\sqrt{22}}} \\approx 0.877[\/latex] rad.<\/center><\/li>\r\n<\/ol>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

            Direction Angles and Direction Cosines<\/h3>\r\n

            The angles a vector makes with the coordinate axes are called direction angles<\/strong>. These are crucial in applications like engineering, where precise angular measurements matter\u2014a small error in a rocket's launch angle can cause it to miss its target by hundreds of miles.<\/p>\r\nDirection angles are often calculated by using the dot product and the cosines of the angles, called the direction cosines. Therefore, we define both these angles and their cosines.\r\n\r\n

            \r\n

            direction angles<\/h3>\r\nThe direction angles<\/strong> of a vector are the angles it makes with the positive [latex]x[\/latex]-, [latex]y[\/latex]-, and [latex]z[\/latex]-axes. The direction cosines<\/strong> are the cosines of these angles.\r\n\r\n<\/section>[caption id=\"attachment_5118\" align=\"aligncenter\" width=\"373\"]\"This Figure 5. Angle [latex]\\alpha[\/latex] is formed by vector [latex]{\\bf{v}}[\/latex] and unit vector [latex]{\\bf{i}}[\/latex]. Angle [latex]\\beta[\/latex] is formed by vector [latex]{\\bf{v}}[\/latex] and unit vector [latex]{\\bf{j}}[\/latex]. Angle [latex]\\gamma[\/latex] is formed by vector [latex]{\\bf{v}}[\/latex] and unit vector [latex]{\\bf{k}}[\/latex].[\/caption] \r\n\r\n
            For vector [latex]\\mathbf{v} = \\langle 2,3,3 \\rangle[\/latex], the direction cosines are [latex]\\cos \\alpha = \\frac{2}{\\sqrt{22}}[\/latex], [latex]\\cos \\beta = \\frac{3}{\\sqrt{22}}[\/latex], and [latex]\\cos \\gamma = \\frac{3}{\\sqrt{22}}[\/latex], giving direction angles [latex]\\alpha = 1.130[\/latex] rad, [latex]\\beta = 0.877[\/latex] rad, and [latex]\\gamma = 0.877[\/latex] rad.<\/section>","rendered":"

            Using the Dot Product to Find the Angle between Two Vectors<\/h2>\n

            When you place two nonzero vectors in standard position (starting from the origin), they form an angle between them. The dot product provides a powerful way to find this angle.<\/p>\n

            \"This
            Figure 1. Let [latex]\u03b8[\/latex] be the angle between two nonzero vectors [latex]{\\bf{u}}[\/latex] and [latex]{\\bf{v}}[\/latex] such that [latex]0\u2264\u03b8\u2264\u03c0[\/latex].<\/figcaption><\/figure>\n

            The dot product has both algebraic and geometric interpretations. We’ve seen the algebraic version (multiply components and add). The geometric version connects the dot product to angles.<\/p>\n

            \n

            geometric dot product formula<\/h3>\n

            The dot product of two vectors is the product of the magnitude of each vector and the cosine of the angle between them:<\/p>\n

            [latex]\\mathbf{u} \\cdot \\mathbf{v} = ||\\mathbf{u} ||\\;||\\mathbf{v} ||\\cos{\\theta}[\/latex].<\/div>\n<\/section>\n
            \n

            Proof<\/strong><\/p>\n

            \n

            Place vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] in standard position and consider the vector [latex]\\mathbf{v} - \\mathbf{u}[\/latex] (Figure 2). These three vectors form a triangle with side lengths [latex]||\\mathbf{u}||[\/latex],\u00a0[latex]||\\mathbf{v}||[\/latex], and\u00a0[latex]||\\mathbf{v} - \\mathbf{u}||[\/latex].<\/p>\n

            \"This
            Figure 2. The lengths of the sides of the triangle are given by the magnitudes of the vectors that form the triangle.<\/figcaption><\/figure>\n

            The law of cosines describes the relationship among the side lengths of the triangle and the angle [latex]\\theta[\/latex]. Applying the law of cosines here gives<\/p>\n

            [latex]||\\mathbf{v} - \\mathbf{u}||^2 = ||\\mathbf{u}||^2 + ||\\mathbf{v}||^2 - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}[\/latex].<\/div>\n

            The dot product provides a way to rewrite the left side of this equation:<\/p>\n

            \n[latex]\\begin{align*} ||\\mathbf{v} - \\mathbf{u}||^2 &= (\\mathbf{v} - \\mathbf{u}) \\cdot (\\mathbf{v} - \\mathbf{u})\\\\ &= (\\mathbf{v} - \\mathbf{u}) \\cdot \\mathbf{v}-(\\mathbf{v} - \\mathbf{u}) \\cdot \\mathbf{u}\\\\ &= \\mathbf{v} \\cdot \\mathbf{v} - \\mathbf{u} \\cdot \\mathbf{v} - \\mathbf{v} \\cdot \\mathbf{u}+\\mathbf{u} \\cdot \\mathbf{u}\\\\ &= \\mathbf{v} \\cdot \\mathbf{v} - \\mathbf{u} \\cdot \\mathbf{v} - \\mathbf{u} \\cdot \\mathbf{v}+\\mathbf{u} \\cdot \\mathbf{u}\\\\ &= ||\\mathbf{v}||^2 - 2 \\mathbf{u} \\cdot \\mathbf{v} + ||\\mathbf{u}||^2.\\\\ \\end{align*}[\/latex]<\/div>\n

            Substituting into the law of cosines yields<\/p>\n

            [latex]\\begin{align*} ||\\mathbf{v} - \\mathbf{u}||^2 &= ||\\mathbf{u}||^2 + ||\\mathbf{v}||^2 - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}\\\\ ||\\mathbf{v}||^2 - 2 \\mathbf{u} \\cdot \\mathbf{v} + ||\\mathbf{u}||^2 &= ||\\mathbf{u}||^2 + ||\\mathbf{v}||^2 - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}\\\\ - 2 \\mathbf{u} \\cdot \\mathbf{v} &= - 2||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}\\\\ \\mathbf{u} \\cdot \\mathbf{v} &= ||\\mathbf{u}||\\;||\\mathbf{v}||\\cos{\\theta}.\\\\ \\end{align*}[\/latex]<\/div>\n

            [latex]_\\blacksquare[\/latex]<\/p>\n<\/section>\n

            Finding Angles Between Vectors<\/h3>\n

            We can rearrange the geometric formula to solve for the angle:<\/p>\n

            [latex]\\cos \\theta = \\frac{\\mathbf{u} \\cdot \\mathbf{v}}{||\\mathbf{u}|| , ||\\mathbf{v}||}[\/latex]<\/p>\n

            Since we measure the smallest angle between vectors, we have [latex]0\u00b0 \\leq \\theta \\leq 180\u00b0[\/latex] (or [latex]0 \\leq \\theta \\leq \\pi[\/latex] radians). The inverse cosine function is unique over this range, so we can determine [latex]\\theta[\/latex] precisely.<\/p>\n

            \n

            Find the measure of the angle between each pair of vectors.<\/p>\n

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