{"id":902,"date":"2025-10-14T15:43:31","date_gmt":"2025-10-14T15:43:31","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=902"},"modified":"2025-10-22T13:41:59","modified_gmt":"2025-10-22T13:41:59","slug":"cross-product-and-applications-learn-it-2","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/cross-product-and-applications-learn-it-2\/","title":{"raw":"Cross Product and Applications: Learn It 2","rendered":"Cross Product and Applications: Learn It 2"},"content":{"raw":"

Determinants and the Cross Product<\/h2>\r\nThe cross product formula [latex]\\mathbf{u} \\times \\mathbf{v} = \\langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \\rangle[\/latex] works well, but it's complicated and hard to remember. Fortunately, determinant notation<\/strong> provides an elegant alternative that's easier to use and remember.\r\n

Review: Determinants<\/h3>\r\nBefore using determinants for cross products, let's review how they work.\r\n

2\u00d72 Determinants<\/strong><\/p>\r\n

A [latex]2 \\times 2[\/latex] determinant is calculated as:<\/p>\r\n

[latex]\\begin{vmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end{vmatrix} = a_1b_2 - b_1a_2[\/latex]<\/p>\r\n\r\n

[latex]\\begin{vmatrix} 3 & -2 \\\\ 5 & 1 \\end{vmatrix} = 3(1) - 5(-2) = 3 + 10 = 13[\/latex]<\/center><\/section>\r\n

3\u00d73 Determinants<\/strong><\/p>\r\n

A [latex]3 \\times 3[\/latex] determinant expands along the first row using [latex]2 \\times 2[\/latex] determinants:<\/p>\r\n

[latex]\\begin{vmatrix} a_1 & a_2 & a_3 \\\\ b_1 & b_2 & b_3 \\\\ c_1 & c_2 & c_3\\end{vmatrix} = a_1\\begin{vmatrix} b_2 & b_3 \\\\ c_2 & c_3 \\end{vmatrix} - a_2 \\begin{vmatrix} b_1 & b_3 \\\\ c_1 & c_3 \\end{vmatrix} + a_3\\begin{vmatrix} b_1 & b_2 \\\\ c_1 & c_2 \\end{vmatrix}[\/latex]<\/p>\r\n\r\n

\r\n

Expansion Pattern<\/strong>: For each term:<\/p>\r\n\r\n

    \r\n \t
  1. Take the element from the first row as the multiplier<\/li>\r\n \t
  2. Create a [latex]2 \\times 2[\/latex] determinant from the entries that remain after crossing out the multiplier's row and column<\/li>\r\n \t
  3. Alternate signs: [latex]+[\/latex], [latex]-[\/latex], [latex]+[\/latex]<\/li>\r\n<\/ol>\r\n<\/section>
    Evaluate the determinant [latex]\\begin{vmatrix} 2 & 5 & -1 \\\\ -1 & 1 & 3 \\\\ -2 & 3 & 4\\end{vmatrix}[\/latex].[reveal-answer q=\"157384568\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"157384568\"]\r\n

    We have<\/p>\r\n\r\n

    \r\n[latex]\r\n\\begin{align*}\r\n\\begin{vmatrix} 2 & 5 & -1 \\\\ -1 & 1 & 3 \\\\ -2 & 3 & 4\\end{vmatrix} &= 2\\begin{vmatrix} 1 & 3 \\\\ 3 & 4 \\end{vmatrix} -5 \\begin{vmatrix} -1 & 3 \\\\ -2 & 4 \\end{vmatrix} -1\\begin{vmatrix} -1 & 1 \\\\ -2 & 3 \\end{vmatrix} \\\\\r\n&= 2(4-9) -5(-4+6) -1(-3+2)\\\\\r\n&= 2(-5) -5(2) -1(-1) = -10 - 10 +1\\\\\r\n&= -19. \\\\\r\n\\end{align*}\r\n[\/latex]<\/center> \r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

    Cross Product Using Determinants<\/h3>\r\n

    Now we can use determinant notation to simplify cross product calculations. The idea is to organize the cross product components using the same expansion pattern we use for [latex]3 \\times 3[\/latex] determinants.<\/p>\r\n

    If we arrange the unit vectors [latex]\\mathbf{i}[\/latex], [latex]\\mathbf{j}[\/latex], [latex]\\mathbf{k}[\/latex] in the first row and the vector components in the second and third rows, then expand using the determinant pattern, we get exactly the cross product formula.<\/p>\r\n\r\n

    \r\n

    cross product as determinant<\/h3>\r\n

    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

    [latex]\\mathbf{u} \\times \\mathbf{v} = \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ u_1 & u_2 & u_3 \\\\ v_1 & v_2 & v_3\\end{vmatrix}[\/latex]<\/p>\r\nExpanding along the first row:\r\n\r\n

    [latex]\\mathbf{u} \\times \\mathbf{v} = \\mathbf{i}\\begin{vmatrix} u_2 & u_3 \\\\ v_2 & v_3 \\end{vmatrix} - \\mathbf{j}\\begin{vmatrix} u_1 & u_3 \\\\ v_1 & v_3 \\end{vmatrix} + \\mathbf{k}\\begin{vmatrix} u_1 & u_2 \\\\ v_1 & v_2 \\end{vmatrix}[\/latex]<\/center><\/section>
    Step-by-Step Process<\/strong>\r\n
      \r\n \t
    • Step 1<\/strong>: Set up the \"determinant\" with [latex]\\mathbf{i}[\/latex], [latex]\\mathbf{j}[\/latex], [latex]\\mathbf{k}[\/latex] in the first row<\/li>\r\n \t
    • Step 2<\/strong>: Put [latex]\\mathbf{u}[\/latex] components in the second row<\/li>\r\n \t
    • Step 3<\/strong>: Put [latex]\\mathbf{v}[\/latex] components in the third row<\/li>\r\n \t
    • Step 4<\/strong>: Expand along the first row, remembering to alternate signs<\/li>\r\n<\/ul>\r\n<\/section>
      Why This Works<\/strong>: When you expand the determinant, you get exactly the same component expressions as the original cross product formula: [latex](u_2v_3 - u_3v_2)[\/latex], [latex]-(u_1v_3 - u_3v_1)[\/latex], and [latex](u_1v_2 - u_2v_1)[\/latex].<\/section>This method automatically gives you the correct cross product formula, but in an organized way that's much easier to remember and use.\r\n\r\n
      Technically, determinants contain only numbers, not vectors. This is a mnemonic device<\/strong>\u2014a memory aid that uses determinant notation to organize the cross product calculation.<\/section>
      Let [latex]\\mathbf{p} = \\langle -1,2,5\\rangle[\/latex] and [latex]\\mathbf{q} = \\langle 4,0,-3\\rangle[\/latex]. Find [latex]\\mathbf{p} \\times \\mathbf{q}[\/latex].[reveal-answer q=\"188409830\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"188409830\"]\r\n

      We set up our determinant by putting the standard unit vectors across the first row, the components of [latex]\\textbf u[\/latex] in the second row, and the components of [latex]\\textbf v[\/latex] in the third row. Then, we have<\/p>\r\n\r\n

      \r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{p} \\times \\mathbf{q} &= \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ -1 & 2 & 5 \\\\ 4 & 0 & -3\\end{vmatrix} = \\begin{vmatrix} 2 & 5 \\\\ 0 & -3 \\end{vmatrix}\\mathbf{i} - \\begin{vmatrix} -1 & 5 \\\\ 4 & -3 \\end{vmatrix}\\mathbf{j} +\\begin{vmatrix} -1 & 2 \\\\ 4 & 0 \\end{vmatrix}\\mathbf{k} \\\\\r\n&=(-6-0)\\mathbf{i} -(3-20)\\mathbf{j} +(0-8)\\mathbf{k}\\\\\r\n&= -6\\mathbf{i} + 17\\mathbf{j} -8\\mathbf{k}.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center> \r\n\r\n[\/hidden-answer]\r\n\r\n<\/section>","rendered":"

      Determinants and the Cross Product<\/h2>\n

      The cross product formula [latex]\\mathbf{u} \\times \\mathbf{v} = \\langle u_2v_3 - u_3v_2, u_3v_1 - u_1v_3, u_1v_2 - u_2v_1 \\rangle[\/latex] works well, but it’s complicated and hard to remember. Fortunately, determinant notation<\/strong> provides an elegant alternative that’s easier to use and remember.<\/p>\n

      Review: Determinants<\/h3>\n

      Before using determinants for cross products, let’s review how they work.<\/p>\n

      2\u00d72 Determinants<\/strong><\/p>\n

      A [latex]2 \\times 2[\/latex] determinant is calculated as:<\/p>\n

      [latex]\\begin{vmatrix} a_1 & a_2 \\\\ b_1 & b_2 \\end{vmatrix} = a_1b_2 - b_1a_2[\/latex]<\/p>\n

      \n
      [latex]\\begin{vmatrix} 3 & -2 \\\\ 5 & 1 \\end{vmatrix} = 3(1) - 5(-2) = 3 + 10 = 13[\/latex]<\/div>\n<\/section>\n

      3\u00d73 Determinants<\/strong><\/p>\n

      A [latex]3 \\times 3[\/latex] determinant expands along the first row using [latex]2 \\times 2[\/latex] determinants:<\/p>\n

      [latex]\\begin{vmatrix} a_1 & a_2 & a_3 \\\\ b_1 & b_2 & b_3 \\\\ c_1 & c_2 & c_3\\end{vmatrix} = a_1\\begin{vmatrix} b_2 & b_3 \\\\ c_2 & c_3 \\end{vmatrix} - a_2 \\begin{vmatrix} b_1 & b_3 \\\\ c_1 & c_3 \\end{vmatrix} + a_3\\begin{vmatrix} b_1 & b_2 \\\\ c_1 & c_2 \\end{vmatrix}[\/latex]<\/p>\n

      \n

      Expansion Pattern<\/strong>: For each term:<\/p>\n

        \n
      1. Take the element from the first row as the multiplier<\/li>\n
      2. Create a [latex]2 \\times 2[\/latex] determinant from the entries that remain after crossing out the multiplier’s row and column<\/li>\n
      3. Alternate signs: [latex]+[\/latex], [latex]-[\/latex], [latex]+[\/latex]<\/li>\n<\/ol>\n<\/section>\n
        Evaluate the determinant [latex]\\begin{vmatrix} 2 & 5 & -1 \\\\ -1 & 1 & 3 \\\\ -2 & 3 & 4\\end{vmatrix}[\/latex].<\/p>\n