{"id":903,"date":"2025-10-14T15:44:11","date_gmt":"2025-10-14T15:44:11","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=903"},"modified":"2026-08-25T16:29:48","modified_gmt":"2026-08-25T16:29:48","slug":"cross-product-and-applications-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/cross-product-and-applications-learn-it-3\/","title":{"raw":"Cross Product and Applications: Learn It 3","rendered":"Cross Product and Applications: Learn It 3"},"content":{"raw":"

Using the Cross Product<\/h2>\r\n

The cross product is a powerful tool for solving geometric problems in three dimensions. Its key applications include finding vectors orthogonal to two given vectors, calculating areas of parallelograms and triangles, and determining volumes of three-dimensional shapes called parallelepipeds.<\/p>\r\n\r\n

Finding Orthogonal Vectors<\/h3>\r\nSince [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] is always orthogonal to both [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex], cross products provide an easy way to find perpendicular vectors.\r\n\r\n
Let [latex]\\mathbf{a} = \\langle 5,2,-1\\rangle[\/latex] and\u00a0[latex]\\mathbf{b} = \\langle 0,-1,4\\rangle[\/latex]. Find a unit vector orthogonal to both [latex]\\textbf a[\/latex] and\u00a0[latex]\\textbf b[\/latex].[reveal-answer q=\"27965428\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"27965428\"]\r\n

The cross product [latex]\\mathbf{a} \\times \\mathbf{b}[\/latex] is orthogonal to both vectors [latex]\\textbf a[\/latex] and [latex]\\textbf b[\/latex]. We can calculate it with a determinant:<\/p>\r\n\r\n

\r\n[latex]\r\n\\begin{align*}\r\n\\mathbf{a} \\times \\mathbf{b} &= \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ 5 & 2 & -1 \\\\ 0 & -1 & 4\\end{vmatrix} = \\begin{vmatrix} 2 & -1 \\\\ -1 & 4 \\end{vmatrix}\\mathbf{i} - \\begin{vmatrix} 5 & -1 \\\\ 0 & 4 \\end{vmatrix}\\mathbf{j} +\\begin{vmatrix} 5 & 2 \\\\ 0 & -1 \\end{vmatrix}\\mathbf{k} \\\\\r\n&=(8-1)\\mathbf{i} -(20-0)\\mathbf{j} +(-5-0)\\mathbf{k}\\\\\r\n&= 7\\mathbf{i} - 20\\mathbf{j} -5\\mathbf{k}.\\\\\r\n\\end{align*}\r\n[\/latex]<\/center> \r\n

Normalize this vector to find a unit vector in the same direction:<\/p>\r\n\r\n

[latex]||\\mathbf{a} \\times \\mathbf{b}|| = \\sqrt{(7)^2 +(-20)^2 +(-5)^2} = \\sqrt{474}[\/latex].<\/center>\r\n

Thus, [latex]\\langle \\frac{7}{\\sqrt{474}},\\frac{-20}{\\sqrt{474}},\\frac{-5}{\\sqrt{474}}\\rangle[\/latex] is a unit vector orthogonal to [latex]\\textbf a[\/latex] and\u00a0[latex]\\textbf b[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

Remember that [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] and [latex]\\mathbf{v} \\times \\mathbf{u}[\/latex] point in opposite directions. Both are orthogonal to [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex], but choose the one appropriate for your application.<\/section>
[ohm_question hide_question_numbers=1]328435[\/ohm_question]<\/section>\r\n

Calculating Areas<\/h3>\r\nThe cross product provides an elegant way to calculate areas of parallelograms and triangles formed by vectors.\r\n\r\n
\r\n

area of a parallelogram<\/h3>\r\nIf vectors [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex] form adjacent sides of a parallelogram, then the area of the parallelogram is:\r\n

[latex]\\text{Parallelogram Area } = ||\\mathbf{u} \\times \\mathbf{v}||[\/latex]<\/p>\r\n\r\n<\/section>[caption id=\"attachment_5144\" align=\"aligncenter\" width=\"422\"]\"This Figure 1. The parallelogram with adjacent sides [latex]{\\bf{u}}[\/latex] and [latex]{\\bf{v}}[\/latex] has base [latex]\u2225{\\bf{u}}\u2225[\/latex] and height [latex]\u2225{\\bf{v}}\u2225\\sin\\theta[\/latex].[\/caption]

\r\n

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

\r\n

From basic geometry, the area of a parallelogram equals base times height. Using [latex]\\mathbf{u}[\/latex] as the base:<\/p>\r\n\r\n

    \r\n \t
  • Base = [latex]||\\mathbf{u}||[\/latex]<\/li>\r\n \t
  • Height = [latex]||\\mathbf{v}|| \\sin \\theta[\/latex] (perpendicular distance from [latex]\\mathbf{v}[\/latex] to [latex]\\mathbf{u}[\/latex])<\/li>\r\n<\/ul>\r\n

    Therefore:<\/p>\r\n

    Area = [latex]||\\mathbf{u}|| \\cdot ||\\mathbf{v}|| \\sin \\theta = ||\\mathbf{u} \\times \\mathbf{v}||[\/latex]<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>

    \r\n

    area of a triangle<\/h3>\r\nSince a triangle has half the area of the parallelogram formed by the same two vectors:\r\n

    [latex]\\text{Triangle Area }= \\frac{1}{2}||\\mathbf{u} \\times \\mathbf{v}||[\/latex]<\/p>\r\n\r\n<\/section>\r\n

    \r\n
    \r\n

    This geometric interpretation makes cross products invaluable for area calculations in computer graphics, engineering, and physics applications where you need to find areas from vector data.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

    Let [latex]P=(1, 0, 0)[\/latex], [latex]Q=(0, 1, 0)[\/latex], and [latex]R=(0, 0, 1)[\/latex] be the vertices of a triangle (Figure 2). Find its area.\r\n\r\n[caption id=\"attachment_5147\" align=\"aligncenter\" width=\"501\"]\"This Figure 2. Finding the area of a triangle by using the cross product.[\/caption]\r\n\r\n[reveal-answer q=\"452836520\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"452836520\"]\r\nWe have [latex]\\overrightarrow{PQ} = \\langle 0-1,1-0,0-0\\rangle = \\langle -1,1,0\\rangle[\/latex] and [latex]\\overrightarrow{PR} = \\langle 0-1,0-0,1-0\\rangle = \\langle -1,0,1\\rangle[\/latex]. The area of the parallelogram with adjacent sides [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{PR}[\/latex] is given by\u00a0[latex]||\\overrightarrow{PQ} \\times \\overrightarrow{PR}||[\/latex]:\r\n\r\n
    \r\n[latex]\r\n\\begin{align*}\r\n\\overrightarrow{PQ} \\times \\overrightarrow{PR} &= \\begin{vmatrix} \\mathbf{i} & \\mathbf{j} & \\mathbf{k} \\\\ -1 & 1 & 0 \\\\ -1 & 0 & 1\\end{vmatrix} = (1-0)\\mathbf{i} -(-1-0)\\mathbf{j} +(0-(-1))\\mathbf{k} = \\mathbf{i} +\\mathbf{j} +\\mathbf{k} \\\\\r\n||\\overrightarrow{PQ} \\times \\overrightarrow{PR}||&= ||\\langle 1,1,1\\rangle|| = \\sqrt{1^2 +1^2 +1^2} = \\sqrt{3}.\r\n\\end{align*}\r\n[\/latex]<\/center>\r\n

    The area of [latex]\\Delta PQR[\/latex] is half the area of the parallelogram, or [latex]\\frac{\\sqrt{3}}{2}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>

    [ohm_question hide_question_numbers=1]328436[\/ohm_question]<\/section>","rendered":"

    Using the Cross Product<\/h2>\n

    The cross product is a powerful tool for solving geometric problems in three dimensions. Its key applications include finding vectors orthogonal to two given vectors, calculating areas of parallelograms and triangles, and determining volumes of three-dimensional shapes called parallelepipeds.<\/p>\n

    Finding Orthogonal Vectors<\/h3>\n

    Since [latex]\\mathbf{u} \\times \\mathbf{v}[\/latex] is always orthogonal to both [latex]\\mathbf{u}[\/latex] and [latex]\\mathbf{v}[\/latex], cross products provide an easy way to find perpendicular vectors.<\/p>\n

    Let [latex]\\mathbf{a} = \\langle 5,2,-1\\rangle[\/latex] and\u00a0[latex]\\mathbf{b} = \\langle 0,-1,4\\rangle[\/latex]. Find a unit vector orthogonal to both [latex]\\textbf a[\/latex] and\u00a0[latex]\\textbf b[\/latex].<\/p>\n