{"id":904,"date":"2025-10-14T15:46:08","date_gmt":"2025-10-14T15:46:08","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus3\/?post_type=chapter&p=904"},"modified":"2026-08-25T16:28:59","modified_gmt":"2026-08-25T16:28:59","slug":"cross-product-and-applications-learn-it-4","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus3\/chapter\/cross-product-and-applications-learn-it-4\/","title":{"raw":"Cross Product and Applications: Learn It 4","rendered":"Cross Product and Applications: Learn It 4"},"content":{"raw":"

The Triple Scalar Product<\/h2>\r\n

Since the cross product of two vectors produces a vector, we can take the dot product of this result with a third vector. This combination\u2014the dot product of a vector with the cross product of two other vectors\u2014is called the triple scalar product<\/strong> because the final result is a scalar.<\/p>\r\n\r\n

\r\n

triple scalar product<\/h3>\r\nFor vectors [latex]\\mathbf{u}[\/latex], [latex]\\mathbf{v}[\/latex], and [latex]\\mathbf{w}[\/latex]:\r\n

[latex]\\text{Triple scalar product }= \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})[\/latex]<\/p>\r\n\r\n<\/section>Just as we used determinants to simplify cross product calculations, we can use them for triple scalar products.\r\n\r\n

\r\n

\u00a0triple scalar product formula<\/h3>\r\n

For vectors [latex]\\mathbf{u} = u_1\\mathbf{i} + u_2\\mathbf{j} + u_3\\mathbf{k}[\/latex], [latex]\\mathbf{v} = v_1\\mathbf{i} + v_2\\mathbf{j} + v_3\\mathbf{k}[\/latex], and [latex]\\mathbf{w} = w_1\\mathbf{i} + w_2\\mathbf{j} + w_3\\mathbf{k}[\/latex]:<\/p>\r\n

[latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = \\begin{vmatrix}u_1&u_2&u_3 \\\\ v_1&v_2&v_3 \\\\w_1&w_2&w_3 \\end{vmatrix}[\/latex]<\/p>\r\n\r\n<\/section>

\r\n

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

\r\n

The calculation is straightforward.<\/p>\r\n\r\n

<\/div>\r\n
[latex]\\large{\\begin{aligned}\r\n{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&=\\langle u_1,u_2,u_3\\rangle\\cdot\\langle v_2w_3-v_3w_2,-v_1w_3+v_3w_1,v_1w_2-v_2w_1\\rangle \\\\\r\n&=u_1(v_2w_3-v_3w_2)+u_2(-v_1w_3+v_3w_1)+u_3(v_1w_2-v_2w_1) \\\\\r\n&=u_1(v_2w_3-v_3w_2)-u_2(v_1w_3-v_3w_1)+u_3(v_1w_2-v_2w_1) \\\\\r\n&=\\begin{vmatrix}u_1&u_2&u_3 \\\\ v_1&v_2&v_3 \\\\w_1&w_2&w_3 \\end{vmatrix}.\r\n\\end{aligned}}[\/latex]<\/div>\r\n
[latex]_\\blacksquare[\/latex]<\/div>\r\n<\/section>\r\n
\u00a0Instead of computing [latex]\\mathbf{v} \\times \\mathbf{w}[\/latex] first and then taking the dot product with [latex]\\mathbf{u}[\/latex], you can simply evaluate a [latex]3 \\times 3[\/latex] determinant.<\/section><\/div>\r\n
Let [latex]{\\bf{u}}=\\langle1,3,5\\rangle[\/latex], [latex]{\\bf{v}}=\\langle2,-1,0\\rangle[\/latex], and [latex]{\\bf{w}}=\\langle-3,0,-1\\rangle[\/latex]. Calculate the triple scalar product [latex]{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})[\/latex].[reveal-answer q=\"348759826\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"348759826\"]Apply\u00a0Theorem: Calculating a Triple Scalar Product directly:\r\n

[latex]\\begin{aligned}\r\n{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&=\\begin{vmatrix}1&3&5 \\\\ 2&-1&0 \\\\ -3&0&-1 \\end{vmatrix} \\\\\r\n&=1\\begin{vmatrix}-1&0 \\\\ 0&-1 \\end{vmatrix}-3\\begin{vmatrix} 2&0 \\\\-3&-1 \\end{vmatrix}+5\\begin{vmatrix} 2&-1 \\\\-3&0 \\end{vmatrix} \\\\\r\n&=(1-0)-3(-2-0)+5(0-3) \\\\\r\n&=1+6-15=-8.\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n

Order Matters<\/h3>\r\n

When you rearrange vectors in the triple scalar product, you're essentially reordering rows in the determinant matrix. Each time you swap two rows, the determinant changes sign.<\/p>\r\n\r\n

Let\u2019s take a look at this with an example:\r\n

[latex]\\begin{aligned}\r\n\\begin{vmatrix}1&2&1 \\\\ -2&0&3 \\\\4&1&-1\\end{vmatrix} &=\\begin{vmatrix}0&3 \\\\1&-1\\end{vmatrix}-2\\begin{vmatrix}-2&3 \\\\4&-1\\end{vmatrix}+\\begin{vmatrix}-2&0 \\\\4&1\\end{vmatrix} \\\\\r\n&=(0-3)-2(2-12)+(-2-0)=-3+20-2=15.\r\n\\end{aligned}[\/latex]<\/p>\r\n

Switching the top two rows we have<\/p>\r\n

[latex]\\begin{aligned}\r\n\\begin{vmatrix}-2&0&3 \\\\1&2&1\\\\ 4&1&-1\\end{vmatrix} &=-2\\begin{vmatrix}2&1 \\\\1&-1\\end{vmatrix}+3\\begin{vmatrix}1&2 \\\\4&1\\end{vmatrix} \\\\\r\n&=-2(-2-1)+3(1-8)=6-21=-15.\r\n\\end{aligned}[\/latex]<\/p>\r\nNotice the sign change!\r\n\r\n<\/section>\r\n

This leads to important identities for triple scalar products.<\/p>\r\n\r\n

\r\n

triple scalar product identities<\/h3>\r\n
    \r\n \t
  • [latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = -\\mathbf{u} \\cdot (\\mathbf{w} \\times \\mathbf{v})[\/latex] (switching cross product order)<\/li>\r\n \t
  • [latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = \\mathbf{v} \\cdot (\\mathbf{w} \\times \\mathbf{u}) = \\mathbf{w} \\cdot (\\mathbf{u} \\times \\mathbf{v})[\/latex] (cyclic permutations)<\/li>\r\n<\/ul>\r\n<\/section>\r\n

    Volume of Parallelepipeds<\/h3>\r\nThe triple scalar product has a powerful geometric interpretation. When three non-coplanar vectors [latex]\\mathbf{u}[\/latex], [latex]\\mathbf{v}[\/latex], and [latex]\\mathbf{w}[\/latex] share the same initial point, they form the edges of a parallelepiped<\/strong>\u2014a 3D prism where all faces are parallelograms.\r\n\r\n
    \r\n

    theorem: volume of a parallelepiped<\/h3>\r\n

    The volume of a parallelepiped with adjacent edges given by vectors [latex]\\mathbf{u}[\/latex], [latex]\\mathbf{v}[\/latex], and [latex]\\mathbf{w}[\/latex] is:<\/p>\r\n

    [latex]V = |\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})|[\/latex]<\/p>\r\n

    We use absolute value because volume is always positive.<\/p>\r\n\r\n<\/section>[caption id=\"attachment_5149\" align=\"aligncenter\" width=\"415\"]\"This Figure 3. The height of the parallelepiped is given by [latex]||\\text{proj}_{{\\bf{v}}\\times{\\bf{w}}}{\\bf{u}}||[\/latex].[\/caption]\u00a0

    \r\n

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

    \r\n

    The area of the base of the parallelepiped is given by [latex]||{\\bf{v}}\\times{\\bf{w}}||[\/latex]. The height of the figure is given by [latex]||\\text{proj}_{{\\bf{v}}\\times{\\bf{w}}}{\\bf{u}}||[\/latex]. The volume of the parallelepiped is the product of the height and the area of the base, so we have<\/p>\r\n

    [latex]\\begin{array}{rcl}\r\nV & = & \\left\\|\\operatorname{proj}_{\\mathbf{v}\\times\\mathbf{w}}\\mathbf{u}\\right\\|\\left\\|\\mathbf{v}\\times\\mathbf{w}\\right\\| \\\\[6pt]\r\n& = & \\left|\\dfrac{\\mathbf{u}\\cdot(\\mathbf{v}\\times\\mathbf{w})}{\\left\\|\\mathbf{v}\\times\\mathbf{w}\\right\\|}\\right|\\left\\|\\mathbf{v}\\times\\mathbf{w}\\right\\| \\\\[6pt]\r\n& = & \\left|\\mathbf{u}\\cdot(\\mathbf{v}\\times\\mathbf{w})\\right|.\r\n\\end{array}[\/latex]<\/p>\r\n[latex]_\\blacksquare[\/latex]\r\n\r\n<\/section>

    Let [latex]{\\bf{u}}=\\langle-1,-2,1\\rangle[\/latex],\u00a0[latex]{\\bf{v}}=\\langle4,3,2\\rangle[\/latex], and\u00a0[latex]{\\bf{w}}=\\langle0,-5,-2\\rangle[\/latex]. Find the volume of the parallelepiped with adjacent edges [latex]\\textbf u[\/latex], [latex]\\textbf v[\/latex], and [latex]\\textbf w[\/latex] (Figure 4).\""This\"[reveal-answer q=\"384527634\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"384527634\"]\r\n

    We have<\/p>\r\n

    [latex]\\begin{aligned}\r\n{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&=\\begin{vmatrix}-1&-2&1 \\\\4&3&2 \\\\0&-5&-2\\end{vmatrix}=(-1)\\begin{vmatrix}3&2 \\\\-5&-2\\end{vmatrix}+2\\begin{vmatrix}4&2 \\\\ 0&-2\\end{vmatrix}+\\begin{vmatrix} 4&3 \\\\ 0&-5\\end{vmatrix} \\\\\r\n&=(-1)(-6+10)+2(-8-0)+(-20-0) \\\\\r\n&=-4-16-20 \\\\\r\n&=-40.\r\n\\end{aligned}[\/latex]<\/p>\r\nThus, the volume of the parallelepiped is [latex]|-40|=40[\/latex] units3<\/sup>.\r\n

    <\/div>\r\n[\/hidden-answer]\r\n\r\n<\/section>
    [ohm_question hide_question_numbers=1]328437[\/ohm_question]<\/section><\/div>","rendered":"

    The Triple Scalar Product<\/h2>\n

    Since the cross product of two vectors produces a vector, we can take the dot product of this result with a third vector. This combination\u2014the dot product of a vector with the cross product of two other vectors\u2014is called the triple scalar product<\/strong> because the final result is a scalar.<\/p>\n

    \n

    triple scalar product<\/h3>\n

    For vectors [latex]\\mathbf{u}[\/latex], [latex]\\mathbf{v}[\/latex], and [latex]\\mathbf{w}[\/latex]:<\/p>\n

    [latex]\\text{Triple scalar product }= \\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w})[\/latex]<\/p>\n<\/section>\n

    Just as we used determinants to simplify cross product calculations, we can use them for triple scalar products.<\/p>\n

    \n

    \u00a0triple scalar product formula<\/h3>\n

    For vectors [latex]\\mathbf{u} = u_1\\mathbf{i} + u_2\\mathbf{j} + u_3\\mathbf{k}[\/latex], [latex]\\mathbf{v} = v_1\\mathbf{i} + v_2\\mathbf{j} + v_3\\mathbf{k}[\/latex], and [latex]\\mathbf{w} = w_1\\mathbf{i} + w_2\\mathbf{j} + w_3\\mathbf{k}[\/latex]:<\/p>\n

    [latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = \\begin{vmatrix}u_1&u_2&u_3 \\\\ v_1&v_2&v_3 \\\\w_1&w_2&w_3 \\end{vmatrix}[\/latex]<\/p>\n<\/section>\n

    \n

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

    \n

    The calculation is straightforward.<\/p>\n

    <\/div>\n
    [latex]\\large{\\begin{aligned} {\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&=\\langle u_1,u_2,u_3\\rangle\\cdot\\langle v_2w_3-v_3w_2,-v_1w_3+v_3w_1,v_1w_2-v_2w_1\\rangle \\\\ &=u_1(v_2w_3-v_3w_2)+u_2(-v_1w_3+v_3w_1)+u_3(v_1w_2-v_2w_1) \\\\ &=u_1(v_2w_3-v_3w_2)-u_2(v_1w_3-v_3w_1)+u_3(v_1w_2-v_2w_1) \\\\ &=\\begin{vmatrix}u_1&u_2&u_3 \\\\ v_1&v_2&v_3 \\\\w_1&w_2&w_3 \\end{vmatrix}. \\end{aligned}}[\/latex]<\/div>\n
    [latex]_\\blacksquare[\/latex]<\/div>\n<\/section>\n
    \n
    \u00a0Instead of computing [latex]\\mathbf{v} \\times \\mathbf{w}[\/latex] first and then taking the dot product with [latex]\\mathbf{u}[\/latex], you can simply evaluate a [latex]3 \\times 3[\/latex] determinant.<\/section>\n<\/div>\n
    \n
    Let [latex]{\\bf{u}}=\\langle1,3,5\\rangle[\/latex], [latex]{\\bf{v}}=\\langle2,-1,0\\rangle[\/latex], and [latex]{\\bf{w}}=\\langle-3,0,-1\\rangle[\/latex]. Calculate the triple scalar product [latex]{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})[\/latex].<\/p>\n