coplanar vector test<\/h3>\r\nThree vectors [latex]\\mathbf{u}[\/latex], [latex]\\mathbf{v}[\/latex], and [latex]\\mathbf{w}[\/latex] are coplanar if and only if [latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = 0[\/latex].\r\n\r\n<\/section>Use the triple scalar product to show that vectors [latex]{\\bf{u}}=\\langle2,0,5\\rangle[\/latex], [latex]{\\bf{v}}=\\langle2,2,4\\rangle[\/latex], and [latex]{\\bf{w}}=\\langle1,-1,3\\rangle[\/latex] are coplanar\u2014that is, show that these vectors lie in the same plane.[reveal-answer q=\"094352634\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"094352634\"]\r\nStart by calculating the triple scalar product to find the volume of the parallelepiped defined by [latex]\\textbf u[\/latex],\u00a0[latex]\\textbf v[\/latex], and\u00a0[latex]\\textbf w[\/latex]:<\/p>\r\n
[latex]\\begin{aligned}\r\n{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&=\\begin{vmatrix}2&0&5\\\\ 2&2&4 \\\\ 1&-1&3\\end{vmatrix} \\\\\r\n&=[2(2)(3)+(0)(4)(1)+5(2)(-1)]-[5(2)(1)+(2)(4)(-1)+(0)(2)(3)] \\\\&=2-2 \\\\\r\n&=0.\r\n\\end{aligned}[\/latex]<\/p>\r\n
The volume of the parallelepiped is [latex]0[\/latex] units3<\/sup>, so one of the dimensions must be zero. Therefore, the three vectors all lie in the same plane.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\nFinding Vectors Orthogonal to Planes<\/h3>\r\n
When you need a vector perpendicular to a plane, cross products provide the solution. Any plane containing two non-parallel vectors can be described using their cross product.<\/p>\r\n\r\nOnly a single plane can pass through any set of three noncolinear points. Find a vector orthogonal to the plane containing points [latex]P=(9, -3, -2)[\/latex],\u00a0[latex]Q=(1, 3, 0)[\/latex], and\u00a0[latex]R=(-2,5,0)[\/latex].[reveal-answer q=\"314652076\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"314652076\"]\r\nThe plane must contain vectors[latex]\\overrightarrow{PQ}[\/latex] and\u00a0[latex]\\overrightarrow{QR}[\/latex]:<\/p>\r\n
[latex]\\begin{aligned}\r\n\\overrightarrow{PQ}&=\\langle1-9,3-(-3),0-(-2)\\rangle =\\langle-8,6,2\\rangle \\\\\r\n\\overrightarrow{QR}&=\\langle-2-1,5-3,0-0\\rangle=\\langle-3,2,0\\rangle.\r\n\\end{aligned}[\/latex]<\/p>\r\n
The cross product [latex]\\overrightarrow{PQ}\\times\\overrightarrow{QR}[\/latex] produces a vector orthogonal to both [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{QR}[\/latex]. Therefore, the cross product is orthogonal to the plane that contains these two vectors:<\/p>\r\n
[latex]\\begin{aligned}\r\n\\overrightarrow{PQ}\\times\\overrightarrow{QR}&=\\begin{vmatrix} {\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\ -8&6&2 \\\\ -3&2&0\\end{vmatrix} \\\\\r\n&=0{\\bf{i}}-6{\\bf{j}}-16{\\bf{k}}-(-18{\\bf{k}}+4{\\bf{i}}+0{\\bf{j}}) \\\\\r\n&=-4{\\bf{i}}-6{\\bf{j}}+2{\\bf{k}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>Finding normal vectors to planes is essential in computer graphics (for lighting calculations), engineering (for surface analysis), and physics (for collision detection). The cross product makes this calculation straightforward.<\/section>\r\nTorque: Force and Rotation<\/h3>\r\n
One of the most important applications of cross products is calculating torque<\/strong>\u2014the rotational effect of a force. When you turn a wrench, tighten a screw, or spin a wheel, you're applying torque.<\/p>\r\n\r\n\r\nTorque<\/h3>\r\n
\u00a0Torque [latex]\\tau[\/latex] (Greek letter tau) measures the tendency of a force to produce rotation about an axis. For a force [latex]\\mathbf{F}[\/latex] applied at position [latex]\\mathbf{r}[\/latex] from the axis of rotation:<\/p>\r\n
[latex]\\tau = \\mathbf{r} \\times \\mathbf{F}[\/latex]<\/p>\r\n\r\n<\/section>\r\n\r\n[caption id=\"attachment_5153\" align=\"aligncenter\" width=\"469\"]![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27211439\/Figure-2.61.jpg\")
Figure 5. Torque measures how a force causes an object to rotate.[\/caption]\r\n\r\n\r\nUnderstanding Torque<\/strong>:<\/p>\r\n\r\n\r\n \t- Magnitude<\/strong>: [latex]|\\tau| = |\\mathbf{r}| \\cdot |\\mathbf{F}| \\sin \\theta[\/latex] where [latex]\\theta[\/latex] is the angle between [latex]\\mathbf{r}[\/latex] and [latex]\\mathbf{F}[\/latex]<\/li>\r\n \t
- Direction<\/strong>: The torque vector points along the axis of rotation (using the right-hand rule)<\/li>\r\n \t
- Maximum effect<\/strong>: Torque is greatest when force is applied perpendicular to the radius vector<\/li>\r\n \t
- No effect<\/strong>: When force points toward or away from the axis ([latex]\\sin 0\u00b0 = 0[\/latex])<\/li>\r\n<\/ul>\r\n<\/section>\r\n
Think about using a wrench to tighten a bolt. The torque [latex]\\tau[\/latex] applied to the bolt depends on how hard we push the wrench (force) and how far up the handle we apply the force (distance). The torque increases with a greater force on the wrench at a greater distance from the bolt. Common units of torque are the newton-meter or foot-pound.<\/p>\r\n\r\nAlthough torque has the same units as work (N\u22c5m or ft\u22c5lb), they're different concepts. Work involves force along a line of motion, while torque involves rotational effects around an axis.<\/section>A bolt is tightened by applying a force of [latex]6[\/latex] N to a [latex]0.15[\/latex]-m wrench (Figure 6). The angle between the wrench and the force vector is [latex]40^{\\small\\circ}[\/latex]. Find the magnitude of the torque about the center of the bolt. Round the answer to two decimal places.\r\n\r\n[caption id=\"attachment_5155\" align=\"aligncenter\" width=\"281\"]![\"This](\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/5667\/2021\/08\/27211539\/Figure-2.62.jpg\")
Figure 6. Torque describes the twisting action of the wrench.[\/caption]\r\n\r\n[reveal-answer q=\"853047651\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"853047651\"]\r\n\r\nSubstitute the given information into the equation defining torque:\r\n[latex]||\\tau||=||{\\bf{r}}\\times{\\bf{F}}||=||{\\bf{r}}|| \\ ||{\\bf{F}}||\\sin\\theta=(0.15\\text{ m})(6\\text{ N})\\sin{40^{\\small\\circ}}\\approx0.58\\text{N}\\cdot\\text{m}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>[ohm_question hide_question_numbers=1]325839[\/ohm_question]<\/section>","rendered":"Applications of the Cross Product<\/h2>\n
The cross product appears throughout mathematics, physics, and engineering. Beyond the geometric applications we’ve seen, cross products solve practical problems involving rotation, planes, and forces. Let’s explore these real-world applications.<\/p>\n
Testing for Coplanar Vectors<\/h3>\n
Three vectors are coplanar<\/strong> if they all lie in the same plane. The triple scalar product provides a simple test: if the volume of the parallelepiped formed by three vectors is zero, then the vectors must be coplanar.<\/p>\n\ncoplanar vector test<\/h3>\n
Three vectors [latex]\\mathbf{u}[\/latex], [latex]\\mathbf{v}[\/latex], and [latex]\\mathbf{w}[\/latex] are coplanar if and only if [latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = 0[\/latex].<\/p>\n<\/section>\nUse the triple scalar product to show that vectors [latex]{\\bf{u}}=\\langle2,0,5\\rangle[\/latex], [latex]{\\bf{v}}=\\langle2,2,4\\rangle[\/latex], and [latex]{\\bf{w}}=\\langle1,-1,3\\rangle[\/latex] are coplanar\u2014that is, show that these vectors lie in the same plane.<\/p>\n
Start by calculating the triple scalar product to find the volume of the parallelepiped defined by [latex]\\textbf u[\/latex],\u00a0[latex]\\textbf v[\/latex], and\u00a0[latex]\\textbf w[\/latex]:<\/p>\r\n
[latex]\\begin{aligned}\r\n{\\bf{u}}\\cdot({\\bf{v}}\\times{\\bf{w}})&=\\begin{vmatrix}2&0&5\\\\ 2&2&4 \\\\ 1&-1&3\\end{vmatrix} \\\\\r\n&=[2(2)(3)+(0)(4)(1)+5(2)(-1)]-[5(2)(1)+(2)(4)(-1)+(0)(2)(3)] \\\\&=2-2 \\\\\r\n&=0.\r\n\\end{aligned}[\/latex]<\/p>\r\n
The volume of the parallelepiped is [latex]0[\/latex] units3<\/sup>, so one of the dimensions must be zero. Therefore, the three vectors all lie in the same plane.<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section>\r\n When you need a vector perpendicular to a plane, cross products provide the solution. Any plane containing two non-parallel vectors can be described using their cross product.<\/p>\r\n\r\n The plane must contain vectors[latex]\\overrightarrow{PQ}[\/latex] and\u00a0[latex]\\overrightarrow{QR}[\/latex]:<\/p>\r\n [latex]\\begin{aligned}\r\n\\overrightarrow{PQ}&=\\langle1-9,3-(-3),0-(-2)\\rangle =\\langle-8,6,2\\rangle \\\\\r\n\\overrightarrow{QR}&=\\langle-2-1,5-3,0-0\\rangle=\\langle-3,2,0\\rangle.\r\n\\end{aligned}[\/latex]<\/p>\r\n The cross product [latex]\\overrightarrow{PQ}\\times\\overrightarrow{QR}[\/latex] produces a vector orthogonal to both [latex]\\overrightarrow{PQ}[\/latex] and [latex]\\overrightarrow{QR}[\/latex]. Therefore, the cross product is orthogonal to the plane that contains these two vectors:<\/p>\r\n [latex]\\begin{aligned}\r\n\\overrightarrow{PQ}\\times\\overrightarrow{QR}&=\\begin{vmatrix} {\\bf{i}}&{\\bf{j}}&{\\bf{k}} \\\\ -8&6&2 \\\\ -3&2&0\\end{vmatrix} \\\\\r\n&=0{\\bf{i}}-6{\\bf{j}}-16{\\bf{k}}-(-18{\\bf{k}}+4{\\bf{i}}+0{\\bf{j}}) \\\\\r\n&=-4{\\bf{i}}-6{\\bf{j}}+2{\\bf{k}}.\r\n\\end{aligned}[\/latex]<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section> One of the most important applications of cross products is calculating torque<\/strong>\u2014the rotational effect of a force. When you turn a wrench, tighten a screw, or spin a wheel, you're applying torque.<\/p>\r\n\r\n \u00a0Torque [latex]\\tau[\/latex] (Greek letter tau) measures the tendency of a force to produce rotation about an axis. For a force [latex]\\mathbf{F}[\/latex] applied at position [latex]\\mathbf{r}[\/latex] from the axis of rotation:<\/p>\r\n [latex]\\tau = \\mathbf{r} \\times \\mathbf{F}[\/latex]<\/p>\r\n\r\n<\/section>\r\n\r\n[caption id=\"attachment_5153\" align=\"aligncenter\" width=\"469\"] Understanding Torque<\/strong>:<\/p>\r\n\r\n Think about using a wrench to tighten a bolt. The torque [latex]\\tau[\/latex] applied to the bolt depends on how hard we push the wrench (force) and how far up the handle we apply the force (distance). The torque increases with a greater force on the wrench at a greater distance from the bolt. Common units of torque are the newton-meter or foot-pound.<\/p>\r\n\r\n [latex]||\\tau||=||{\\bf{r}}\\times{\\bf{F}}||=||{\\bf{r}}|| \\ ||{\\bf{F}}||\\sin\\theta=(0.15\\text{ m})(6\\text{ N})\\sin{40^{\\small\\circ}}\\approx0.58\\text{N}\\cdot\\text{m}[\/latex].<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/section> The cross product appears throughout mathematics, physics, and engineering. Beyond the geometric applications we’ve seen, cross products solve practical problems involving rotation, planes, and forces. Let’s explore these real-world applications.<\/p>\n Three vectors are coplanar<\/strong> if they all lie in the same plane. The triple scalar product provides a simple test: if the volume of the parallelepiped formed by three vectors is zero, then the vectors must be coplanar.<\/p>\n Three vectors [latex]\\mathbf{u}[\/latex], [latex]\\mathbf{v}[\/latex], and [latex]\\mathbf{w}[\/latex] are coplanar if and only if [latex]\\mathbf{u} \\cdot (\\mathbf{v} \\times \\mathbf{w}) = 0[\/latex].<\/p>\n<\/section>\nFinding Vectors Orthogonal to Planes<\/h3>\r\n
Torque: Force and Rotation<\/h3>\r\n
Torque<\/h3>\r\n
Figure 5. Torque measures how a force causes an object to rotate.[\/caption]\r\n\r\n\r\n \t
Figure 6. Torque describes the twisting action of the wrench.[\/caption]\r\n\r\n[reveal-answer q=\"853047651\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"853047651\"]\r\n\r\nSubstitute the given information into the equation defining torque:\r\nApplications of the Cross Product<\/h2>\n
Testing for Coplanar Vectors<\/h3>\n
coplanar vector test<\/h3>\n