~~REVEAL~~ ===== Geometry of the cross product ===== ==== The area of a triangle ==== * Consider a triangle with two sides $\def\c#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}\def\uu{\vec u}\def\vv{\vec v}\def\ww{\vec w}\def\bR{\mathbb R}\vv$ and $\ww$ * Think of $\vv$ as the base. * Length of the base is $b=\|\vv\|$ * Height (at right angles to base) is $h=\|\ww\|\sin \theta$ where $\theta$ is the angle between $\vv$ and $\ww$. {{ :z1.jpg?nolink&300 |}} * Hence area of this triangle is $A=\tfrac12 bh=\tfrac12\|\vv\|\,\|\ww\|\sin\theta$ * So $A=\tfrac12\|\vv\times\ww\|$. ==== The area of a parallelogram ==== Consider a parallelogram, two of whose sides are $\vv$ and $\ww$. {{ :z2b.jpg?nolink&400 |}} * area is double the area of the triangle above * So its area is $A=\|\vv\times\ww\|$. ==== Example ==== A triangle with two sides $\vv=\c13{-1}$ and $\ww=\c21{-2}$ has area * $\tfrac12\|\vv\times\ww\|$ * $=\tfrac12\left\|\c13{-1}\times\c21{-2}\right\|$ * $=\tfrac12\left\|\c{-5}0{-5}\right\|=\tfrac52\left\|\c{-1}0{-1}\right\|=\tfrac52\sqrt2$. The parallelogram with sides $\vv$ and $\ww$ has area * $\|\vv\times\ww\|=5\sqrt2$. ==== The volume of a parallelepiped in $\mathbb R^3$ ==== Consider a [[wp>parallelepiped]], with three sides given by $\uu,\vv,\ww$. {{ :z3b.png?nolink&400 |}} * Volume is $V=\left|\det\left[\begin{smallmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{smallmatrix}\right]\right|$ ==== Proof ==== * Area of base (parallelogram): $A=\|\vv\times\ww\|$ * Volume is $V=Ah$ ($h=$ height, measured at right-angles to the base). * $\vv\times\ww$ is at right-angles to base * So $h$ is the length of $\vec p=\text{proj}_{\vv\times\ww}\uu$ * So $ h=\|\text{proj}_{\vv\times\ww}\uu\|=\left\|\frac{\uu\cdot(\vv\times\ww)}{\|\vv\times\ww\|^2}\vv\times\ww\right\| = \frac{|\uu\cdot(\vv\times\ww)|}{\|\vv\times\ww\|}$ * So $V=Ah=|\uu\cdot (\vv\times\ww)| = \left| \det\left[\begin{smallmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{smallmatrix}\right]\right|.$ ==== Example ==== Find volume of the parallelepiped with vertices including $A=(1,1,1)$, $B=(2,1,3)$, $C=(0,2,2)$ and $D=(3,4,1)$, where $A$ is adjacent to $B$, $C$ and $D$. === Solution === * $\vec{AB}=\c102$, $\vec{AC}=\c{-1}11$ and $\vec{AD}=\c230$ are edges * Volume is $ V=\left|\det\left[ \begin{smallmatrix}1&0&2\\-1&1&1\\2&3&0\end{smallmatrix}\right] \right| = | 1(0-3)-0+2(-3-2)|$ * $ = |-13| = 13.$ ===== Planes and lines in $\mathbb{R}^3$ ===== ==== Planes in $\mathbb{R}^3$ ==== * Typical plane in $\bR^3$ has equation $ ax+by+cz=d$ * $a,b,c,d$ are constants * Same as $\def\nn{\vec n}\nn\cdot \c xyz=d$ where $\def\nn{\vec n}\nn=\c abc$ ==== Normal vectors to a plane ==== * Consider a plane $\Pi$ given by $\nn\cdot \c xyz=d$ * Take $A=\def\cc#1{(x_{#1},y_{#1},z_{#1})}\cc1$, $B=\cc2$ in $\Pi$ * Say vector $\vec{AB}$ is **in $\Pi$** or **parallel to** $\Pi$ * $ \vec n\cdot \vec{AB}=\nn\cdot\def\cp#1{\c{x_{#1}}{y_{#1}}{z_{#1}}}\left(\cp2-\cp1\right)$ * $ = \nn\cdot\cp2-\nn\cdot\cp1=d-d=0$ * So $\nn\cdot\vv=0$ for every vector $\vv$ in $\Pi$. * So $\nn$ is orthogonal to every vector in $\Pi$. * We call $\nn$ a **normal** vector to $\Pi$. ==== Normal vectors to a plane ==== {{ :z4b.jpg?nolink&600 |}} ==== Example ==== Find a unit normal vector to the plane $x+y-3z=4$. === Solution === * $\nn=\c11{-3}$ is a normal vector to this plane * So $\vv=\frac1{\|\nn\|}\nn=\frac1{\sqrt{11}}\c11{-3}$ is a unit normal vector to this plane. ==== Example 2 ==== Find the equation of the plane with normal vector $\c1{-3}2$ which contains the point $(1,-2,1)$, and find three other points in the plane. === Solution=== * Equation is $x-3y+2z=d$ for some $d$ * Find $d$ by subbing in $(x,y,z)=(1,-2,1)$ * $1-3(-2)+2(1)=d$, so $d=9$ * Equation $x-3y+2z=9$. * By inspection, some other points in this plane are $(9,0,0)$, $(0,1,6)$, $(1,1,\tfrac{11}2)$ ==== Example 3 ==== What's the equation of the plane parallel to $\c111$ and $\c1{-1}1$ containing the point $(3,0,1)$? * A normal vector is $\nn=\c111\times\c1{-1}1=\def\cp#1#2#3#4#5#6{\left|\begin{smallmatrix}\vec\imath&\vec\jmath&\vec k\\#1\\#4\end{smallmatrix}\right|}\cp1111{-1}1=\c{2}0{-2}$ * So the equation is $2x+0y-2z=2(3)-2(1)=4$, * or $2x-2z=4$ * or $x-z=2$. ==== Example 4 ==== Find the equation of the plane $\Pi$ containing $A=(1,2,0)$, $B=(3,0,1)$ and $C=(4,3,-2)$. === Solution === * $\vec{AB}=\c2{-2}1$ and $\vec{AC}=\c31{-2}$ are both vectors in $\Pi$ * Need $\nn$, orthogonal to both. Use cross product! * $\nn=\vec{AB}\times\vec{AC}=\cp2{-2}131{-2}=\c378$ * Equation is $3x+7y+8z=d$; find $d=17$ by subbing in $A=(1,2,0)$ * Answer: $ 3x+7y+8z=17$. ==== Parallel planes ==== Let $\Pi_1$ be a plane with normal vector $\nn_1$, and let $\Pi_2$ be a plane with normal vector $\nn_2$. * $\Pi_1$ and $\Pi_2$ are //parallel// planes if $\nn_1$ and $\nn_2$ are are in the same direction * If $\Pi_1$ has equation $ax+by+cz=d_1$ then any parallel plane $\Pi_2$ has an equation which may be written with the same left hand side: $ax+by+cz=d_2$. * i.e., we can assume that $\nn_1=\nn_2=\c abc$. ==== Example ==== The plane parallel to $2x-4y+5z=8$ passing through $(1,2,3)$ is * $2x-4y+5z=2(1)-4(2)+5(3) = 10$ * i.e., $2x-4y+5z=10$. ==== Orthogonal planes ==== Let~$\Pi_1$ be a plane with normal vector $\nn_1$ and let $\Pi_2$ be a plane with normal vector $\nn_2$. $\Pi_1$ and $\Pi_2$ are //orthogonal// or //perpendicular// planes if they meet at right angles. The following conditions are equivalent: - $\Pi_1$ and $\Pi_2$ are orthogonal planes; - $\nn_1\cdot\nn_2=0$; - $\nn_1$ is a vector in $\Pi_2$; - $\nn_2$ is a vector in $\Pi_1$. ==== Example 1 ==== Find the equation of the plane $\Pi$ passing through $A=(1,3,-3)$ and $B=(4,-2,1)$ which is orthogonal to the plane $x-y+z=5$. * $x-y+z=5$ has normal $\c1{-1}1$; this is in $\Pi$. * $\vec{AB}=\def\c#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}\c3{-5}4$ is also a vector in $\Pi$ * Normal for $\Pi$: $\def\nn{\vec n}\nn=\c1{-1}1\times\c3{-5}4=\cp1{-1}13{-5}4=\c1{-1}{-2}$. * Sub in $A$ (or $B$): get $x-y-2z=4$. ==== Example 2 ==== Find the equation of the plane $\Pi$ which contains the line of intersection of the planes \[ \Pi_1: x-y+2z=1\quad\text{and}\quad \Pi_2: 3x+2y-z=4,\] and is perpendicular to the plane $\Pi_3:2x+y+z=3$. * First find the line of intersection of $\Pi_1$ and $\Pi_2$ * Solve $x-y+2z=1$, $3x+2y-z=4$ * $\left[\begin{smallmatrix}1&-1&2&1\\3&2&-1&4\end{smallmatrix}\right]\to_{EROs}\left[\begin{smallmatrix}1&0&3/5&6/5\\0&1&-7/5&1/5\end{smallmatrix}\right]$ * Line $L$ of intersection is $\c xyz=\c{6/5}{1/5}0+t\c{-3/5}{7/5}1$, $t\in\mathbb{R}$. ==== ==== * $\Pi$ contains $L:\c xyz=\c{6/5}{1/5}0+t\c{-3/5}{7/5}1$, $t\in\mathbb{R}$, orthogonal to $\Pi_3: 2x+y+z=3$ * $5\c{-3/5}{7/5}1=\c{-3}75$ is a vector along $L$, so in $\Pi$ * $\Pi_3$ has normal vector $\nn_3=\c211$, which is in $\Pi$. * Normal vector for $\Pi$: $\nn=\c211\times\c{-3}75 = \cp211{-3}75=\c{-2}{-13}{17}$ * $\Pi$ has equation $-2x-13y+17z=d$ ==== ==== * $\Pi$ has equation $-2x-13y+17z=d$ and contains $L:\c xyz=\c{6/5}{1/5}0+t\c{-3/5}{7/5}1$, $t\in\mathbb{R}$ * Take $t=2$: $(0,3,2)$ in $L$, so in $\Pi$ * Sub in: $d=0-13(3)+17(2)=-39+34=-5$ * Answer: $-2x-13y+17z=-5$, or $2x+13y-17z=5$.