Differences

This shows you the differences between two versions of the page.

--- lecture_21 [2016/04/14 09:51] – rupert
+++ lecture_21 [2017/04/18 09:40] (current) – rupert
@@ Line 1: / Line 1: @@
-. Find the equation of the plane containing the points $
+. Find the equation of the plane containing the points $A=(1,2,0)$, $B=(3,0,1)$ and $C=(4,3,-2)$.
-==== The area of a parallelogram ====
-Consider a parallelogram, two of whose sides are $\def\bR{\mathbb{R}}\def\vv{\vec v}\def\ww{\vec w}\vv$ and $\ww$.
+Solution: $\def\nn{\vec n}\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\vec{AB}=\c2{-2}1$ and $\vec{AC}=\c31{-2}$ are both vectors in this plane. We want to find a normal vector $\nn$ which is be orthogonal to both of these. The cross product of two vectors is orthogonal to both, so we can take the cross product of $\vec{AB}$ and $\vec{AC}$:
-{{ :z2b.jpg?nolink&500 |}}
-This has double the area of the triangle considered above, so its area is $\|\vv\times\ww\|$.
-=== Example ===
-A triangle with two sides $\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\vv=\c13{-1}$ and $\ww=\c21{-2}$ has area $\tfrac12\|\vv\times\ww\|=\tfrac12\left\|\c13{-1}\times\c21{-2}\right\|=\tfrac12\left\|\c{-5}0{-5}\right\|=\tfrac52\left\|\c{-1}0{-1}\right\|=\tfrac52\sqrt2$, and the parallelogram with sides $\vv$ and $\ww$ has area $\|\vv\times\ww\|=5\sqrt2$.
-==== The volume of a parallelepiped in $\mathbb R^3$ ====
-Let $\def\uu{\vec u}\uu$, $\vv$ and $\ww$ be vectors in $\bR^3$.
-Consider a [[wp>parallelepiped]], with three sides given by $\uu$, $\vv$ and $\ww$.
-{{ :z3b.png?nolink&800 |}}
-Call the face with sides $\vv$ and $\ww$ the base of the parallelpiped. The area of the base is $A=\|\vv\times\ww\|$, and the volume of the parallelpiped is $Ah$ where $h$ is the height, measured at right-angles to the base.
-One vector which is at right-angles to the base is $\vv\times\ww$. It follows that $h$ is the length of $\vec p=\text{proj}_{\vv\times\ww}\uu$, so
-\[ h=\|\text{proj}_{\vv\times\ww}\uu\|=\left\|\frac{\uu\cdot(\vv\times\ww)}{\|\vv\times\ww\|^2}\vv\times\ww\right\| = \frac{\uu\cdot(\vv\times\ww)}{\|\vv\times\ww\|^2}\|\vv\times\ww\| = \frac{|\uu\cdot(\vv\times\ww)|}{\|\vv\times\ww\|}\]
-so the volume is
-\[ V=Ah=\|\vv\times\ww\|\frac{|\uu\cdot(\vv\times\ww)|}{\|\vv\times\ww\|}\]
-or
-\[ V=|\uu\cdot(\vv\times\ww)|.\]
-Now $\uu\cdot(\vv\times \ww)=\det\left(\begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{bmatrix}\right)$, so $V$ is the absolute value of this determinant:
-\[ V=\left|\quad\det\left( \begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{bmatrix}\right)\quad \right|.\]
-=== Example ===
-Find volume of a parallelepiped whose vertices include $A=(1,1,1)$, $B=(2,1,3)$, $C=(0,2,2)$ and $D=(3,4,1)$, where $A$ is an adjacent vertex to $B$, $C$ and $D$.
-== Solution ==
-The vectors $\vec{AB}=\c102$, $\vec{AC}=\c{-1}11$ and $\vec{AD}=\c230$ are all edges of this parallepiped, so the volume is
-\[ V=\left|\quad \begin{vmatrix}1&0&2\\-1&1&1\\2&3&0\end{vmatrix}\quad  \right|  = | 1(0-3)-0+2(-3-2)| = |-13| = 13.\]
-===== Planes and lines in $\mathbb{R}^3$ =====
-Recall that a typical plane in $\bR^3$ has equation
-\[ ax+by+cz=d\]
-where $a,b,c,d$ are constants. If we write
-\[ \def\nn{\vec n}\nn=\c abc\]
-then we can rewrite the equation of this plane in the form
-\[ \nn\cdot \c xyz=d.\]
-If $A=\def\cc#1{(x_{#1},y_{#1},z_{#1})}\cc1$ and $B=\cc2$ are both points in this plane, then the vector $\vec{AB}$ is said to be **in the plane**, or to be **parallel to** the plane. Observe that
-\[ \vec n\cdot \vec{AB}=\nn\cdot\def\cp#1{\c{x_{#1}}{y_{#1}}{z_{#1}}}\left(\cp2-\cp1\right) = \nn\cdot\cp2-\nn\cdot\cp1=d-d=0,\]
-so \[\nn\cdot\vv=0\]
-for every vector $\vv$ in the plane. In other words: the vector $\nn$ is orthogonal to every vector in the plane.
-{{ :z4b.jpg?nolink&500 |}}
-We call a vector with this property a **normal** vector to the plane.
-==== Examples ====
-. Find a unit normal vector to the plane $x+y-3z=4$.
-Solution: The vector $\nn=\c11{-3}$ 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. Indeed, $\vv$ is a unit vector and it's in the same direction as the normal vector $\nn$, so $\vv$ is also a normal vector.
-. Find the equation of the plane with normal vector $\def\nn{\vec n}\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c1{-3}2$ which contains the point $(1,-2,1)$. Then find three other points in this plane.
-Solution: the equation is $x-3y+2z=d$, and we can find $d$ by subbing in $(x,y,z)=(1,-2,1)$: $1-3(-2)+2(1)=d$, so $d=9$ and the equation of the plane is
-\[ x-3y+2z=9.\]
-Some other points in this plane are $(9,0,0)$, $(0,1,6)$, $(1,1,\tfrac{11}2)$. (We can find these by inspection).
-. Find the equation of the plane parallel to the vectors $\c111$ and $\c1{-1}1$ containing the point $(3,0,1)$.
-Solution: a normal vector is $\nn=\c111\times\c1{-1}1=\cp1111{-1}1=\c{2}0{-2}$, so the equation is $2x+0y-2z=2(3)-2(1)=4$, or $2x-2z=4$, or $x-z=2$.
-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 this plane. We want to find a normal vector $\nn$ which is be orthogonal to both of these. The cross product of two vectors is orthogonal to both, so we can take the cross product of $\vec{AB}$ and $\vec{AC}$:
 \[ \nn=\vec{AB}\times\vec{AC}=\def\cp#1#2#3#4#5#6{\begin{vmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{vmatrix}}\cp2{-2}131{-2}=\c378\]
 so the equation of the plane is $3x+7y+8z=d$, and we find $d$ by subbing in a point in the plane, say $A=(1,2,0)$, which gives $d=3(1)+7(2)+8(0)=17$. So the equation is
@@ Line 129: / Line 54: @@
 hence $\Pi$ has equation $-2x-13y+17z=d$, and subbing in the point $(0,3,2)$ gives $d=0-13(3)+17(2)=-39+34=-5$, so $\Pi$ has equation $-2x-13y+17z=-5$, or
 \[ 2x+13y-17z=5.\]