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--- lecture_20 [2016/04/12 09:42] – rupert
+++ lecture_20 [2016/04/14 09:50] (current) – rupert
@@ Line 27: / Line 27: @@
 \[ V=Ah=\|\vv\times\ww\|\frac{|\uu\cdot(\vv\times\ww)|}{\|\vv\times\ww\|}\]
 or
-\[ V=|\uu\cdot(\vv\times\ww)|,\]
+\[ V=|\uu\cdot(\vv\times\ww)|.\]
-so $V$ is the absolute value of the determinant $\begin{vmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{vmatrix}$:
+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:
-\[ V=\left|\quad \begin{vmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{vmatrix}\quad \right|.\]
+\[ V=\left|\quad\det\left( \begin{bmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{bmatrix}\right)\quad \right|.\]
 === Example ===
@@ Line 39: / Line 39: @@
 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 $x-z=2$.