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--- lecture_21 [2016/04/19 08:55] – rupert
+++ lecture_21 [2017/04/18 09:40] (current) – rupert
@@ Line 54: / 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
 \[ 2x+13y-17z=5.\]
-====== The distance to a plane ======
-===== The distance from a point to a plane =====
-Let $\Pi$ be a plane in $\def\rt{\mathbb{R}^3}\rt$ with equation $ax+by+cz=d$, so that $\def\nn{\vec n}\nn=\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c abc$ is a normal vector to $\Pi$. Also let $A$ be any point in $\rt$.
-The shortest path from $A$ to a point in $\Pi$ goes in the same direction as $\nn$. Let $B$ be any point in the plane $\Pi$.
-{{ :dpp.jpg?nolink&600 |}}
-From the diagram, we see that the shortest distance from $A$ to $\Pi$ is given by
-\[ \text{dist}(A,\Pi)=\|\def\pp{\vec p}\pp\|\]
-where
-\[ \pp=\text{proj}_{\nn}{\vec{AB}}.\]
-Using the formula for $\text{proj}_{\vec w}\vec v$ and the fact that $\|c\vec v\|=|c|\,\|\vec v\|$ where $c$ is a scalar and $\vec v$ is a vector, we obtain the formula
-\[ \text{dist}(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}.\]
-==== Example ====
-To find the distance from  $A=(1,-4,3)$ to the plane $\Pi:2x-3y+6z=1$, choose any point $B$ in $\Pi$; for example, let $B=(2,1,0)$. Then $\nn=\c2{-3}6$ and $\vec{AB}=\c15{-3}$, so
-\[ \def\dist{\text{dist}}\dist(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}=\frac{|2(1)+(-3)5+6(-3)|}{\sqrt{2^2+(-3)^2+6^2}}=\frac{|-31|}{\sqrt{49}}=\frac{31}7.\]