Differences

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

--- lecture_22_slides [2016/04/20 12:24] – [The distance between skew lines in $\mathbb{R}^3$] rupert
+++ lecture_22_slides [2017/04/18 09:33] (current) – rupert
@@ Line 1: / Line 1: @@
 ~~REVEAL~~
-==== Last time ====
+====== The distance to a plane ======
+===== Distance from $A$ to $\Pi$ =====
+  * $\def\dist{\text{dist}}\def\cp#1#2#3#4#5#6{\left|\begin{smallmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{smallmatrix}\right|}\def\nn{\vec n}\def\c#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}\def\uu{\vec u}\def\vv{\vec v}\def\ww{\vec w}\def\bR{\mathbb R}\def\rt{\bR^3}A$: any point in $\def\rt{\mathbb R^3}\rt$. $\Pi$: a plane with normal vector $\nn$.
+  * $\nn$ is direction of shortest path from $A$ to $\Pi$
+  * Let $B$ be any point in the plane $\Pi$.{{ :dpp.jpg?nolink&600 |}}
+  * (shortest) distance from $A$ to $\Pi$ is $\text{dist}(A,\Pi)=\|\def\pp{\vec p}\pp\|$
+  * where $\pp=\text{proj}_{\nn}{\vec{AB}}$.
+  * Do some algebra: we get $\text{dist}(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}$.
+==== Example ====
+Find the distance from $A=(1,-4,3)$ to the plane $\Pi:2x-3y+6z=1$.
+  * choose any point $B$ in $\Pi$, e.g. $B=(2,1,0)$
+  * $\nn=\c2{-3}6$ and $\vec{AB}=\c15{-3}$
+  * So $\def\dist{\text{dist}}\dist(A,\Pi)=\frac{|\nn\cdot\vec{AB}|}{\|\nn\|}=\frac{|2(1)+(-3)5+6(-3)|}{\sqrt{2^2+(-3)^2+6^2}}=\frac{|-31|}{\sqrt{49}}=\frac{31}7$.
-  * Take a plane $\Pi$ with normal vector $\def\dist{\text{dist}}\def\cp#1#2#3#4#5#6{\left|\begin{smallmatrix}\vec\imath&\vec\jmath&\vec k\\#1&#2&#3\\#4&#5&#6\end{smallmatrix}\right|}\def\nn{\vec n}\def\c#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}\def\uu{\vec u}\def\vv{\vec v}\def\ww{\vec w}\def\bR{\mathbb R}\def\rt{\bR^3}\nn$
-  * Take a point $A$
-  * $\dist(A,\Pi)=\frac{|\vec{AB}\cdot \nn|}{\|n\|}$ where $B$ is any point in $\Pi$
 ==== The distance from the origin to a plane ====