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--- lecture_17_slides [2016/04/04 16:17] – [Proof of the Theorem] rupert
+++ lecture_17_slides [2017/04/03 15:34] (current) – [Corollary] rupert
@@ Line 1: / Line 1: @@
 ~~REVEAL~~
-==== Last time ====
-  * a vector $\vec v$ moves $\vec x$ to $\vec x+\vec v$
+====== Chapter 3: Vectors and geometry ======
-  * e.g. if $\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\def\vv#1#2{\mat{#1\\#2}}A=(-1,3)$, $B=(5,-4)$, which vector moves $A$ to $B$?
-  * Answer: $\vec{AB}=\vv 6{-7}$.
+==== Vectors ====
-  * Reason: $A+\vec {AB}=B$, so $\vec{AB}=B-A=\vv 5{-4}-\vv{-1}3=\vv6{-7}$.
+  * $\def\m#1{\begin{bmatrix}#1\end{bmatrix}}\m{4\\3}$ is a $2\times 1$ column vector
+  * i.e., a pair of numbers written in a column
+  * We also use pairs of numbers to write points in the plane $\mathbb R^2$
+  * e.g., $(4,3)$ is a point
+    * you get there by starting from the origin, moving $4$ units to the right and $3$ units up.
+  * We think $\vec v=\m{4\\3}$ as an instruction to move $4$ units to the right and $3$ units up.
+  * This movement is called "translation by $\vec v$".
+==== Translation by $\vec v$ ====
+The vector $\vec v=\m{4\\3}$ moves:
+  * $(0,0)$ to $(4,3)$
+  * $(-2,6)$ to $(2,9)$
+  * $(x,y)$ to $(x+4,y+3)$.
+  * We're not too fussy about the difference between points like $(4,3)$ and vectors like $\m{4\\3}$.
+  * If we write points as column vectors, we can perform algebra (addition, subtraction, scalar multiplication) using points and column vectors.
+==== ====
+For example, we could rewrite the examples above by saying that $\vec v=\m{4\\3}$ moves:
+  * $\m{0\\0}$ to $\m{0\\0}+\m{4\\3}=\m{4\\3}$
+  * $\m{-2\\6}$ to $\m{-2\\6}+\m{4\\3}=\m{2\\9}$
+  * $\m{x\\y}$ to $\m{x\\y}+\m{4\\3}=\m{x+4\\y+3}$.
+  * More generally: a column vector $\vec v$ moves a point $\vec x$ to $\vec x+\vec v$.
+==== Example ====
+Which vector $\vec v$ moves the point $A=(-1,3)$ to $B=(5,-4)$?
+  * We need a vector $\vec v$ with $A+\vec v=B$
+  * So $\vec v=B-A = \m{5\\-4}-\m{-1\\3}=\m{6\\-7}$.
+  * We write $\vec{AB}=\m{6\\-7}$, since this is the vector which moves $A$ to $B$.
 ==== Definition of $\vec{AB}$ ====
@@ Line 21: / Line 57: @@
   * if $A=(3,-4,5)$
   * and $B=(11,6,-2)$
-  * then $\vec{AB}=\def\m#1{\mat{#1}}\m{11\\6\\-2}-\m{3\\-4\\5}=\m{8\\10\\-7}$.
+  * then $\vec{AB}=\def\mat#1{\left[\begin{smallmatrix}#1\end{smallmatrix}\right]}\def\m#1{\mat{#1}}\m{11\\6\\-2}-\m{3\\-4\\5}=\m{8\\10\\-7}$.
 ==== The uses of vectors ====
@@ Line 50: / Line 86: @@
 That is, multiplying a vector by a scalar $c$ scales its length by $|c|$, the absolute value of $c$.
-  * Hints: $\sqrt{xy}=\sqrt{x}\sqrt{y}$, and $\sqrt{c^2}=|c|$.
+  * Hints: $\sqrt{xy}=\sqrt{x}\sqrt{y}$ whenever $x,y\ge0$, and $\sqrt{c^2}=|c|$ for any $c\in \mathbb R$.
 ==== Distance between two points ====
@@ Line 145: / Line 181: @@
 ==== Example ====
-  * Let $\vec v=\m{1\\2}$ and $\vec w=\m{-2\\1}$
+What's the angle between $\vec v=\m{1\\2}$ and $\vec w=\m{-2\\1}$?
-  * Then $\dp vw=1(-2)+2(1)=-2+2=0$.
+  * $\dp vw=1(-2)+2(1)=-2+2=0$.
-  * On the other hand, $\|\vec v\|=\sqrt5=\|\vec w\|$
+  * $\|\vec v\|=\sqrt5=\|\vec w\|$
   * So the angle $\theta$ between $\vec v$ and $\vec w$ has $ 0=\dp vw=\sqrt 5\times \sqrt 5 \times \cos\theta$
   * So $5\cos\theta=0$, so $\cos\theta=0$,
@@ Line 183: / Line 219: @@
   * $ \|\vv-\ww\|^2=\|\vv\|\,\|\ww\|-2\vv\cdot\ww\cos\theta.$
   * We know that $\|\vec x\|^2=\vec x\cdot\vec x$
-  * So $\|\vv-\ww\|^2=(\vv-\ww)\cdot(\vv-\ww)=\vv\cdot\vv+\ww\cdot\ww-\ww\cdot\vv-\vv\cdot\ww$
+  * So $\|\vv-\ww\|^2=(\vv-\ww)\cdot(\vv-\ww)$
-  * $=\|\vv\|^2+\|\ww\|^2-2\vv\cdot\ww$.
+    * $=\vv\cdot\vv+\ww\cdot\ww-\ww\cdot\vv-\vv\cdot\ww$
+    * $=\|\vv\|^2+\|\ww\|^2-2\vv\cdot\ww$.
   * So $\|\vv\|\,\|\ww\|-2\vv\cdot\ww\cos\theta=\|\vv\|^2+\|\ww\|^2-2\vv\cdot\ww$
   * So $\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$. ■
-==== Corollary ====
+==== Corollary 1 ====
 If $\vv$ and $\ww$ are non-zero vectors and $\theta$ is the angle between them, then $\cos\theta=\displaystyle\frac{\vv\cdot\ww}{\|\vv\|\,\|\ww\|}$.
-==== Corollary ====
+==== Corollary 2 ====
 If $\vv$ and $\ww$ are non-zero vectors with $\vv\cdot\ww=0$, then $\vv$ and $\ww$ are **orthogonal**: they are at right-angles.
-==== Examples ====
+==== Example 1 ====
+What is the angle $\theta$ between $\def\c#1#2{\left[\begin{smallmatrix}{#1}\\{#2}\end{smallmatrix}\right]}\c12$ and $\c3{-4}$?
+  * $ \cos\theta=\frac{\c12\cdot\c3{-4}}{\left\|\c12\right\|\,\left\|\c3{-4}\right\|} =\frac{3-8}{\sqrt5\sqrt{25}}=-\frac1{\sqrt5}$
+  * So $\theta=\cos^{-1}(-1/\sqrt5) \approx 2.03\,\text{radians}\approx 116.57^\circ$.
+==== Example 2 ====
+Prove that $A=(2,3)$, $B=(3,6)$ and $C=(-4,5)$ are the vertices of a right-angled triangle.
+  * $\vec{AB}=\c36-\c23=\c13$
+  * $\vec{AC}=\c{-4}5-\c23=\c{-6}2$
+  * So $\vec{AB}\cdot\vec{AC}=\c13\cdot\c{-6}2=1(-6)+3(2)=0$
+  * So the sides $AB$ and $AC$ are at right-angles.
+  * So $ABC$ is a right-angled triangle.
+==== Example 3 ====
-  - The angle $\theta$ between $\def\c#1#2{\begin{bmatrix}#1\\#2\end{bmatrix}}\c12$ and $\c3{-4}$ has \[ \cos\theta=\frac{\c12\cdot\c3{-4}}{\left\|\c12\right\|\,\left\|\c3{-4}\right\|} =\frac{3-8}{\sqrt5\sqrt{25}}=-\frac1{\sqrt5},\] so $\theta=\cos^{-1}(-1/\sqrt5) \approx 2.03\,\text{radians}\approx 116.57^\circ$.
+Find a unit vector orthogonal to the vector $\vv=\c12$.
-  - The points $A=(2,3)$, $B=(3,6)$ and $C=(-4,5)$ are the vertices of a right-angled triangle. Indeed, we have $\vec{AB}=\c36-\c23=\c13$ and $\vec{AC}=\c{-4}5-\c23=\c{-6}2$, so $\vec{AB}\cdot\vec{AC}=\c13\cdot\c{-6}2=1(-6)+3(2)=0$, so the sides $AB$ and $AC$ are at right-angles.
+  * Observe that $\ww=\c{-2}1$ has $\vv\cdot\ww=0$
-  - To find a unit vector orthogonal to the vector $\vv=\c12$, we can first observe that $\ww=\c{-2}1$ has $\vv\cdot\ww=0$, so $\vv$ and $\ww$ are orthogonal; and then consider the vector $\vec u=\frac1{\|\ww\|}\ww$, which is a unit vector in the same direction as $\ww$, so is also orthogonal to $\vv$. Hence $\vec u=\frac1{\sqrt5}\c{-2}1=\c{-2/\sqrt5}{1/\sqrt5}$ is a unit vector orthogonal to $\vv=\c12$.
+  * So $\vv$ and $\ww$ are orthogonal
+  * $\vec u=\frac1{\|\ww\|}\ww$ is a unit vector in the same direction as $\ww$, (so is also orthogonal to $\vv$).
+  * So $\vec u=\frac1{\sqrt5}\c{-2}1=\c{-2/\sqrt5}{1/\sqrt5}$ is a unit vector orthogonal to $\vv=\c12$.