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--- lecture_17_slides [2017/04/03 15:26] – rupert
+++ lecture_17_slides [2017/04/03 15:34] (current) – [Corollary] rupert
@@ Line 57: / 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 86: / 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 181: / 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 225: / Line 225: @@
   * 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.