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--- lecture_18_slides [2017/04/05 15:34] – [Example 1] rupert
+++ lecture_18_slides [2017/04/10 16:30] (current) – [Proof of the geometric dot product formula] rupert
@@ Line 75: / Line 75: @@
   * Consider a triangle with two sides $\vv$ and $\ww$. By the triangle rule for vector addition, the third side is $\vv-\ww$:{{ :t2.png?nolink&400|}}
   * Use the cosine rule with $A=\theta$, $a=\|\vv-\ww\|$, $b=\|\vv\|$, $c=\|\ww\|$
-  * $ \|\vv-\ww\|^2=\|\vv\|\,\|\ww\|-2\|\vv\|\,\|\ww\|\,\cos\theta$
+  * $ \|\vv-\ww\|^2=\|\vv\|^2+\|\ww\|^2-2\|\vv\|\,\|\ww\|\,\cos\theta$
 ==== Proof of the geometric dot product formula, continued====
@@ Line 98: / Line 98: @@
 What is the angle $\theta$ between $\def\c#1#2{\left[\begin{smallmatrix}{#1}\\{#2}\end{smallmatrix}\right]}\c12$ and $\c3{-4}$?
+  * Use the angle formula $\cos\theta=\frac{\vv\cdot\ww}{\|\vv\|\,\|\ww\|}$:
   * $ \cos\theta=\frac{\c12\cdot\c3{-4}}{\left\|\c12\right\|\,\left\|\c3{-4}\right\|} =\frac{3-8}{\sqrt5\sqrt{25}}=\frac{-5}{5\sqrt 5}=-\frac1{\sqrt5}$
   * So $\theta=\cos^{-1}(-\tfrac1{\sqrt5}) \approx 2.03\,\text{radians}\approx 116.57^\circ$.
@@ Line 103: / Line 104: @@
 ==== Corollary 2: orthogonal vectors ====
 If $\vv$ and $\ww$ are non-zero vectors with $\vv\cdot\ww=0$, then $\vv$ and $\ww$ are **orthogonal**: they are at right-angles.
+  *Proof: the angle formula gives $\cos \theta=\frac{\vv\cdot\ww}{\|\vv\|\,\|\ww\|}=0$
+  *So $\cos\theta=0$
+  *So $\theta$ is a right-angle.
 ==== Example 2 ====
 Prove that $A=(2,3)$, $B=(3,6)$ and $C=(-4,5)$ are the vertices of a right-angled triangle.
@@ Line 117: / Line 120: @@
   * Observe that $\ww=\c{-2}1$ has $\vv\cdot\ww=0$
   * 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 $\vv$).
+  * $\vec u=\frac1{\|\ww\|}\ww$ is a unit vector in the same direction as $\ww$
+    * so it is also orthogonal to $\vv$.
   * So $\vec u=\frac1{\sqrt5}\c{-2}1=\c{-2/\sqrt5}{1/\sqrt5}$ is a unit vector orthogonal to $\vv=\c12$.
-===== Orthogonal projection =====
-Let $\def\pp{\vec p}\def\ww{\vec w}\def\vv{\vec v}\def\nn{\vec n}\ww$ non-zero, and $\vv$ any vector.
-$\pp$ is the **orthogonal projection of $\vv$ onto $\ww$** if:
-  - $\pp$ is in the same direction as $\ww$; and
-  - the vector $\nn=\vv-\pp$ is orthogonal to $\ww$.
-  * {{ :t3.png?nolink&400 |}}
-  * We write $\pp=\def\ppp{\text{proj}_{\ww}\vv}\ppp$.
-  * $\nn=\vv-\pp$ is **the component of $\vv$ orthogonal to $\ww$**.
-==== Formula for $\pp=\ppp$ ====
-  * $\pp$ same direction as $\ww$, so $\pp=c\ww$, some scalar $c$
-  * $\nn=\vv-\pp$ is orthogonal to $\ww$, so $\nn\cdot \ww=0$.
-    * so $(\vv-\pp)\cdot \ww=0$
-    * so $\vv\cdot\ww-\pp\cdot\ww=0$
-    * so $\pp\cdot\ww=\vv\cdot\ww$
-    * so $c\ww\cdot \ww=\vv\cdot\ww$
-    * so $c\|\ww\|^2=\vv\cdot\ww$
-    * so $c=\frac{\vv\cdot\ww}{\|\ww\|^2}$.
-  * So $\color{blue}{\pp=\ppp=\frac{\vv\cdot\ww}{\|\ww\|^2}\ww}.$
-==== Example ====
-$\vv=\def\c#1#2#3{\left[\begin{smallmatrix}#1\\#2\\#3\end{smallmatrix}\right]}\c12{-1}$ and $\ww=\c2{-1}4$
-  * $\ppp=\frac{\vv\cdot\ww}{\|\ww\|^2}\ww$
-    * $=\frac{2-2-4}{2^2+(-1)^2+4^2}\c2{-1}4$
-    * $=-\frac4{21}\c2{-1}4$
-  * component of $\vv$ orthogonal to $\ww$ is $\nn=\vv-\ppp$
-    * $\nn=\c12{-1}-\left(-\frac4{21}\right)\c2{-1}4=\frac1{21}\c{29}{38}{-5}$.