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--- lecture_18 [2016/04/07 14:32] – [Theorem: the relationship between angle and the dot product] rupert
+++ lecture_18 [2017/04/06 10:04] (current) – rupert
@@ Line 66: / Line 66: @@
   - 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.
   - 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 $\vv=\c12$.
-===== The orthogonal projection of one vector onto another =====
-Let $\ww$ be a non-zero vector, and let $\vv$ be any vector. We call a vector $\def\pp{\vec p}\def\nn{\vec{n}}\pp$ the **orthogonal projection of $\vv$ onto $\ww$**, and write $\pp=\def\ppp{\text{proj}_{\ww}\vv}\ppp$, if
-  - $\pp$ is in the same direction as $\ww$; and
-  - the vector $\nn=\vv-\pp$ joining the end of $\pp$ to the end of $\vv$ is orthogonal to $\ww$.
-{{ :t3.png?nolink&300 |}}
-We can use these properties of $\pp$ to find a formula for $\pp$ in terms of $\vv$ and $\ww$.
-  - Since  $\pp$ is in the same direction as $\ww$, we have $\pp=c\ww$ for some scalar $c\in \mathbb{R}$.
-  - Since $\nn=\vv-\pp$ is orthogonal to $\ww$, we have $\nn\cdot \ww=0$. Hence \begin{align*}&&(\vv-\pp)\cdot \ww&=0\\&\implies& \vv\cdot\ww-\pp\cdot\ww&=0\\&\implies& \pp\cdot\ww&=\vv\cdot\ww\\&\implies& c\ww\cdot \ww&=\vv\cdot\ww\\&\implies& c\|\ww\|^2&=\vv\cdot\ww\\&\implies& c&=\frac{\vv\cdot\ww}{\|\ww\|^2}.\end{align*}
-So
-\[ \pp=\ppp=\frac{\vv\cdot\ww}{\|\ww\|^2}\ww.\]
-We call $\nn=\vv-\ppp$ the component of $\vv$ orthogonal to $\ww$.