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--- lecture_18 [2016/04/07 14:26] – rupert
+++ lecture_18 [2017/04/06 10:04] (current) – rupert
@@ Line 49: / Line 49: @@
 Consider a triangle with two sides $\vv$ and $\ww$. By the triangle rule for vector addition, the third side $\vec x$ has $\ww+\vec x=\vv$, so $\vec x=\vv-\ww$:{{ :t2.png?nolink&300 |}}
-Applying the cosine rule gives \[ \|\vv-\ww\|^2=\|\vv\|\,\|\ww\|-2\vv\cdot\ww\cos\theta.\]
+Applying the cosine rule gives \[ \|\vv-\ww\|^2=\|\vv\|^2+\|\ww\|^2-2\|\vv\|\,\|\ww\|\,\cos\theta.\]
 On the other hand, we know that $\|\vec x\|^2=\vec x\cdot\vec x$, so
 \begin{align*}\|\vv-\ww\|^2&=(\vv-\ww)\cdot(\vv-\ww)\\&=\vv\cdot\vv+\ww\cdot\ww-\ww\cdot\vv-\vv\cdot\ww\\&=\|\vv\|^2+\|\ww\|^2-2\vv\cdot\ww.\end{align*}
-So $\|\vv\|\,\|\ww\|-2\vv\cdot\ww\cos\theta=\|\vv\|^2+\|\ww\|^2-2\vv\cdot\ww$, hence $\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$. ■
+So \[\|\vv\|^2+\|\ww\|^2-2\|\vv\|\,\|\ww\|\,\cos\theta=\|\vv\|^2+\|\ww\|^2-2\vv\cdot\ww\cos\theta.\] Subtracting $\|\vv\|^2+\|\ww\|^2$ from both sides and dividing by $-2$ gives $\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$. ■
 ==== Corollary ====
@@ 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 $\ww$.
-=== Example ===
-If $\vv=\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c12{-1}$ and $\ww=\c2{-1}4$, then
-\begin{align*}\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\end{align*}
-and the component of $\vv$ orthogonal to $\ww$ is
-\begin{align*}\nn&=\vv-\ppp\\&=\c12{-1}-\left(-\frac4{21}\right)\c2{-1}4\\&=\c12{-1}+\c{8/21}{-4/21}{16/21}\\&=\c{29/21}{38/21}{-5/21}.\end{align*}