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lecture_19

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lecture_19 [2017/04/11 09:47] – [Theorem] rupertlecture_19 [2017/05/06 10:14] (current) – rupert
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   - 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*}   - 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 +{{anchor:proj}}So we obtain the **orthogonal projection formula**:
 \[ \pp=\ppp=\frac{\vv\cdot\ww}{\|\ww\|^2}\ww.\] \[ \pp=\ppp=\frac{\vv\cdot\ww}{\|\ww\|^2}\ww.\]
  
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   - Observe that $\uu\cdot (\vv\times \ww)=\vm{u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3}$. The determinant of a matrix with a repeated row is zero, so \[\vv\cdot (\vv\times \ww)=\vm{v_1&v_2&v_3\\v_1&v_2&v_3\\w_1&w_2&w_3}=0\] so $\vv$ is orthogonal to $\vv\times\ww$; and similarly, \[\ww\cdot(\vv\times \ww)=\vm{w_1&w_2&w_3\\v_1&v_2&v_3\\w_1&w_2&w_3}=0\] so $\ww$ is orthogonal to $\vv\times\ww$. ■    - Observe that $\uu\cdot (\vv\times \ww)=\vm{u_1&u_2&u_3\\v_1&v_2&v_3\\w_1&w_2&w_3}$. The determinant of a matrix with a repeated row is zero, so \[\vv\cdot (\vv\times \ww)=\vm{v_1&v_2&v_3\\v_1&v_2&v_3\\w_1&w_2&w_3}=0\] so $\vv$ is orthogonal to $\vv\times\ww$; and similarly, \[\ww\cdot(\vv\times \ww)=\vm{w_1&w_2&w_3\\v_1&v_2&v_3\\w_1&w_2&w_3}=0\] so $\ww$ is orthogonal to $\vv\times\ww$. ■ 
  
-==== Theorem ====+==== Theorem: the dot product/cross product length formula ====
 For any vectors $\vv$ and $\ww$ in $\mathbb{R}^3$, we have For any vectors $\vv$ and $\ww$ in $\mathbb{R}^3$, we have
 \[ \|\vv\times\ww\|^2+(\vv\cdot\ww)^2=\|\vv\|^2\,\|\ww\|^2.\] \[ \|\vv\times\ww\|^2+(\vv\cdot\ww)^2=\|\vv\|^2\,\|\ww\|^2.\]
  
 === Proof === === Proof ===
-This is a tedious but elementary calculation, which we choose to describe rather than write out. 
- 
 Let $D$ be the sum of $v_i^2w_j^2$ over all $i,j\in\{1,2,3\}$ with $i=j$. (So $D=v_1^2w_1^2+v_2^2w_2^2+v_3^2w_3^2$.) Let $D$ be the sum of $v_i^2w_j^2$ over all $i,j\in\{1,2,3\}$ with $i=j$. (So $D=v_1^2w_1^2+v_2^2w_2^2+v_3^2w_3^2$.)
  
-Let $F$ be the sum of $v_i^2w_j^2$ over all $i,j\in\{1,2,3\}$ with $i\ne j$. (So $F=v_1^2w_2^2+v_2^2w_1^2+\dots+v_3^2w_2^2$.)+Let $F$ be the sum of $v_i^2w_j^2$ over all $i,j\in\{1,2,3\}$ with $i\ne j$. (So $F=v_1^2w_2^2+v_2^2w_1^2+\dots+v_3^2w_2^2$, with 6 terms on the right hand side.)
  
-Let $C$ be the sum of $v_iw_iv_jw_j$ over all $i,j\in\{1,2,3\}$ with $i\ne j$. (So $C=v_1w_1v_2w_2+\dots+v_3w_3v_2w_2$.)+Let $C$ be the sum of $v_iw_iv_jw_j$ over all $i,j\in\{1,2,3\}$ with $i<j$. (So $C=v_1w_1v_2w_2+v_1w_1v_3w_3+v_2w_2v_3w_3$.)
  
 Then $D+F$ is the sum of $v_i^2w_j^2$ over all $i,j\in \{1,2,3\}$. Then $D+F$ is the sum of $v_i^2w_j^2$ over all $i,j\in \{1,2,3\}$.
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 Expanding the formulae for $\|\vv\|^2$ and $\|\ww\|^2$, we get $\|\vv\|^2\|\ww\|^2=D+F$. Expanding the formulae for $\|\vv\|^2$ and $\|\ww\|^2$, we get $\|\vv\|^2\|\ww\|^2=D+F$.
  
-Expanding the formulae for the cross product, we get $\|\vv\times\ww\|^2=F-2C$.+Expanding the formula for the cross product, we get $\|\vv\times\ww\|^2=F-2C$.
  
-Expanding the formulae for the dot product, we get $(v\cdot w)^2=D+2C$.+Expanding the formula for the dot product, we get $(v\cdot w)^2=D+2C$.
  
 So $\|\vv\times\ww\|^2+(v\cdot w)^2=F-2C+D+2C=F+D=\|\vv\|^2\|\ww\|^2.■$ So $\|\vv\times\ww\|^2+(v\cdot w)^2=F-2C+D+2C=F+D=\|\vv\|^2\|\ww\|^2.■$
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 For any vectors $\vv$ and $\ww$ in $\mathbb{R}^3$, we have For any vectors $\vv$ and $\ww$ in $\mathbb{R}^3$, we have
 \[ \|\vv\times\ww\|=\|\vv\|\,\|\ww\|\,\sin\theta\] \[ \|\vv\times\ww\|=\|\vv\|\,\|\ww\|\,\sin\theta\]
-where $\theta$ is the angle between $\vv$ and $\ww$ (with $0\le\theta<\pi$).+where $\theta$ is the angle between $\vv$ and $\ww$ (with $0\le\theta\le\pi$).
  
 === Proof === === Proof ===
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 &=\|\vv\|^2\,\|\ww\|^2(1-\cos^2\theta)\\ &=\|\vv\|^2\,\|\ww\|^2(1-\cos^2\theta)\\
 &=\|\vv\|^2\,\|\ww\|^2\sin^2\theta.\end{align*} &=\|\vv\|^2\,\|\ww\|^2\sin^2\theta.\end{align*}
-Since $\sqrt {a^2}=a$ if $a\ge0$ and $\sin\theta\ge0$ for $0\le\theta<\pi$, taking square roots of both sides gives+Since $\sqrt {a^2}=a$ if $a\ge0$ and $\sin\theta\ge0$ for $0\le\theta\le\pi$, taking square roots of both sides gives
 \[ \|\vv\times\ww\|=\|\vv\|\,\|\ww\|\,\sin\theta. ■ \] \[ \|\vv\times\ww\|=\|\vv\|\,\|\ww\|\,\sin\theta. ■ \]
  
lecture_19.1491904043.txt.gz · Last modified: by rupert
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