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--- lecture_19 [2017/04/11 09:55] – [Theorem] rupert
+++ lecture_19 [2017/05/06 10:14] (current) – rupert
@@ Line 14: / Line 14: @@
   - 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**:
 \[ \pp=\ppp=\frac{\vv\cdot\ww}{\|\ww\|^2}\ww.\]
@@ Line 80: / Line 80: @@
   - 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
 \[ \|\vv\times\ww\|^2+(\vv\cdot\ww)^2=\|\vv\|^2\,\|\ww\|^2.\]
@@ Line 106: / Line 106: @@
 For any vectors $\vv$ and $\ww$ in $\mathbb{R}^3$, we have
 \[ \|\vv\times\ww\|=\|\vv\|\,\|\ww\|\,\sin\theta\]
-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 ===
@@ Line 115: / Line 115: @@
 &=\|\vv\|^2\,\|\ww\|^2(1-\cos^2\theta)\\
 &=\|\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
 \[ \|\vv\times\ww\|=\|\vv\|\,\|\ww\|\,\sin\theta. ■ \]