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--- lecture_19 [2016/04/12 09:41] – [Corollary: the length of $\vec v\times\vec w$] rupert
+++ lecture_19 [2017/05/06 10:14] (current) – rupert
@@ Line 1: / Line 1: @@
+===== The orthogonal projection of one vector onto another =====
+Let $\def\ww{\vec{w}}\def\vv{\vec{v}}\def\uu{\vec{u}}\ww$ 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*}
+{{anchor:proj}}So we obtain the **orthogonal projection formula**:
+\[ \pp=\ppp=\frac{\vv\cdot\ww}{\|\ww\|^2}\ww.\]
+We call $\nn=\vv-\ppp$ the component of $\vv$ orthogonal to $\ww$.
 === Example ===
@@ Line 59: / 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
 \[ \|\vv\times\ww\|^2+(\vv\cdot\ww)^2=\|\vv\|^2\,\|\ww\|^2.\]
-The proof is a tedious but elementary calculation, which we leave as an exercise.
+=== Proof ===
+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 $F=v_1^2w_2^2+v_2^2w_1^2+\dots+v_3^2w_2^2$, 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 $i<j$. (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 $i,j\in \{1,2,3\}$.
+Expanding the formulae for $\|\vv\|^2$ and $\|\ww\|^2$, we get $\|\vv\|^2\|\ww\|^2=D+F$.
+Expanding the formula for the cross product, we get $\|\vv\times\ww\|^2=F-2C$.
+Expanding the formula for the dot product, we get $(v\cdot w)^2=D+2C$.
+So $\|\vv\times\ww\|^2+(v\cdot w)^2=F-2C+D+2C=F+D=\|\vv\|^2\|\ww\|^2.■$
 ==== Corollary: the length of $\vec v\times\vec w$ ====
@@ Line 69: / 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 78: / 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
 \[ \|\vv\times\ww\|=\|\vv\|\,\|\ww\|\,\sin\theta. ■ \]
@@ Line 92: / Line 129: @@
 Hence the area of this triangle is $\tfrac12 bh=\tfrac12\|\vv\|\,\|\ww\|\sin\theta$, which is equal to $\tfrac12\|\vv\times\ww\|$ (by the formula for $\|\vv\times\ww\|$ which appears above).
-==== The area of a parallelogram ====
-Consider a parallelogram, two of whose sides are $\vv$ and $\ww$.
-{{ :z2b.jpg?nolink&500 |}}
-This has double the area of the triangle considered above, so its area is $\|\vv\times\ww\|$.
-=== Example ===
-A triangle with two sides $\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\vv=\c13{-1}$ and $\ww=\c21{-2}$ has area $\tfrac12\|\vv\times\ww\|=\tfrac12\left\|\c13{-1}\times\c21{-2}\right\|=\tfrac12\left\|\c{-5}0{-5}\right\|=\tfrac52\left\|\c{-1}0{-1}\right\|=\tfrac52\sqrt2$, and the parallelogram with sides $\vv$ and $\ww$ has area $\|\vv\times\ww\|=5\sqrt2$.
-==== The volume of a parallelepiped in $\mathbb R^3$ ====
-Let $\def\uu{\vec u}\uu$, $\vv$ and $\ww$ be vectors in $\bR^3$.
-Consider a [[wp>parallelepiped]], with three sides given by $\uu$, $\vv$ and $\ww$.
-{{ :z3b.png?nolink&800 |}}
-Call the face with sides $\vv$ and $\ww$ the base of the parallelpiped. The area of the base is $A=\|\vv\times\ww\|$, and the volume of the parallelpiped is $Ah$ where $h$ is the height, measured at right-angles to the base.
-One vector which is at right-angles to the base is $\vv\times\ww$. It follows that $h$ is the length of $\vec p=\text{proj}_{\vv\times\ww}\uu$, so
-\[ h=\|\text{proj}_{\vv\times\ww}\uu\|=\left\|\frac{\uu\cdot(\vv\times\ww)}{\|\vv\times\ww\|^2}\vv\times\ww\right\| = \frac{\uu\cdot(\vv\times\ww)}{\|\vv\times\ww\|^2}\|\vv\times\ww\| = \frac{|\uu\cdot(\vv\times\ww)|}{\|\vv\times\ww\|}\]
-so the volume is
-\[ V=Ah=\|\vv\times\ww\|\frac{|\uu\cdot(\vv\times\ww)|}{\|\vv\times\ww\|}\]
-or
-\[ V=|\uu\cdot(\vv\times\ww)|,\]
-so $V$ is the absolute value of the determinant $\begin{vmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{vmatrix}$:
-\[ V=\left|\quad \begin{vmatrix}u_1&u_2&u_3\\v_1&v_2&v_3\\w_1& w_2&w_3\end{vmatrix}\quad \right|.\]
-=== Example ===
-Find volume of a parallelepiped whose vertices include $A=(1,1,1)$, $B=(2,1,3)$, $C=(0,2,2)$ and $D=(3,4,1)$, where $A$ is an adjacent vertex to $B$, $C$ and $D$.
-== Solution ==
-The vectors $\vec{AB}=\c102$, $\vec{AC}=\c{-1}11$ and $\vec{AD}=\c230$ are all edges of this parallepiped, so the volume is
-\[ V=\left|\quad \begin{vmatrix}1&0&2\\-1&1&1\\2&3&0\end{vmatrix}\quad  \right|  = | 1(0-3)-0+2(-3-2)| = |-13| = 13.\]