Differences

This shows you the differences between two versions of the page.

--- lecture_19 [2017/04/04 13:52] – 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 ===
+If $\def\vv{\vec v}\def\pp{\vec p}\def\ppp{\text{proj}_{\ww}\vv}\def\ww{\vec w}\def\nn{\vec n}\vv=\def\c#1#2#3{\begin{bmatrix}#1\\#2\\#3\end{bmatrix}}\c12{-1}$ 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*}
 ===== The cross product of vectors in $\mathbb{R}^3$ =====
@@ Line 53: / 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 63: / 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 72: / 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
 \[ \|\vv\times\ww\|=\|\vv\|\,\|\ww\|\,\sin\theta. ■ \]