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--- lecture_18 [2016/04/05 09:26] – rupert
+++ lecture_18 [2017/04/06 10:04] (current) – rupert
@@ Line 39: / Line 39: @@
 {{ :orth1.png?nolink&300 |}}
+=== Proof of the Theorem ===
+We wish to show that $\def\vv{\vec v}
+\def\ww{\vec w}\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$ where $\theta$ is the angle between $\vv$ and $\ww$.
+Recall the [[wp>cosine rule]]:{{ :t1.png?nolink&300 |}}
+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 \[ \|\vv-\ww\|^2=\|\vv\|^2+\|\ww\|^2-2\|\vv\|\,\|\ww\|\,\cos\theta.\]
+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 \[\|\vv\|^2+\|\ww\|^2-2\|\vv\|\,\|\ww\|\,\cos\theta=\|\vv\|^2+\|\ww\|^2-2\vv\cdot\ww\cos\theta.\] Subtracting $\|\vv\|^2+\|\ww\|^2$ from both sides and dividing by $-2$ gives $\vv\cdot\ww=\|\vv\|\,\|\ww\|\cos\theta$. ■
+==== Corollary ====
+If $\vv$ and $\ww$ are non-zero vectors and $\theta$ is the angle between them, then $\cos\theta=\displaystyle\frac{\vv\cdot\ww}{\|\vv\|\,\|\ww\|}$.
+==== Corollary ====
+If $\vv$ and $\ww$ are non-zero vectors with $\vv\cdot\ww=0$, then $\vv$ and $\ww$ are **orthogonal**: they are at right-angles.
+==== Examples ====
+  - The angle $\theta$ between $\def\c#1#2{\begin{bmatrix}#1\\#2\end{bmatrix}}\c12$ and $\c3{-4}$ has \[ \cos\theta=\frac{\c12\cdot\c3{-4}}{\left\|\c12\right\|\,\left\|\c3{-4}\right\|} =\frac{3-8}{\sqrt5\sqrt{25}}=-\frac1{\sqrt5},\] so $\theta=\cos^{-1}(-1/\sqrt5) \approx 2.03\,\text{radians}\approx 116.57^\circ$.
+  - 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 $\vv=\c12$.